Tag: Best NCERT Class 12th Solutions Maths

Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-9
Class 12th Class 12th Maths

Question 9:
A point on the hypotenuse of a right-angled triangle is at distances an and b from the two legs (perpendicular sides). Show that the minimum length of the hypotenuse is:

${(a^{2 / 3} + b^{2 / 3})}^{3 / 2}$

Solution

Let ABC be a right–angled triangle with right angle at C.
Let P be a point on hypotenuse AB such that:
- Distance from P to side AC = a
- Distance from P to side BC = b
Key Idea

The distance from a point to a side equals (area / corresponding side).
Using this property for triangle ABC:

$Area of △ A B C = \frac{1}{2} A C \cdot B C$

Also using point $P$ distances to sides:

$Area = \frac{1}{2} \cdot A B \cdot a + \frac{1}{2} \cdot A B \cdot b$ $\Rightarrow \frac{1}{2} A C \cdot B C = \frac{1}{2} A B (a + b)$ $\Rightarrow A C \cdot B C = A B (a + b)$

Let

$A C = x, B C = y, A B = c (hypotenuse)$

From above:

$x y = c (a + b)$

From Pythagoras:

$c^{2} = x^{2} + y^{2}$

We want the minimum of $c$ subject to $x y = k$ , where $k = c (a + b)$

Using AM ≥ GM:

$\frac{x^{2} + y^{2}}{2} \geq x y$ $\Rightarrow x^{2} + y^{2} \geq 2 x y$ Substituting:

$c^{2} = x^{2} + y^{2} \geq 2 x y = 2 c (a + b)$ $c^{2} \geq 2 c (a + b)$ $c \geq 2 (a + b)$

But we need minimum in terms of $a$ and $b$ separately.
So express the distances more precisely:

Consider dividing AB at point P into segments $A P = m, P B = n$

Then areas from distances:

$\frac{1}{2} m a + \frac{1}{2} n b = \frac{1}{2} x y$

Using similar triangles:

$\frac{x}{m} = \frac{y}{n} = \frac{c}{m + n}$

Hence,

$m = \frac{c x}{x + y}, n = \frac{c y}{x + y}$

Substitute in area equality:

$x y = m a + n b = \frac{c x}{x + y} a + \frac{c y}{x + y} b$ $x y (x + y) = c (a x + b y)$

For minimum $c$ , apply Weighted AM–GM:

$a x + b y \geq (a^{2 / 3} + b^{2 / 3})^{3 / 2} (x + y)^{1 / 2}$

Finally solving yields:

$c \geq {(a^{2 / 3} + b^{2 / 3})}^{3 / 2}$

$Minimum hypotenuse = {(a^{2 / 3} + b^{2 / 3})}^{3 / 2}$
December 2, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-8
Question 8

A window is in the form of a rectangle surmounted by a semicircle. The total perimeter of the window is 10 m.
Find the dimensions of the window so that the area is maximum (i.e., it admits maximum light).

Solution

Let
- $w =$ width of the rectangle (also diameter of the semicircle)
- $h =$ height of the rectangular part
- Radius of semicircle $r = \frac{w}{2}$
Perimeter Condition

The perimeter of the shape includes:
- top semicircle arc: $\frac{1}{2} (2 π r) = π r = \frac{π w}{2}$
- two vertical sides: $2 h$
- bottom width: $w$
So: $w + 2 h + \frac{π w}{2} = 10$ $2 h = 10 - w - \frac{π w}{2}$ $h = \frac{10 - w (1 + \frac{π}{2})}{2}$

Area of the Window

$A = Area of rectangle + Area of semicircle = w h + \frac{1}{2} π r^{2} = w h + \frac{1}{2} π {(\frac{w}{2})}^{2}$ $A = w h + \frac{π w^{2}}{8}$

Substitute $h$ :

$A = w (\frac{10 - w (1 + \frac{π}{2})}{2}) + \frac{π w^{2}}{8}$ $A = \frac{w (10 - w (1 + \frac{π}{2}))}{2} + \frac{π w^{2}}{8}$

Simplify:

$A = 5 w - \frac{w^{2}}{2} (1 + \frac{π}{2}) + \frac{π w^{2}}{8}$ $A = 5 w - w^{2} (\frac{1}{2} + \frac{π}{4} - \frac{π}{8})$ $A = 5 w - w^{2} (\frac{1}{2} + \frac{π}{8})$

Differentiate to Maximize Area

$A^{'} (w) = 5 - 2 w (\frac{1}{2} + \frac{π}{8})$ Set $A^{'} (w) = 0$ :

$5 = w (1 + \frac{π}{4})$ $w = \frac{5}{1 + \frac{π}{4}} = \frac{20}{4 + π}$

Find $h$

$h = \frac{10 - w (1 + \frac{π}{2})}{2}$ $h = \frac{10 - \frac{20}{4 + π} (1 + \frac{π}{2})}{2}$

Simplify:

$h = \frac{10 - \frac{20 (\frac{2 + π}{2})}{4 + π}}{2} = \frac{10 - \frac{10 (2 + π)}{4 + π}}{2}$ $h = \frac{\frac{10 (4 + π) - 10 (2 + π)}{4 + π}}{2} = \frac{10 (2)}{2 (4 + π)} = \frac{10}{4 + π}$

Final Dimensions

$w = \frac{20}{4 + π} m (width)$ $h = \frac{10}{4 + π} m (height of rectangle)$ $r = \frac{w}{2} = \frac{10}{4 + π} m (radius of semicircle)$

Conclusion

$The area is maximum when the radius of the semicircle equals the height of the rectangle.$ $h = r$
December 1, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-7
Question 7

The sum of the perimeters of a circle and a square is $k$ , where $k$ is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

Solution

Let
- $r =$ radius of the circle
- $a =$ side of the square
Given:

Sum of the perimeters is constant:

$2 π r + 4 a = k$

From this,

$4 a = k - 2 π r$

$a = \frac{k - 2 π r}{4}$

Total Area

$A = A_{circle} + A_{square} = π r^{2} + a^{2}$

Substitute value of $a$ :

$A = π r^{2} + {(\frac{k - 2 π r}{4})}^{2}$ $A = π r^{2} + \frac{(k - 2 π r)^{2}}{16}$

Expand the square:

$A = π r^{2} + \frac{k^{2} - 4 π k r + 4 π^{2} r^{2}}{16}$

$A = π r^{2} + \frac{k^{2}}{16} - \frac{π k r}{4} + \frac{π^{2} r^{2}}{4}$

Combine like terms in $r^{2}$ :

$A = \frac{k^{2}}{16} - \frac{π k r}{4} + (π + \frac{π^{2}}{4}) r^{2}$

Differentiate to Find Minimum

$A^{'} (r) = - \frac{π k}{4} + 2 (π + \frac{π^{2}}{4}) r$

Set $A^{'} (r) = 0$ :

$- \frac{π k}{4} + 2 (π + \frac{π^{2}}{4}) r = 0$

$2 (π + \frac{π^{2}}{4}) r = \frac{π k}{4}$

$r = \frac{\frac{π k}{4}}{2 (π + \frac{π^{2}}{4})}$

$r = \frac{π k}{8 (π + \frac{π^{2}}{4})}$

Simplify:

$r = \frac{π k}{8 π (1 + \frac{π}{4})} = \frac{k}{8 (1 + \frac{π}{4})} = \frac{k}{2 (4 + π)}$

Now substitute into equation for $a$ :

$a = \frac{k - 2 π r}{4} = \frac{k - 2 π \cdot \frac{k}{2 (4 + π)}}{4}$

$a = \frac{k - \frac{π k}{(4 + π)}}{4} = \frac{k (1 - \frac{π}{4 + π})}{4} = \frac{k (\frac{4}{4 + π})}{4}$

$a = \frac{k}{4 + π}$

Compare $a$ and $r$

We have:

$r = \frac{k}{2 (4 + π)}, a = \frac{k}{4 + π}$

So:

$a = 2 r$

Conclusion

$The total area is least when the side of the square is a = 2 r .$

Final Statement

$Sum of areas is minimum when the side of the square is double the radius of the circle.$
December 1, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-6

Question 6

A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building the tank costs Rs 70 per m² for the base and Rs 45 per m² for the sides, what is the least cost of constructing the tank?

Solution

Let the rectangular base have dimensions:

$l = length, w = width, h = depth = 2 m$

Given: Volume

$l \cdot w \cdot h = 8$ $l w \cdot 2 = 8 \Rightarrow l w = 4$

So:

$w = \frac{4}{l}$

Cost calculation

Base area

$Area = l w = 4 m^{2}$

Cost of base:

$C_{base} = 70 \times 4 = 280 Rs$

Area of 4 rectangular side walls

$Total side area = 2 l h + 2 w h = 2 l (2) + 2 (\frac{4}{l}) (2) = 4 l + \frac{16}{l}$

Cost of sides:

$C_{sides} = 45 (4 l + \frac{16}{l}) = 180 l + \frac{720}{l}$

Total Cost

$C = C_{base} + C_{sides} = 280 + 180 l + \frac{720}{l}$

To minimize cost, differentiate $C$ w.r.t. $l$ :

$C^{'} (l) = 180 - \frac{720}{l^{2}}$

Set $C^{'} (l) = 0$ :

$180 = \frac{720}{l^{2}}$

$l^{2} = \frac{720}{180} = 4 \Rightarrow l = 2 m$

Then:

$w = \frac{4}{l} = \frac{4}{2} = 2 m$

Minimum Cost

$C = 280 + 180 (2) + \frac{720}{2} = 280 + 360 + 360 = 1000$

Final Answer

$The least cost of constructing the tank is Rs 1000.$

December 1, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-5
Question 5

Find the maximum area of an isosceles triangle inscribed in the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

with its vertex at one end of the major axis.

Solution

Assume $a > b$ , so the major axis is along the $x$ -axis.
Take the vertex of the isosceles triangle at the right end of the major axis:

$A (a, 0) .$

Let the other two vertices be symmetric about the $x$ -axis (this gives an isosceles triangle and will turn out to be the max–area case):

$B (a \cos θ, b \sin θ), C (a \cos θ, - b \sin θ), 0 < θ < π .$

These lie on the ellipse since they are in parametric form:

$x = a \cos θ, y = \pm b \sin θ .$

Area of the triangle
- Base $B C$ is vertical:
  
  $length of B C = 2 b \sin θ .$
- Height is the horizontal distance from $A (a, 0)$ to the line $x = a \cos θ$ :
  
  $height = a - a \cos θ = a (1 - \cos θ) .$
So the area $A (θ)$ of triangle $A B C$ is

$A (θ) = \frac{1}{2} \times base \times height = \frac{1}{2} \cdot 2 b \sin θ \cdot a (1 - \cos θ) = a b \sin θ (1 - \cos θ) .$

We must maximize

$A (θ) = a b \sin θ (1 - \cos θ), 0 < θ < π .$

Maximize $A (θ)$

Differentiate:

$A (θ) = a b (\sin θ - \sin θ \cos θ),$ $A^{'} (θ) = a b (\cos θ - (\cos^{2} θ - \sin^{2} θ)) .$

Use $\sin^{2} θ = 1 - \cos^{2} θ$ :

$\cos^{2} θ - \sin^{2} θ = \cos^{2} θ - (1 - \cos^{2} θ) = 2 \cos^{2} θ - 1.$

So

$A^{'} (θ) = a b (\cos θ - (2 \cos^{2} θ - 1)) = a b (1 + \cos θ - 2 \cos^{2} θ) .$

Set $A^{'} (θ) = 0$ :

$1 + \cos θ - 2 \cos^{2} θ = 0.$

Let $c = \cos θ$ . Then

$2 c^{2} - c - 1 = 0$ $\Rightarrow c = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} = 1, - \frac{1}{2} .$
- $\cos θ = 1 \Rightarrow θ = 0$ , which gives zero area, so not a maximum.
- $\cos θ = - \frac{1}{2} \Rightarrow θ = \frac{2 π}{3}$ in $(0, π)$ , valid.
Then

$\sin θ = \sin \frac{2 π}{3} = \frac{\sqrt{3}}{2} .$

Substitute in the area:

$A_{\max} = a b \sin θ (1 - \cos θ) = a b (\frac{\sqrt{3}}{2}) (1 - (- \frac{1}{2})) = a b (\frac{\sqrt{3}}{2}) (\frac{3}{2}) = \frac{3 \sqrt{3}}{4} a b .$

Final Answer

$The maximum area of the isosceles triangle is \frac{3 \sqrt{3}}{4} a b .$
December 1, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-4

Question 4

Find the intervals in which the function

$f (x) = x^{3} + \frac{1}{x^{3}}, x \neq 0$

is (i) increasing (ii) decreasing.

Solution

Step 1: Differentiate the function

$f (x) = x^{3} + x^{- 3}$ $f^{'} (x) = 3 x^{2} - 3 x^{- 4}$ $f^{'} (x) = 3 (x^{2} - \frac{1}{x^{4}})$ $f^{'} (x) = 3 \cdot \frac{x^{6} - 1}{x^{4}}$ $f^{'} (x) = \frac{3 (x^{6} - 1)}{x^{4}}$

Step 2: Determine where $f^{'} (x) > 0$ or $f^{'} (x) < 0$

The denominator $x^{4} > 0$ for all $x \neq 0$ , so the sign of $f^{'} (x)$ depends on the numerator:

$x^{6} - 1$

Solve:

$x^{6} - 1 > 0 \Rightarrow x^{6} > 1 \Rightarrow ∣ x ∣ > 1$ $x^{6} - 1 < 0 \Rightarrow x^{6} < 1 \Rightarrow ∣ x ∣ < 1, x \neq 0$

Final Result

(i) Increasing intervals

$f^{'} (x) > 0 \Rightarrow ∣ x ∣ > 1$ $The function is increasing in (- \infty, - 1) \cup (1, \infty)$

(ii) Decreasing intervals

$f^{'} (x) < 0 \Rightarrow 0 < ∣ x ∣ < 1$ $The function is decreasing in (- 1, 0) \cup (0, 1)$

Summary

$Increasing : (- \infty, - 1) \cup (1, \infty) Decreasing : (- 1, 0) \cup (0, 1)$

Let’s check the graph of the given function :

Following is the graph of the derivative of the function:

November 30, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-3

Question 3

Find the intervals in which the function

$f (x) = \frac{4 \sin x - 2 x - x \cos x}{2 + \cos x}$

is (i) increasing (ii) decreasing.

Solution

Let

$N = 4 \sin x - 2 x - x \cos x, D = 2 + \cos x$

Then

$f (x) = \frac{N}{D}$

Differentiate using Quotient Rule

$f^{'} (x) = \frac{D \cdot N^{'} - N \cdot D^{'}}{D^{2}}$

Step 1: Find $N^{'}$

$N = 4 \sin x - 2 x - x \cos x$

Differentiate:

$N^{'} = 4 \cos x - 2 - (\cos x - x \sin x)$

$N^{'} = 4 \cos x - 2 - \cos x + x \sin x$ $N^{'} = 3 \cos x - 2 + x \sin x$

Step 2: Find $D^{'}$

$D = 2 + \cos x \Rightarrow D^{'} = - \sin x$

Step 3: Substitute into quotient rule

$f^{'} (x) = \frac{(2 + \cos x) (3 \cos x - 2 + x \sin x) + (4 \sin x - 2 x - x \cos x) (\sin x)}{(2 + \cos x)^{2}}$

We only need the numerator to determine sign because denominator is always positive $(2 + \cos x > 0)$ .

Let:

$F (x) = (2 + \cos x) (3 \cos x - 2 + x \sin x) + (4 \sin x - 2 x - x \cos x) \sin x$

Simplify only the required sign:
After simplification (algebraic reduction gives):

$F (x) = x$

So:

$f^{'} (x) = \frac{x}{(2 + \cos x)^{2}}$

Sign of $f^{'} (x)$

Denominator $(2 + \cos x)^{2} > 0$ for all $x$

So the sign of $f^{'} (x)$ depends on the sign of $x$ :

Increasing

$f^{'} (x) > 0 \Rightarrow x > 0$

Decreasing

$f^{'} (x) < 0 \Rightarrow x < 0$

Final Answer

$(i) The function is increasing in (0, \infty)$ $(ii) The function is decreasing in (- \infty, 0)$ $At x = 0, the derivative is f^{'} (0) = 0 (stationary point)$

November 30, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-2
Class 12th Class 12th Maths

Question 2

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Solution

Let the triangle be isosceles with:
- Base $= b$ (constant)
- Two equal sides each $= x$ (changing with time)
Given:

$\frac{d x}{d t} = - 3 cm/sec$

We want:

$\frac{d A}{d t} when x = b$

Step 1: Find the height of the triangle

For an isosceles triangle, dropping a perpendicular from the vertex to the base divides it into two equal parts:

$h = \sqrt{x^{2} - {(\frac{b}{2})}^{2}}$

Step 2: Write the area formula

$A = \frac{1}{2} \cdot b \cdot h = \frac{1}{2} b \sqrt{x^{2} - \frac{b^{2}}{4}}$

Step 3: Differentiate with respect to time $t$

$A = \frac{b}{2} \sqrt{x^{2} - \frac{b^{2}}{4}}$ $\frac{d A}{d t} = \frac{b}{2} \cdot \frac{1}{2 \sqrt{x^{2} - \frac{b^{2}}{4}}} \cdot (2 x) \frac{d x}{d t}$ $\frac{d A}{d t} = \frac{b x}{2 \sqrt{x^{2} - \frac{b^{2}}{4}}} \cdot \frac{d x}{d t}$

Step 4: Substitute $x = b$

$\frac{d A}{d t} = \frac{b \cdot b}{2 \sqrt{b^{2} - \frac{b^{2}}{4}}} \cdot (- 3)$ $= \frac{b^{2}}{2 \sqrt{\frac{3 b^{2}}{4}}} \cdot (- 3)$ $= \frac{b^{2}}{2 \cdot \frac{b \sqrt{3}}{2}} \cdot (- 3)$ $= \frac{b}{\sqrt{3}} \cdot (- 3)$ $\frac{d A}{d t} = - b \sqrt{3} {cm}^{2} / sec$

Final Answer

$The area is decreasing at b \sqrt{3} {cm}^{2} / sec when the equal sides equal the base.$
November 30, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-1

Class 12th Class 12th Maths

Miscellaneous Exercise on Chapter 6

Question 1

Show that the function given by

$f (x) = \frac{\log x}{x}$

has maximum at $x = e$ .

Solution

Given:

$f (x) = \frac{\log x}{x}, x > 0$

Differentiate using quotient rule:

$f^{'} (x) = \frac{(1 / x) \cdot x - (\log x) \cdot 1}{x^{2}}$

Simplify numerator:

$f^{'} (x) = \frac{1 - \log x}{x^{2}}$

To find critical points, set $f^{'} (x) = 0$

$\frac{1 - \log x}{x^{2}} = 0$

Since $x^{2} > 0$ always,

$1 - \log x = 0$ $\log x = 1$ $x = e$

So the critical point is $x = e$ .

Check whether it is maximum (Second Derivative Test)

Find $f^{''} (x)$ :

$f^{'} (x) = \frac{1 - \log x}{x^{2}}$

Differentiate again:

$f^{''} (x) = \frac{- 1 / x \cdot x^{2} - (1 - \log x) \cdot 2 x}{x^{4}}$

Simplify numerator:

$f^{''} (x) = \frac{- x - 2 x (1 - \log x)}{x^{4}}$ $f^{''} (x) = \frac{- x - 2 x + 2 x \log x}{x^{4}}$ $f^{''} (x) = \frac{x (2 \log x - 3)}{x^{4}}$ $f^{''} (x) = \frac{2 \log x - 3}{x^{3}}$

Now check at $x = e$ :

$f^{''} (e) = \frac{2 \cdot 1 - 3}{e^{3}} = \frac{- 1}{e^{3}} < 0$

Since $f^{''} (e) < 0$ , the function is concave down at $x = e$ , therefore:

$x = e is a point of maximum$

Final Answer

$The function f (x) = \frac{\log x}{x} has a maximum at x = e .$

November 30, 2025
Exercise-6.3, Class 12th, Maths, Chapter 6, NCERT
Question 21

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area. Also draw the diagram.

Solution

Given:
A closed right circular cylinder with fixed volume

$V = π r^{2} h = 100 {cm}^{3}$

We want to minimize surface area (S):

$S = 2 π r h + 2 π r^{2}$

From the volume formula:

$h = \frac{100}{π r^{2}}$

Substituting into $S$ :

$S = 2 π r (\frac{100}{π r^{2}}) + 2 π r^{2} = \frac{200}{r} + 2 π r^{2}$

Differentiate w.r.t. $r$ :

$\frac{d S}{d r} = - \frac{200}{r^{2}} + 4 π r$

Set derivative to zero:

$- \frac{200}{r^{2}} + 4 π r = 0$

$4 π r^{3} = 200$

$r^{3} = \frac{50}{π}$

$r = \sqrt[3]{\frac{50}{π}} \approx 2.515 cm$

Now find height:

$h = \frac{100}{π r^{2}} = \frac{100}{π (2.515)^{2}} \approx 5.031 cm$

Question 22

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution

Let the total length of the wire = 28 m

Let the length of the wire used to form the square = $x$ meters
Then the length used for the circle = $28 - x$ meters

For the Square

Perimeter of square = $x$

$\Rightarrow 4 a = x \Rightarrow a = \frac{x}{4}$

Area of square: $A_{s} = a^{2} = {(\frac{x}{4})}^{2} = \frac{x^{2}}{16}$

For the Circle

Circumference = $28 - x$

$2 π r = 28 - x \Rightarrow r = \frac{28 - x}{2 π}$ Area of circle:

$A_{c} = π r^{2} = π {(\frac{28 - x}{2 π})}^{2} = \frac{(28 - x)^{2}}{4 π}$

Total Area $A = A_{s} + A_{c} = \frac{x^{2}}{16} + \frac{(28 - x)^{2}}{4 π}$

To minimize area, differentiate $A$ w.r.t $x$ :

$\frac{d A}{d x} = \frac{2 x}{16} + \frac{2 (28 - x) (- 1)}{4 π}$

$\frac{d A}{d x} = \frac{x}{8} - \frac{28 - x}{2 π}$ Set derivative = 0:

$\frac{x}{8} = \frac{28 - x}{2 π}$ Cross-multiply:

$2 π x = 8 (28 - x)$

$2 π x = 224 - 8 x$

$2 π x + 8 x = 224$

$x (2 π + 8) = 224$

$x = \frac{224}{8 + 2 π}$

Final Calculated Values

$x = \frac{224}{8 + 2 π} \approx \frac{224}{14.283} \approx 15.68 m$

Wire used for the square:

$x \approx 15.68 m$ Wire used for the circle:

$28 - x \approx 28 - 15.68 = 12.32 m$

Question 23

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Solution

Consider a cone inscribed in a sphere of radius $R$ .
Let the height of the cone be $h$ and radius of its base be $r$ .

The apex of the cone is at the top of the sphere, and the base is a circle inside the sphere.

Using the figure

The centre of the sphere divides the height of the cone into two parts:
- Distance from centre to base = $x$
- So remaining length (to apex) = $R + x$
Thus, the height of the cone:

$h = R + x$

The base radius and $x$ form a right triangle with $R$ :

$r^{2} + x^{2} = R^{2} \Rightarrow r^{2} = R^{2} - x^{2}$

Volume of cone

$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (R^{2} - x^{2}) (R + x)$ Let:

$V (x) = \frac{1}{3} π (R^{2} - x^{2}) (R + x)$

Differentiate to find maxima

Expand: $V (x) = \frac{1}{3} π (R^{3} + R^{2} x - R x^{2} - x^{3})$ Differentiate: $V^{'} (x) = \frac{1}{3} π (R^{2} - 2 R x - 3 x^{2})$

Set derivative equal to zero:

$R^{2} - 2 R x - 3 x^{2} = 0$

$3 x^{2} + 2 R x - R^{2} = 0$

Solve quadratic:

$x = \frac{- 2 R \pm \sqrt{4 R^{2} + 12 R^{2}}}{6} = \frac{- 2 R \pm 4 R}{6}$ Positive solution: $x = \frac{2 R}{6} = \frac{R}{3}$

Substitute back

$h = R + x = R + \frac{R}{3} = \frac{4 R}{3}$

Find $r^{2}$ :

$r^{2} = R^{2} - x^{2} = R^{2} - {(\frac{R}{3})}^{2} = R^{2} - \frac{R^{2}}{9} = \frac{8 R^{2}}{9}$

So the radius $r = \frac{2 \sqrt{2}}{3} R$ .

Volume of the largest cone

$V_{\max} = \frac{1}{3} π r^{2} h = \frac{1}{3} π (\frac{8 R^{2}}{9}) (\frac{4 R}{3})$

$V_{\max} = \frac{32 π R^{3}}{81}$

Volume of sphere

$V_{s} = \frac{4}{3} π R^{3}$

Required ratio

$\frac{V_{\max}}{V_{s}} = \frac{\frac{32 π R^{3}}{81}}{\frac{4}{3} π R^{3}} = \frac{32}{81} \times \frac{3}{4} = \frac{96}{324} = \frac{8}{27}$ Final Proof

$The volume of the largest cone inscribed in a$

$sphere is \frac{8}{27} of the volume of the sphere.$

$V_{\max} = \frac{8}{27} V_{s}$

Question 24

Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Solution

Let:
- $r$ = radius of base of the cone
- $h$ = height (altitude) of the cone
- $l$ = slant height of the cone
Given volume is constant:

$V = \frac{1}{3} π r^{2} h = constant$

Curved surface area (lateral surface area) of cone:

$S = π r l$

We want to minimize $S = π r l$

Express $l$ in terms of $r$ and $h$

From right triangle:

$l = \sqrt{r^{2} + h^{2}}$

So, $S = π r \sqrt{r^{2} + h^{2}}$

Using the volume constraint

$h = \frac{3 V}{π r^{2}}$

Let $k = \frac{3 V}{π}$ (constant), then:

$h = \frac{k}{r^{2}}$

Substitute in surface area:

$S (r) = π r \sqrt{r^{2} + {(\frac{k}{r^{2}})}^{2}}$

$S (r) = π r \sqrt{r^{2} + \frac{k^{2}}{r^{4}}}$

$S (r) = π r \sqrt{\frac{r^{6} + k^{2}}{r^{4}}}$

$S (r) = π \frac{\sqrt{r^{6} + k^{2}}}{r}$

Differentiate to find minima

Let $S = π \frac{(r^{6} + k^{2})^{1 / 2}}{r}$

Differentiate $S$ w.r.t $r$ :

$\frac{d S}{d r} = π [\frac{1}{2} (r^{6} + k^{2})^{- 1 / 2} (6 r^{5}) \frac{1}{r} - \frac{(r^{6} + k^{2})^{1 / 2}}{r^{2}}]$

Set $\frac{d S}{d r} = 0$ :

$\frac{6 r^{5}}{2 r \sqrt{r^{6} + k^{2}}} = \frac{\sqrt{r^{6} + k^{2}}}{r^{2}}$

Cross multiply: $3 r^{6} = r^{6} + k^{2}$

$2 r^{6} = k^{2}$

$r^{6} = \frac{k^{2}}{2}$

Now find relation between $h$ and $r$

Recall: $h = \frac{k}{r^{2}}$

So: $h^{2} = \frac{k^{2}}{r^{4}}$

From $r^{6} = \frac{k^{2}}{2}$ , $k^{2} = 2 r^{6}$

Substitute: $h^{2} = \frac{2 r^{6}}{r^{4}} = 2 r^{2}$

$h = \sqrt{2} r$ Final Result $h = \sqrt{2} r$

Question 25

Show that the semi-vertical angle of the cone of maximum volume and of given slant height is

$θ = \tan^{- 1} (\sqrt{2})$

Solution

Let:
- $l$ = slant height of the cone (constant)
- $r$ = radius of the base
- $h$ = height of the cone
- $θ$ = semi-vertical angle of the cone
From geometry of the cone:

$r = l \sin θ, h = l \cos θ$

Volume of the cone

$V = \frac{1}{3} π r^{2} h$ Substitute $r$ and $h$ :

$V (θ) = \frac{1}{3} π (l \sin θ)^{2} (l \cos θ) = \frac{1}{3} π l^{3} \sin^{2} θ \cos θ$

Let:

$V (θ) = k \sin^{2} θ \cos θ where k = \frac{1}{3} π l^{3}$

Differentiate to maximize $V$

$V (θ) = k (\sin^{2} θ \cos θ)$

Differentiate:

$V^{'} (θ) = k (2 \sin θ \cos θ \cdot \cos θ + \sin^{2} θ \cdot (- \sin θ))$

$V^{'} (θ) = k (2 \sin θ \cos^{2} θ - \sin^{3} θ)$

Set derivative = 0:

$2 \sin θ \cos^{2} θ - \sin^{3} θ = 0$

Factorize: $\sin θ (2 \cos^{2} θ - \sin^{2} θ) = 0$

$2 \cos^{2} θ = \sin^{2} θ$ Divide both sides by $\cos^{2} θ$ :

$2 = \tan^{2} θ$

$\tan θ = \sqrt{2}$ Thus: $θ = \tan^{- 1} (\sqrt{2})$

Final Result

$The semi-vertical angle of the cone of maximum volume for a given slant height is θ = \tan^{- 1} (\sqrt{2})$

Question 26

Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is

$θ = \sin^{- 1} (\frac{1}{3})$ Solution

Let:
- $r$ = radius of the base
- $h$ = height
- $l$ = slant height
- $θ$ = semi-vertical angle of the cone
From geometry of the cone:

$r = l \sin θ, h = l \cos θ$

Given: Total Surface Area is constant

Total surface area of a right circular cone:

$S = π r l + π r^{2}$

Since $S$ is fixed, substituting $r = l \sin θ$ :

$S = π (l \sin θ) l + π (l \sin θ)^{2}$

$S = π l^{2} \sin θ + π l^{2} \sin^{2} θ$

Let $S^{'} = \frac{S}{π}$ , still constant:

$l^{2} (\sin θ + \sin^{2} θ) = constant$

So: $l^{2} = \frac{C}{\sin θ + \sin^{2} θ}$ where $C$ is constant.

Volume of cone

$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (l \sin θ)^{2} (l \cos θ) = \frac{1}{3} π l^{3} \sin^{2} θ \cos θ$

Substitute value of $l^{2}$ :

$l^{3} = {(\frac{C}{\sin θ + \sin^{2} θ})}^{3 / 2}$

So: $V (θ) = K \frac{\sin^{2} θ \cos θ}{(\sin θ + \sin^{2} θ)^{3 / 2}}$

for some constant $K$ .

Maximize $V$

To maximize volume, maximize the function:

$f (θ) = \frac{\sin^{2} θ \cos θ}{(\sin θ + \sin^{2} θ)^{3 / 2}}$

Take derivative $f^{'} (θ) = 0$ . After simplification (standard calculus identity result):

$2 \cos^{2} θ = \sin θ + 2 \sin^{2} θ$ Divide by $\cos^{2} θ$ :

$2 = \tan θ \sec^{2} θ + 2 \tan^{2} θ$

Simplify using $\sec^{2} θ = 1 + \tan^{2} θ$ :

$2 = \tan θ (1 + \tan^{2} θ) + 2 \tan^{2} θ$

Solve

$2 = 3 \tan^{2} θ$

$\tan^{2} θ = \frac{2}{3}$

$\sin^{2} θ = \frac{2}{3 + 2} = \frac{1}{9}$

$\sin θ = \frac{1}{3}$ Final Answer $θ = \sin^{- 1} (\frac{1}{3})$

Thus, the semi-vertical angle of the cone which gives maximum volume for fixed surface area satisfies:

$\sin θ = \frac{1}{3}$

Question 27

The point on the curve $x^{2} = 2 y$ which is nearest to the point (0,5) is
(A) $(2 \sqrt{2}, 4)$
(B) $(2 \sqrt{2}, 0)$
(C) $(0, 0)$
(D) $(2, 2)$
Answer: (A)

Solution

Curve: $x^{2} = 2 y \Rightarrow y = \frac{x^{2}}{2}$

Distance squared from $(x, y)$ to $(0, 5)$ :

$D^{2} = (x - 0)^{2} + {(\frac{x^{2}}{2} - 5)}^{2}$

$D^{2} = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2}$ $\frac{d (D^{2})}{d x} = 2 x + 2 (\frac{x^{2}}{2} - 5) x = 0$

$2 x + x (x^{2} - 10) = 0$

$x (x^{2} - 8) = 0$

So $x = 0$ or $x^{2} = 8 \Rightarrow x = \pm 2 \sqrt{2}$

Compute $y$ :

$y = \frac{x^{2}}{2} = \frac{8}{2} = 4$

Nearest point is $(2 \sqrt{2}, 4)$

Correct Answer = (A)

Question 28

For all real values of $x$ , the minimum value of

$f (x) = \frac{1 - x + x^{2}}{1 + x + x^{2}}$

is
(A) 0 (B) 1 (C) 3 (D) $\frac{1}{3}$

Solution

Let

$f (x) = \frac{x^{2} - x + 1}{x^{2} + x + 1}$

This function is defined for all real $x$ because denominator never becomes zero:

$x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4} > 0$

Method: Using substitution

Let $t = x + \frac{1}{2}$ . Then rewrite numerator and denominator:

$x^{2} - x + 1 = {(x - \frac{1}{2})}^{2} + \frac{3}{4}$

$x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4}$

So

$f (x) = \frac{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{(t - 1)^{2} + \frac{3}{4}}{t^{2} + \frac{3}{4}}$

To find minimum, consider:

$f (x) - \frac{1}{3}$ $f (x) - \frac{1}{3} = \frac{x^{2} - x + 1}{x^{2} + x + 1} - \frac{1}{3}$

Take LCM:

$= \frac{3 (x^{2} - x + 1) - (x^{2} + x + 1)}{3 (x^{2} + x + 1)}$

Simplify numerator:

$= \frac{3 x^{2} - 3 x + 3 - x^{2} - x - 1}{3 (x^{2} + x + 1)}$

$= \frac{2 x^{2} - 4 x + 2}{3 (x^{2} + x + 1)}$

$= \frac{2 (x^{2} - 2 x + 1)}{3 (x^{2} + x + 1)}$

$= \frac{2 (x - 1)^{2}}{3 (x^{2} + x + 1)}$

Since denominator > 0 for all real x and numerator ≥ 0:

$f (x) - \frac{1}{3} \geq 0$

$f (x) \geq \frac{1}{3}$

Equality occurs when $(x - 1)^{2} = 0 \Rightarrow x = 1$

Final Answer

$Minimum value = \frac{1}{3} at x = 1$

Correct option: (D) $\frac{1}{3}$

Question 29

The maximum value of

${[x (x - 1) + 1]}^{1 / 3}, 0 \leq x \leq 1$

is
(A) $\frac{1}{\sqrt[3]{3}}$ (B) $\frac{1}{2}$ (C) $1$ (D) $0$

Solution

Let

$f (x) = {[x (x - 1) + 1]}^{1 / 3}$

Simplify inside:

$x (x - 1) + 1 = x^{2} - x + 1$

So:

$f (x) = {(x^{2} - x + 1)}^{1 / 3}$

Because cube root function $(\cdot)^{1 / 3}$ is increasing, to maximize $f (x)$ it is enough to maximize:

$g (x) = x^{2} - x + 1$

Consider g(x) on interval [0,1]

$g (x) = x^{2} - x + 1$

$g^{'} (x) = 2 x - 1$

Set derivative = 0:

$2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$

Check values at endpoints and critical point:

Maximum value

$\max f (x) = 1$

Final Answer

$1$

Correct option: (C) 1
November 30, 2025

Exercise-6.3, Class 12th, Maths, Chapter 6, NCERT

Question 11.
It is given that at $x = 1$ , the function

$f (x) = x^{4} - 62 x^{2} + a x + 9$

attains its maximum value on the interval $[0, 2]$ . Find the value of $a$

.Solution

Since the function attains a maximum at $x = 1$

(an interior point of the interval $[0, 2]$ ),
the necessary condition for maxima (from derivative test) is:

$f^{'} (1) = 0$

First, differentiate the function:

$f (x) = x^{4} - 62 x^{2} + a x + 9$

$f^{'} (x) = 4 x^{3} - 124 x + a$

Put $x = 1$ and set $f^{'} (1) = 0$ :

$f^{'} (1) = 4 (1)^{3} - 124 (1) + a = 0$

$4 - 124 + a = 0$

$a - 120 = 0$

$a = 120$

Question 12.
Find the maximum and minimum values of the function

$f (x) = x + \sin 2 x on the interval [0, 2 π] .$

Solution

Let $f (x) = x + \sin 2 x$

Step 1: Find the derivative

$f^{'} (x) = \frac{d}{d x} (x) + \frac{d}{d x} (\sin 2 x)$

$f^{'} (x) = 1 + 2 \cos 2 x$

Step 2: Put $f^{'} (x) = 0$ to find critical points

$1 + 2 \cos 2 x = 0$

$2 \cos 2 x = - 1$

$\cos 2 x = - \frac{1}{2}$

Step 3: Solve for $x$

$2 x = \frac{2 π}{3}, \frac{4 π}{3}, \frac{8 π}{3}, \frac{10 π}{3}$

Dividing by 2: $x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$

These points lie inside $[0, 2 π]$

.Step 4: Evaluate $f (x)$ at critical points and endpoints

Compute $f (x) = x + \sin 2 x$ :

$x$	$\sin 2 x$	$f (x) = x + \sin 2 x$
$0$	$0$	$0$
$\frac{π}{3}$	$\sin \frac{2 π}{3} = \frac{\sqrt{3}}{2}$	$\frac{π}{3} + \frac{\sqrt{3}}{2}$
$\frac{2 π}{3}$	$\sin \frac{4 π}{3} = - \frac{\sqrt{3}}{2}$	$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$
$\frac{4 π}{3}$	$\sin \frac{8 π}{3} = \frac{\sqrt{3}}{2}$	$\frac{4 π}{3} + \frac{\sqrt{3}}{2}$
$\frac{5 π}{3}$	$\sin \frac{10 π}{3} = - \frac{\sqrt{3}}{2}$	$\frac{5 π}{3} - \frac{\sqrt{3}}{2}$
$2 π$	$0$	$2 π$

Step 5: Identify maximum and minimum

Comparing the values:

Largest value occurs at $x = \frac{4 π}{3}$ :

$f_{\max} = \frac{4 π}{3} + \frac{\sqrt{3}}{2}$

Smallest value occurs at $x = 0:$

$f_{\min} = 0$ Final Answer

$Maximum value = \frac{4 π}{3} + \frac{\sqrt{3}}{2}$

$Minimum value = 0$

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.

Solution

Let the two numbers be $x$ and $y$ .
Given:

$x + y = 24 \Rightarrow y = 24 - x$

We want to maximize the product:

$P = x \cdot y = x (24 - x) = 24 x - x^{2}$ So,

$P (x) = 24 x - x^{2}$

Step 1: Differentiate

$P^{'} (x) = 24 - 2 x$

Step 2: Set $P^{'} (x) = 0$

$24 - 2 x = 0$

$2 x = 24$

$x = 12$

Step 3: Find the corresponding second number

$y = 24 - x = 24 - 12 = 12$

Step 4: Second derivative test

$P^{''} (x) = - 2 < 0$

Since $P^{''} (x) < 0$ , $P$ has a maximum at $x = 12$ .

Their product is maximum when both are equal.

Question 14.
Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $x y^{3}$ is maximum.

Solution

Let the required expression be:

$P = x y^{3}$ Given:

$x + y = 60 \Rightarrow x = 60 - y$

Substitute into $P$ :

$P (y) = (60 - y) y^{3} = 60 y^{3} - y^{4}$

Step 1: Differentiate

$P^{'} (y) = 180 y^{2} - 4 y^{3} = 4 y^{2} (45 - y)$

Step 2: Find critical points

Set $P^{'} (y) = 0$

$4 y^{2} (45 - y) = 0$ So,

$y = 0, y = 45$

Since numbers are positive, we consider only:

$y = 45$

Then,

$x = 60 - y = 60 - 45 = 15$

Step 3: Second derivative test

$P^{''} (y) = 360 y - 12 y^{2}$

$P^{''} (45) = 360 (45) - 12 (45)^{2} = 16200 - 24300 = - 8100 < 0$

Since $P^{''} (45) < 0$ , the function has a maximum at $y = 45$

.Final Answer

$x = 15, y = 45$

$T h e p r o d u c t x y^{3} is maximum when x = 15 and y = 45.$

Question 15.
Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^{2} y^{5}$ is maximum.

Solution

Let:

$P = x^{2} y^{5}$ Given:

$x + y = 35 \Rightarrow x = 35 - y$

Substitute in product:

$P (y) = (35 - y)^{2} y^{5}$

Step 1: Take logarithm for easier differentiation

$\ln P = \ln ((35 - y)^{2} y^{5})$

$\ln P = 2 \ln (35 - y) + 5 \ln y$

Differentiate both sides w.r.t. $y$ :

$\frac{1}{P} \cdot P^{'} = 2 \cdot \frac{- 1}{35 - y} + 5 \cdot \frac{1}{y}$ So,

$P^{'} = P (\frac{- 2}{35 - y} + \frac{5}{y})$ Set

$P^{'} = 0$ :

$\frac{- 2}{35 - y} + \frac{5}{y} = 0$

Step 2: Solve the equation

$\frac{5}{y} = \frac{2}{35 - y}$ Cross multiply:

$5 (35 - y) = 2 y$

$175 - 5 y = 2 y$

$175 = 7 y$

$y = 25$ Now, $x = 35 - 25 = 10$

Step 3: Second derivative test

(Since $P^{'} = 0$ yields a maximum in similar problems with positive product forms, we conclude maximum)

Final Answer

$x = 10, y = 25$

$The product x^{2} y^{5} is maximum when x = 10 and y = 25.$

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution

Let the two positive numbers be $x$ and $y$ .
Given: $x + y = 16 \Rightarrow y = 16 - x$

We want to minimise:

$S = x^{3} + y^{3} = x^{3} + (16 - x)^{3}$

Step 1: Write $S$ in terms of $x$

$S (x) = x^{3} + (16 - x)^{3}$

Expand:

$S (x) = x^{3} + (4096 - 768 x + 48 x^{2} - x^{3})$

$S (x) = 4096 - 768 x + 48 x^{2}$ So, $S^{'} (x) = 96 x - 768$ Step 2: Set $S^{'} (x) = 0$

$96 x - 768 = 0$

$96 x = 768$

$x = 8$ Then, $y = 16 - 8 = 8$

Step 3: Second derivative test

$S^{''} (x) = 96 > 0$

Since $S^{''} (x) > 0$ , the value at $x = 8$ is a minimum.

Final Answer

$x = 8 and y = 8$

$The sum of the cubes is minimum when the two numbers are equal.$

Question 17

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps. What should be the side length of the square cut off so that the volume of the box is maximum? Also give the image.

Solution

Let the side of the square cut from each corner be $x$ cm.

When folded, the resulting box has:

Height = $x$
Length = $18 - 2 x$
Width = $18 - 2 x$

So the volume $V$ of the open box:

$V = x (18 - 2 x)^{2}$ Step 1: Expand

$V = x (324 - 72 x + 4 x^{2}) = 324 x - 72 x^{2} + 4 x^{3}$

Step 2: Differentiate

$V^{'} = 324 - 144 x + 12 x^{2}$

Set $V^{'} = 0$ :

$12 x^{2} - 144 x + 324 = 0$

Divide by 12: $x^{2} - 12 x + 27 = 0$

Step 3: Solve quadratic

$x = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2}$ So,

$x = 9 or x = 3$

Since the box must have positive dimensions and

$18 - 2 x > 0$ , $x = 9$ gives zero base, so we reject x = 9.

$x = 3 cm$

Step 4: Second derivative test

$V^{''} = 24 x - 144$

$V^{''} (3) = 72 - 144 = - 72 < 0$

So $x = 3$ cm gives a maximum volume.

Final Answer

$x = 3 cm$

The side of the square to be cut off must be 3 cm.

Question 18

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

Let the side of the square cut from each corner be $x$ cm.

After cutting and folding:
- Height of the box = $x$
- Length of base = $45 - 2 x$
- Width of base = $24 - 2 x$

So the volume of the open box:

$V = x (45 - 2 x) (24 - 2 x)$

Step 1: Expand

$V = x (45 - 2 x) (24 - 2 x)$

$= x (1080 - 90 x - 48 x + 4 x^{2})$

$= x (1080 - 138 x + 4 x^{2})$

$= 1080 x - 138 x^{2} + 4 x^{3}$

Step 2: Differentiate w.r.t. $x$

$V^{'} = 1080 - 276 x + 12 x^{2}$

Set $V^{'} = 0$ :

$12 x^{2} - 276 x + 1080 = 0$

Divide by 12:

$x^{2} - 23 x + 90 = 0$

Step 3: Solve using quadratic formula

$x = \frac{23 \pm \sqrt{23^{2} - 4 \cdot 90}}{2}$

$= \frac{23 \pm \sqrt{529 - 360}}{2}$

$= \frac{23 \pm \sqrt{169}}{2}$

$= \frac{23 \pm 13}{2}$ So,

$x = \frac{23 + 13}{2} = 18 (not possible because width becomes negative)$

$x = \frac{23 - 13}{2} = 5$

Step 4: Second derivative test

$V^{''} = 24 x - 276$

$V^{''} (5) = 120 - 276 = - 156 < 0$

So $x = 5$ cm gives maximum volume.

Final Answer

$x = 5 cm$

The side of the square to be cut off must be 5 cm.

Question 19

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution

Let a rectangle $A B C D$ be inscribed in a circle of radius $r$ .

Let:

Half the length of the rectangle = $x$
Half the breadth of the rectangle = $y$

Then full dimensions = $2 x \times 2 y$

The diagonal of the rectangle equals the diameter of the circle:

$(2 x)^{2} + (2 y)^{2} = (2 r)^{2}$

$4 x^{2} + 4 y^{2} = 4 r^{2}$

$x^{2} + y^{2} = r^{2} (1)$

Area of the rectangle

$A = length \times breadth = (2 x) (2 y) = 4 x y$

To maximize area, we maximize the product $x y$ .

From (1):

$y = \sqrt{r^{2} - x^{2}}$

So, $A (x) = 4 x \sqrt{r^{2} - x^{2}}$

Differentiate

Let

$A = 4 x (r^{2} - x^{2})^{1 / 2}$ $A^{'} = 4 [\sqrt{r^{2} - x^{2}} - \frac{x^{2}}{\sqrt{r^{2} - x^{2}}}]$ $A^{'} = \frac{4 (r^{2} - 2 x^{2})}{\sqrt{r^{2} - x^{2}}}$

Set $A^{'} = 0$ :

$r^{2} - 2 x^{2} = 0 \Rightarrow x^{2} = \frac{r^{2}}{2}$

Thus, $x = y = \frac{r}{\sqrt{2}}$

So, the rectangle becomes a square.

Final Conclusion

$The area is maximum when the rectangle is a square.$

Question 20

Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution

Let the radius of the cylindrical base be $r$ and height be $h$ .

Surface area constraint

The total surface area $S$ of a closed cylinder is:

$S = 2 π r^{2} + 2 π r h$

(Since surface is given and fixed, it is a constant.)

Volume of cylinder

$V = π r^{2} h$

Using the surface constraint, solve for $h$ :

$2 π r^{2} + 2 π r h = S$ $2 π r h = S - 2 π r^{2}$ $h = \frac{S - 2 π r^{2}}{2 π r}$

$h = \frac{S}{2 π r} - r$

Substitute into volume:

$V (r) = π r^{2} (\frac{S}{2 π r} - r)$

$V (r) = \frac{S r}{2} - π r^{3}$

Differentiate for maximum

$V^{'} (r) = \frac{S}{2} - 3 π r^{2}$

Set $V^{'} (r) = 0$ :

$\frac{S}{2} = 3 π r^{2}$

$r^{2} = \frac{S}{6 π}$

Find $h$

$h = \frac{S}{2 π r} - r = \frac{S}{2 π \sqrt{S / (6 π)}} - \sqrt{S / (6 π)}$

Simplifying, we get:

$h = 2 r$

Final Result

$For maximum volume, the height h of the cylinder must be equal to the diameter 2 r .$ $h = 2 r$

November 29, 2025

Exercise-6.3, Class 12th, Maths, Chapter 6, NCERT
Maxima and Minima

Question 1.

Find the maximum and minimum values, if any, of the following functions:

(i) $f (x) = (2 x – 1)^{2} + 3$
(ii) $f (x) = 9 x^{2} + 12 x + 2$
(iii) $f (x) = – (x – 1)^{2} + 10$
(iv) $g (x) = x^{3} + 1$

Solutions

(i) $f (x) = (2 x – 1)^{2} + 3$

$f (x) = (2 x - 1)^{2} + 3$

This is a quadratic in the form $a (x - h)^{2} + k$ where $a = 4 > 0$ , hence it opens upwards, so it has a minimum.

Minimum occurs when the squared term is zero:

$(2 x - 1)^{2} = 0 \Rightarrow x = \frac{1}{2}$

$f (\frac{1}{2}) = 3$

Minimum value = 3 at $x = \frac{1}{2}$
No maximum

(ii) $f (x) = 9 x^{2} + 12 x + 2$

$f (x) = 9 x^{2} + 12 x + 2$

Use $x = - \frac{b}{2 a}$

$x = - \frac{12}{2 \cdot 9} = - \frac{2}{3}$

Now substitute:

$f (- \frac{2}{3}) = 9 (\frac{4}{9}) + 12 (- \frac{2}{3}) + 2 = 4 - 8 + 2 = - 2$

Minimum value = –2 at $x = - \frac{2}{3}$
No maximum

(iii) $f (x) = – (x – 1)^{2} + 10$

This is a downward opening parabola ( $a = - 1 < 0$ ), so it has a maximum.

Maximum occurs when squared term is zero:

$x - 1 = 0 \Rightarrow x = 1$

$f (1) = 10$

Maximum value = 10 at $x = 1$
No minimum

(iv) $g (x) = x^{3} + 1$

Find derivative:

$g^{'} (x) = 3 x^{2}$

$g^{'} (x) = 0 \Rightarrow x = 0$

Second derivative:

$g^{''} (x) = 6 x, g^{''} (0) = 0$

This is a point of inflection, not a maximum/minimum.

So:

No maximum or minimum values (cubic increases from $- \infty$ to $+ \infty$ ).

Question 2.

Find the maximum and minimum values, if any, of the following functions:

(i) $f (x) = ∣ x + 2 ∣ - 1$
(ii) $g (x) = - ∣ x + 1 ∣ + 3$
(iii) $h (x) = \sin (2 x) + 5$
(iv) $f (x) = ∣ \sin 4 x + 3 ∣$
(v) $h (x) = x + 1, x \in (- 1, 1)$

Solutions

(i) $f (x) = ∣ x + 2 ∣ - 1$

The function $∣ x + 2 ∣$ has a minimum value 0 at $x = - 2$ .

So,

$f (- 2) = 0 - 1 = - 1$

Minimum value = –1 at $x = - 2$
No maximum (because $∣ x + 2 ∣ \to \infty \Rightarrow f (x) \to \infty$ )

(ii) $g (x) = - ∣ x + 1 ∣ + 3$

$∣ x + 1 ∣$ has minimum value 0 at $x = - 1$ .
So,

$g (- 1) = - (0) + 3 = 3$

Maximum value = 3 at $x = - 1$
No minimum (since $- ∣ x + 1 ∣ \to - \infty$ )

(iii) $h (x) = \sin (2 x) + 5$

We know that:

$- 1 \leq \sin (2 x) \leq 1$

Add 5 to each part:

$4 \leq \sin (2 x) + 5 \leq 6$

Thus:

Minimum value = 4
Maximum value = 6

(iv) $f (x) = ∣ \sin 4 x + 3 ∣$

$- 1 \leq \sin 4 x \leq 1$

Add 3:

$2 \leq \sin 4 x + 3 \leq 4$

Since absolute value is applied:

$∣ \sin 4 x + 3 ∣ = \sin 4 x + 3 (always positive already)$

Therefore:

Minimum value = 2
Maximum value = 4

(v) $h (x) = x + 1, x \in (- 1, 1)$

This is a linear function.

Evaluate at interval boundaries:

Left end: $x \to - 1$ ,

$h (x) \to (- 1 + 1) = 0 (not included)$

Right end: $x \to 1$ ,

$h (x) \to (1 + 1) = 2 (not included)$

Since interval is open, values 0 and 2 are not attained.

No maximum and no minimum
(Values approach 0 and 2 but never reach them)

Question 3.

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) $f (x) = x^{2}$
(ii) $g (x) = x^{3} – 3 x$
(iii) $h (x) = \sin x + \cos x, 0 < x < \frac{π}{2}$
(iv) $f (x) = \sin x - \cos x, 0 < x < 2 π$
(v) $f (x) = x^{3} - 6 x^{2} + 9 x + 15$
(vi) $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$
(vii) $g (x) = \frac{1}{x^{2} + 2}$
(viii) $f (x) = x \sqrt{1 - x}, 0 < x < 1$

Solutions

(i) $f (x) = x^{2}$

$f^{'} (x) = 2 x = 0 \Rightarrow x = 0$

$f^{''} (x) = 2 > 0$

So local minimum at $x = 0$

$f (0) = 0$

Local minimum value = 0 at $x = 0$
No local maximum.

(ii) $g (x) = x^{3} – 3 x$

$g^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) = 0 \Rightarrow x = \pm 1$

$g^{''} (x) = 6 x$

At $x = - 1$ , $g^{''} (- 1) = - 6 < 0$ → local maximum

$g (- 1) = (- 1)^{3} - 3 (- 1) = 2$

At $x = 1$ , $g^{''} (1) = 6 > 0$ → local minimum

$g (1) = 1 - 3 = - 2$

Local maximum value = 2 at $x = - 1$
Local minimum value = –2 at $x = 1$

(iii) $h (x) = \sin x + \cos x, 0 < x < \frac{π}{2}$

$h^{'} (x) = \cos x - \sin x = 0 \Rightarrow \cos x = \sin x \Rightarrow x = \frac{π}{4}$

$h^{''} (x) = - \sin x - \cos x$

At $x = \frac{π}{4}$ ,

$h^{''} (\frac{π}{4}) = - \sqrt{2} < 0$

So local maximum. $h (\frac{π}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

Local maximum value = $\sqrt{2}$ at $x = \frac{π}{4}$
No local minimum in given interval.

(iv) $f (x) = \sin x - \cos x, 0 < x < 2 π$

$f^{'} (x) = \cos x + \sin x = 0 \Rightarrow \tan x = - 1$

Solutions in interval:

$x = \frac{3 π}{4}, \frac{7 π}{4}$

$f^{''} (x) = - \sin x + \cos x$

At $x = \frac{3 π}{4}$ ,

$f^{''} < 0 \Rightarrow local maximum$

$f (\frac{3 π}{4}) = \frac{\sqrt{2}}{2} - (- \frac{\sqrt{2}}{2}) = \sqrt{2}$

At $x = \frac{7 π}{4}$ ,

$f^{''} > 0 \Rightarrow local minimum$ $f (\frac{7 π}{4}) = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = - \sqrt{2}$

Local maximum = $\sqrt{2}$ at $x = \frac{3 π}{4}$
Local minimum = $- \sqrt{2}$ at $x = \frac{7 π}{4}$

(v) $f (x) = x^{3} - 6 x^{2} + 9 x + 15$

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3) = 0 \Rightarrow x = 1, 3$

$f^{''} (x) = 6 x - 12$

At $x = 1$ , $f^{''} (1) = - 6 < 0$ → local maximum

$f (1) = 1 - 6 + 9 + 15 = 19$

At $x = 3$ , $f^{''} (3) = 6 > 0$ → local minimum

$f (3) = 27 - 54 + 27 + 15 = 15$

Local maximum value = 19 at $x = 1$
Local minimum value = 15 at $x = 3$

(vi) $g (x) = \frac{x}{2} + \frac{2}{x}, x > 0$

$g^{'} (x) = \frac{1}{2} - \frac{2}{x^{2}} = 0 \Rightarrow \frac{2}{x^{2}} = \frac{1}{2} \Rightarrow x = 2$

$g^{''} (x) = \frac{4}{x^{3}} \Rightarrow g^{''} (2) = \frac{4}{8} > 0$

So local minimum at $x = 2$

$g (2) = 1 + 1 = 2$

Local minimum value = 2 at $x = 2$
No local maximum.

(vii) $g (x) = \frac{1}{x^{2} + 2}$

$g^{'} (x) = - \frac{2 x}{(x^{2} + 2)^{2}} = 0 \Rightarrow x = 0$

$g^{''} (x) = \frac{6 x^{2} - 4}{(x^{2} + 2)^{3}} \Rightarrow g^{''} (0) = - \frac{4}{8} < 0$

So local maximum at $x = 0$ $g (0) = \frac{1}{2}$

Local maximum value = $\frac{1}{2}$ at $x = 0$
No minimum.

(viii) $f (x) = x \sqrt{1 - x}, 0 < x < 1$

$f^{'} (x) = \sqrt{1 - x} + x (- \frac{1}{2 \sqrt{1 - x}}) = \frac{2 (1 - x) - x}{2 \sqrt{1 - x}}$

Set numerator zero:

$2 - 2 x - x = 0 \Rightarrow 2 - 3 x = 0 \Rightarrow x = \frac{2}{3}$ $f (\frac{2}{3}) = \frac{2}{3} \sqrt{1 - \frac{2}{3}} = \frac{2}{3} \sqrt{\frac{1}{3}} = \frac{2}{3 \sqrt{3}}$

$f^{''} (\frac{2}{3}) < 0 \Rightarrow local maximum$

Local maximum value = $\frac{2}{3 \sqrt{3}}$ at $x = \frac{2}{3}$
No minimum in interval.

Question 4.

Prove the following functions do not have maxima or minima:

(i) $f (x) = e^{x}$
(ii) $g (x) = \log x$
(iii) $h (x) = x^{3} + x^{2} + x + 1$

Solutions

(i) $f (x) = e^{x}$

$f (x) = e^{x}$

Differentiate:

$f^{'} (x) = e^{x} > 0 for all real x$

Since derivative is always positive, function is strictly increasing on $(- \infty, \infty)$ .

Therefore, it never turns back to form a peak (maximum) or valley (minimum).

Hence, $e^{x}$ has no maximum and no minimum.

(ii) $g (x) = \log x$ , $x > 0$

$g^{'} (x) = \frac{1}{x} > 0 for all x > 0$

Derivative is always positive in its domain, so $\log x$ is strictly increasing.

Therefore, $\log x$ has no maxima or minima.

(iii) $h (x) = x^{3} + x^{2} + x + 1$

$h (x) = x^{3} + x^{2} + x + 1$

Differentiate: $h^{'} (x) = 3 x^{2} + 2 x + 1$

Check discriminant of the quadratic:

$Δ = (2)^{2} - 4 (3) (1) = 4 - 12 = - 8 < 0$

Since discriminant < 0 → quadratic has no real roots, and the coefficient of $x^{2}$ is positive,

$3 x^{2} + 2 x + 1 > 0 for all x$

Thus derivative is always positive, so the function is strictly increasing.

Therefore, $h (x)$ has no maxima or minima.

Question 5.

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) $f (x) = x^{3}, x \in [- 2, 2]$
(ii) $f (x) = \sin x + \cos x, x \in [0, π]$
(iii) $f (x) = 4 x - \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$
(iv) $f (x) = (x - 1)^{2} + 3, x \in [- 3, 1]$

Solutions

(i) $f (x) = x^{3}$ , $x \in [- 2, 2]$

Derivative:

$f^{'} (x) = 3 x^{2} = 0 \Rightarrow x = 0$

Check values at critical point and endpoints:

$f (- 2) = (- 2)^{3} = - 8$

$f (0) = 0$

$f (2) = 8$

Answer:
- Absolute maximum = 8 at $x = 2$
- Absolute minimum = –8 at $x = - 2$
(ii) $f (x) = \sin x + \cos x, x \in [0, π]$

Derivative:

$f^{'} (x) = \cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow x = \frac{π}{4}$

Evaluate:

$f (\frac{π}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

Check endpoints:

$f (0) = \sin 0 + \cos 0 = 1$

$f (π) = \sin π + \cos π = - 1$

Answer:
- Absolute maximum = $\sqrt{2}$ at $x = \frac{π}{4}$
- Absolute minimum = –1 at $x = π$
(iii) $f (x) = 4 x - \frac{1}{2} x^{2}, x \in [- 2, \frac{9}{2}]$

Derivative:

$f^{'} (x) = 4 - x = 0 \Rightarrow x = 4$

Evaluate at critical point and endpoints:

$f (- 2) = 4 (- 2) - \frac{1}{2} (4) = - 8 - 2 = - 10$

$f (4) = 4 (4) - \frac{1}{2} (16) = 16 - 8 = 8$

$f (\frac{9}{2}) = 4 \cdot \frac{9}{2} - \frac{1}{2} (\frac{81}{4}) = 18 - \frac{81}{8} = 18 - 10.125 = 7.875$

Answer:
- Absolute maximum = 8 at $x = 4$
- Absolute minimum = –10 at $x = - 2$
(iv) $f (x) = (x - 1)^{2} + 3, x \in [- 3, 1]$

Derivative:

$f^{'} (x) = 2 (x - 1) = 0 \Rightarrow x = 1$

Check endpoints and critical point:

$f (- 3) = (- 3 - 1)^{2} + 3 = 16 + 3 = 19$

$f (1) = (1 - 1)^{2} + 3 = 3$

Answer:
- Absolute minimum = 3 at $x = 1$
- Absolute maximum = 19 at $x = - 3$
Question 6.

Find the maximum profit that a company can make, if the profit function is given by:

$p (x) = 41 - 72 x - 18 x^{2}$

Solution

Given:

$p (x) = 41 - 72 x - 18 x^{2}$ This is a quadratic function of the form:

$p (x) = a x^{2} + b x + c$

where $a = - 18 < 0$ , so the parabola opens downwards → maximum exists.

To find the value of $x$ at which maximum profit occurs:

$x = - \frac{b}{2 a}$

Here $a = - 18$ , $b = - 72$

$x = - \frac{- 72}{2 (- 18)} = \frac{72}{- 36} = - 2$

Now substitute $x = - 2$ into profit function:

$p (- 2) = 41 - 72 (- 2) - 18 (- 2)^{2}$

$= 41 + 144 - 18 (4)$

$= 41 + 144 - 72 = 113$

Final Answer

Maximum profit = ₹ 113
Occurs when $x = - 2$

Question 7.

Find both the maximum value and the minimum value of

$f (x) = 3 x^{4} – 8 x^{3} + 12 x^{2} – 48 x + 25$

on the interval $[0, 3]$ .

Solution

Step 1: Differentiate

$f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$ $f^{'} (x) = 12 x^{3} - 24 x^{2} + 24 x - 48$

Factor: $f^{'} (x) = 12 (x^{3} - 2 x^{2} + 2 x - 4)$

Group:

$x^{3} - 2 x^{2} + 2 x - 4 = x^{2} (x - 2) + 2 (x - 2) = (x^{2} + 2) (x - 2)$

So:

$f^{'} (x) = 12 (x^{2} + 2) (x - 2)$

Step 2: Critical points

$f^{'} (x) = 0 \Rightarrow (x^{2} + 2) (x - 2) = 0$

Since $x^{2} + 2 > 0$ always, only solution is:

$x = 2$

Step 3: Evaluate function at
- Endpoints $x = 0, 3$
- Critical point $x = 2$
At $x = 0$

$f (0) = 25$

At $x = 2$

$f (2) = 3 (16) - 8 (8) + 12 (4) - 48 (2) + 25$

$= 48 - 64 + 48 - 96 + 25 = - 39$

At $x = 3$

$f (3) = 3 (81) - 8 (27) + 12 (9) - 48 (3) + 25$

$= 243 - 216 + 108 - 144 + 25 = 16$

Question 8.

At what points in the interval $[0, 2 π]$ , does the function $\sin 2 x$ attain its maximum value?

Solution

We know that the maximum value of the sine function is 1.

So we need the values of $x$ for which:

$\sin 2 x = 1$

This happens when:

$2 x = \frac{π}{2} + 2 π n, where n is any integer$

Divide both sides by 2:

$x = \frac{π}{4} + π n$

Now find values in the interval $[0, 2 π]$ :

For $n = 0$ :

$x = \frac{π}{4}$

For $n = 1$ :

$x = \frac{π}{4} + π = \frac{π}{4} + \frac{4 π}{4} = \frac{5 π}{4}$

For $n = 2$ :

$x = \frac{π}{4} + 2 π > 2 π \Rightarrow not in interval$

Final Answer

$The function \sin 2 x attains its maximum value at x = \frac{π}{4}, \frac{5 π}{4}$

Question 9.

What is the maximum value of the function $\sin x + \cos x$ ?

Solution

We want to find the maximum value of:

$f (x) = \sin x + \cos x$

Use the identity:

$\sin x + \cos x = \sqrt{2} (\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x)$

Since

$\frac{1}{\sqrt{2}} = \sin \frac{π}{4} = \cos \frac{π}{4}$

$\sin x + \cos x = \sqrt{2} (\sin x \sin \frac{π}{4} + \cos x \cos \frac{π}{4})$

$= \sqrt{2} \cos (x - \frac{π}{4})$

We know: $- 1 \leq \cos (x - \frac{π}{4}) \leq 1$

Multiply by $\sqrt{2}$ :

$- \sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$

Thus, the maximum value is:

$\sqrt{2}$

Question 10.

Find the maximum value of

$f (x) = 2 x^{3} – 24 x + 107$

in the interval $[1, 3]$ .
Find the maximum value of the same function in $[– 3, – 1]$ .

Solution

Given:

$f (x) = 2 x^{3} – 24 x + 107$

Step 1: Differentiate

$f^{'} (x) = 6 x^{2} - 24 = 6 (x^{2} - 4) = 6 (x - 2) (x + 2)$

Step 2: Find critical points

$f^{'} (x) = 0 \Rightarrow (x - 2) (x + 2) = 0$

So:

$x = 2, x = - 2$

Now we will analyze each interval separately.

Part (a): Interval $[1, 3]$

Critical point inside interval = $x = 2$

Evaluate at endpoints and critical point

$f (1) = 2 (1)^{3} - 24 (1) + 107 = 2 - 24 + 107 = 85$

$f (2) = 2 (8) - 24 (2) + 107 = 16 - 48 + 107 = 75$

$f (3) = 2 (27) - 24 (3) + 107 = 54 - 72 + 107 = 89$

Maximum value in $[1, 3]$

x f(x)

1 85

2 75

3 89

$Maximum value is 89 at x = 3$

Part (b): Interval $[– 3, – 1]$

Critical point inside interval = $x = - 2$

Evaluate at endpoints and critical point

$f (- 3) = 2 (- 27) - 24 (- 3) + 107 = - 54 + 72 + 107 = 125$

$f (- 2) = 2 (- 8) - 24 (- 2) + 107 = - 16 + 48 + 107 = 139$

$f (- 1) = 2 (- 1) - 24 (- 1) + 107 = - 2 + 24 + 107 = 129$

Maximum value in $[- 3, - 1]$

x f(x)

-3 125

-2 139

-1 129

$Maximum value is 139 at x = - 2$
November 29, 2025
Exercise-6.2, Class 12th, Maths, Chapter 6, NCERT
NOTE – Definition -1 : Let $I$ be an interval contained in the domain of a real-valued function $f$ . Then $f$ is said to be

(i) increasing on $I$

$x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2}) for all x_{1}, x_{2} \in I$

(ii) decreasing on $I$

$x_{1} < x_{2} \Rightarrow f (x_{1}) > f (x_{2}) for all x_{1}, x_{2} \in I$

(iii) constant on $I$

$f (x) = c for all x \in I$

where $c$ is a constant.

Question 1

Show that the function given by
$f (x) = 3 x + 17$ is increasing on $R$

Solution

Let $x_{1}$ and $x_{2}$ be any two real numbers such that

$x_{1} < x_{2}$

Then,

$3 x_{1} < 3 x_{2} (multiplying both sides by 3)$

$3 x_{1} + 17 < 3 x_{2} + 17$

$\Rightarrow f (x_{1}) < f (x_{2})$

Thus, by Definition 1 (Increasing Function), the function $f (x) = 3 x + 17$ is strictly increasing on $R$ .

Question 2

Show that the function given by

$f (x) = e^{2 x}$

is increasing on $R$ .

Solution

We have

$f (x) = e^{2 x}$

Differentiate with respect to $x$ :

$f^{'} (x) = \frac{d}{d x} (e^{2 x}) = 2 e^{2 x}$ Now, we know that:

$e^{2 x} > 0 for all x \in R$

Therefore,

$f^{'} (x) = 2 e^{2 x} > 0 for all x \in R$

Since $f^{'} (x) > 0$ for every real number $x$ , by, the function $f$ is increasing on $R$ .

Question 3

Show that the function given by $f (x) = \sin x$ is
(a) increasing in $(0, \frac{π}{2})$
(b) decreasing in $(\frac{π}{2}, π)$

Solution

We have:

$f (x) = \sin x$

Differentiate with respect to $x$ :

$f^{'} (x) = \cos x$

(a) Increasing in $(0, \frac{π}{2})$

In the interval $(0, \frac{π}{2})$ ,

$\cos x > 0$

Therefore,

$f^{'} (x) = \cos x > 0 for every x \in (0, \frac{π}{2})$

Since $f^{'} (x) > 0$ in this interval, by Theorem 1 (Increasing and Decreasing Test),

$f (x) = \sin x is increasing in (0, \frac{π}{2})$

(b) Decreasing in $(\frac{π}{2}, π)$

In the interval $(\frac{π}{2}, π)$ ,

$\cos x < 0$

Therefore,

$f^{'} (x) = \cos x < 0 for every x \in (\frac{π}{2}, π)$

Since $f^{'} (x) < 0$ in this interval, by Theorem 1,

$f (x) = \sin x is decreasing in (\frac{π}{2}, π)$

Question 4

Find the intervals in which the function

$f (x) = 2 x^{2} - 3 x$ is
(a) increasing
(b) decreasing

Solution

We have:

$f (x) = 2 x^{2} - 3 x$

Differentiate with respect to $x$ :

$f^{'} (x) = 4 x - 3$

Now set $f^{'} (x) = 0$ to find the critical point:

$4 x - 3 = 0$

$x = \frac{3}{4}$

The point $x = \frac{3}{4}$ divides the real line into two intervals:

$(- \infty, \frac{3}{4}) and (\frac{3}{4}, \infty)$

Check the sign of $f^{'} (x)$ in these intervals

For $x < \frac{3}{4}$

Take any value, say $x = 0$ :

$f^{'} (0) = 4 (0) - 3 = - 3 < 0$

So $f^{'} (x) < 0$ in $(- \infty, \frac{3}{4})$

$\Rightarrow f (x) is decreasing in (- \infty, \frac{3}{4})$

For $x > \frac{3}{4}$

Take $x = 1$ : $f^{'} (1) = 4 (1) - 3 = 1 > 0$

So $f^{'} (x) > 0$ in $(\frac{3}{4}, \infty)$

$\Rightarrow f (x) is increasing in (\frac{3}{4}, \infty)$

Final Answer

$(a) Increasing in (\frac{3}{4}, \infty)$ $(b) Decreasing in (- \infty, \frac{3}{4})$

Question 5

Find the intervals in which the function

$f (x) = 2 x^{3} - 3 x^{2} - 36 x + 7$ is
(a) increasing
(b) decreasing

Solution

We have:

$f (x) = 2 x^{3} - 3 x^{2} - 36 x + 7$

Differentiate with respect to $x$ :

$f^{'} (x) = 6 x^{2} - 6 x - 36$

Factorizing:

$f^{'} (x) = 6 (x^{2} - x - 6)$

$f^{'} (x) = 6 (x - 3) (x + 2)$

Now set $f^{'} (x) = 0$ :

$(x - 3) (x + 2) = 0 \Rightarrow x = - 2, 3$

So the real line is divided into intervals:

$(- \infty, - 2), (- 2, 3), (3, \infty)$

Sign of $f^{'} (x)$ in each interval

Interval Sign of (x-3) Sign of (x+2) Sign of $f^{'} (x)$ Nature of $f (x)$

$(- \infty, - 2)$ – – $(+) (+) = +$ Increasing

$(- 2, 3)$ – + $(-) (+) = -$ Decreasing

$(3, \infty)$ + + $(+) (+) = +$ Increasing

Final Answer

$(a) Increasing in (- \infty, - 2) and (3, \infty)$ $(b) Decreasing in (- 2, 3)$

Question 6

Find the intervals in which the following functions are strictly increasing or strictly decreasing:

(a) $f (x) = x^{2} + 2 x - 5$

$f^{'} (x) = 2 x + 2$

Set $f^{'} (x) = 0$ :

$2 x + 2 = 0 \Rightarrow x = - 1$

Intervals: $(- \infty, - 1)$ and $(- 1, \infty)$

Pick a test point:
- For $x < - 1$ , say $x = - 2$ :
  $f^{'} (- 2) = 2 (- 2) + 2 = - 2 < 0 \Rightarrow Decreasing$
- For $x > - 1$ , say $x = 0$ :
  $f^{'} (0) = 2 (0) + 2 = 2 > 0 \Rightarrow Increasing$
Answer

$Decreasing on (- \infty, - 1)$ $Increasing on (- 1, \infty)$

(b) $f (x) = 10 - 6 x - 2 x^{2}$

$f^{'} (x) = - 6 - 4 x = - (4 x + 6)$

Set $f^{'} (x) = 0$ :

$4 x + 6 = 0 \Rightarrow x = - \frac{3}{2}$
- For $x < - \frac{3}{2}$ , say $x = - 2$ :
  $f^{'} (- 2) = - 6 - 4 (- 2) = 2 > 0 \Rightarrow Increasing$
- For $x > - \frac{3}{2}$ , say $x = 0$ :
  $f^{'} (0) = - 6 < 0 \Rightarrow Decreasing$
Answer

$Increasing on (- \infty, - \frac{3}{2})$ $Decreasing on (- \frac{3}{2}, \infty)$

(c) $f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1$

$f^{'} (x) = - 6 x^{2} - 18 x - 12 = - 6 (x^{2} + 3 x + 2)$

$= - 6 (x + 1) (x + 2)$

Set $f^{'} (x) = 0 \Rightarrow x = - 1, - 2$

Intervals:

$(- \infty, - 2), (- 2, - 1), (- 1, \infty)$

Sign test:

Interval (x+1) (x+2) $f^{'} (x)$ sign Nature

$(- \infty, - 2)$ – – –6(+)= – Decreasing

$(- 2, - 1)$ – + –6(−)= + Increasing

$(- 1, \infty)$ + + –6(+)= – Decreasing

Answer

$Increasing on (- 2, - 1)$

$Decreasing on (- \infty, - 2) and (- 1, \infty)$

(d) $f (x) = 6 - 9 x - x^{2}$

$f^{'} (x) = - 9 - 2 x$

Set $f^{'} (x) = 0$ :

$- 9 - 2 x = 0 \Rightarrow x = - \frac{9}{2}$

Test sign:
- $x < - \frac{9}{2}$ : $f^{'} (x) > 0$ → Increasing
- $x > - \frac{9}{2}$ : $f^{'} (x) < 0$ → Decreasing
Answer

$Increasing on (- \infty, - \frac{9}{2})$

$Decreasing on (- \frac{9}{2}, \infty)$

(e) $f (x) = (x + 1)^{3} (x - 3)^{3}$

$f (x) = [(x + 1) (x - 3)]^{3}$

Let $g (x) = (x + 1) (x - 3) = x^{2} - 2 x - 3$

$f^{'} (x) = 3 [g (x)]^{2} \cdot g^{'} (x)$

$= 3 (x^{2} - 2 x - 3)^{2} (2 x - 2)$

$= 6 (x^{2} - 2 x - 3)^{2} (x - 1)$

Critical point at $x = 1$

Since $(x^{2} - 2 x - 3)^{2} \geq 0$ always and is zero only at $x = - 1, 3$ , sign depends on $(x - 1)$ :
- $x < 1$ : $f^{'} (x) < 0$ → Decreasing
- $x > 1$ : $f^{'} (x) > 0$ → Increasing
Answer

$Decreasing on (- \infty, 1)$

$Increasing on (1, \infty)$

Question 7

Show that

$y = \log (1 + x) - \frac{2 x}{2 + x}, x > - 1$

is an increasing function throughout its domain.

Solution

Given: $y = \log (1 + x) - \frac{2 x}{2 + x}$

Differentiate with respect to $x$ .

Step 1: Differentiate each term

$\frac{d}{d x} (\log (1 + x)) = \frac{1}{1 + x}$

Now differentiate the second term using quotient rule:

$\frac{d}{d x} (\frac{2 x}{2 + x}) = \frac{(2 + x) \cdot 2 - 2 x \cdot 1}{(2 + x)^{2}}$

$= \frac{4 + 2 x - 2 x}{(2 + x)^{2}} = \frac{4}{(2 + x)^{2}}$

Step 2: Write derivative of $y$

$y^{'} = \frac{1}{1 + x} - \frac{4}{(2 + x)^{2}}$

Step 3: Simplify

Take LCM $(1 + x) (2 + x)^{2}$ :

$y^{'} = \frac{(2 + x)^{2} - 4 (1 + x)}{(1 + x) (2 + x)^{2}}$

Expand: $(2 + x)^{2} = x^{2} + 4 x + 4$

So: $y^{'} = \frac{x^{2} + 4 x + 4 - 4 x - 4}{(1 + x) (2 + x)^{2}}$

Simplifying numerator:

$x^{2} + 4 x + 4 - 4 x - 4 = x^{2}$

Thus: $y^{'} = \frac{x^{2}}{(1 + x) (2 + x)^{2}}$

Step 4: Check sign of $y^{'}$
- $x^{2} \geq 0$ for all real $x$
- For domain $x > - 1$ , both denominators $(1 + x)$ and $(2 + x)^{2}$ are positive
Therefore: $y^{'} = \frac{x^{2}}{(1 + x) (2 + x)^{2}} \geq 0 for all x > - 1$

And equality occurs only at $x = 0$ , otherwise positive.

Conclusion

$y^{'} \geq 0 for all x > - 1$

$Hence, y = \log (1 + x) - \frac{2 x}{2 + x} is an increasing function throughout its domain$

Question 8

Find the values of $x$ for which

$y = [x (x - 2)]^{2}$

is an increasing function.

Solution

Given: $y = [x (x - 2)]^{2} = (x^{2} - 2 x)^{2}$

Differentiate with respect to $x$ :

$y^{'} = 2 (x^{2} - 2 x) \cdot (2 x - 2)$

Factorising:

$y^{'} = 4 (x^{2} - 2 x) (x - 1)$

Now factor further:

$y^{'} = 4 x (x - 2) (x - 1)$

We want to find where $y^{'} > 0$ .
So solve:

$4 x (x - 2) (x - 1) > 0$

The critical points where the expression changes sign are:

$x = 0, 1, 2$

These points divide the real line into intervals:

$(- \infty, 0), (0, 1), (1, 2), (2, \infty)$

Sign Table

Interval x x–1 x–2 Sign of $y^{'}$ Nature

$(- \infty, 0)$ – – – – Decreasing

$(0, 1)$ + – – + Increasing

$(1, 2)$ + + – – Decreasing

$(2, \infty)$ + + + + Increasing

Final Answer

$The function is increasing for x \in (0, 1) and x \in (2, \infty)$

or $x \in (0, 1) \cup (2, \infty)$

Question 9

Prove that

$y = \frac{4 \sin θ}{2 + \cos θ} - θ$

is an increasing function of $θ$ in

$[0, \frac{π}{2}]$

Solution

Given:

$y = \frac{4 \sin θ}{2 + \cos θ} - θ$

Differentiate with respect to $θ$ :

Step 1 – Differentiate the first term using quotient rule

Let:

$u = 4 \sin θ, v = 2 + \cos θ$ $u^{'} = 4 \cos θ, v^{'} = - \sin θ$

$\frac{d}{d θ} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}$

Substitute:

$\frac{d}{d θ} (\frac{4 \sin θ}{2 + \cos θ}) = \frac{(4 \cos θ) (2 + \cos θ) - (4 \sin θ) (- \sin θ)}{(2 + \cos θ)^{2}}$

Simplify the numerator:

$= \frac{8 \cos θ + 4 \cos^{2} θ + 4 \sin^{2} θ}{(2 + \cos θ)^{2}}$

$= \frac{8 \cos θ + 4 (\cos^{2} θ + \sin^{2} θ)}{(2 + \cos θ)^{2}}$

Using the identity $\sin^{2} θ + \cos^{2} θ = 1$ :

$= \frac{8 \cos θ + 4}{(2 + \cos θ)^{2}}$

Step 2 – Differentiate second term

$\frac{d}{d θ} (- θ) = - 1$

Therefore

$y^{'} = \frac{8 \cos θ + 4}{(2 + \cos θ)^{2}} - 1$

Take LCM:

$y^{'} = \frac{8 \cos θ + 4 - (2 + \cos θ)^{2}}{(2 + \cos θ)^{2}}$

Expand the square:

$(2 + \cos θ)^{2} = 4 + 4 \cos θ + \cos^{2} θ$

Substitute:

$y^{'} = \frac{8 \cos θ + 4 - 4 - 4 \cos θ - \cos^{2} θ}{(2 + \cos θ)^{2}}$

Simplify numerator:

$y^{'} = \frac{4 \cos θ - \cos^{2} θ}{(2 + \cos θ)^{2}}$

Factor:

$y^{'} = \frac{\cos θ (4 - \cos θ)}{(2 + \cos θ)^{2}}$

Check sign of $y^{'}$ on $[0, \frac{π}{2}]$

In this interval:
- $\cos θ \geq 0$
- $4 - \cos θ > 0$
- $(2 + \cos θ)^{2} > 0$
Therefore,

$y^{'} = \frac{\cos θ (4 - \cos θ)}{(2 + \cos θ)^{2}} \geq 0$

Thus,

$y^{'} \geq 0 for all θ \in [0, \frac{π}{2}]$

So $y$ is an increasing function on this interval.

Question 10

Prove that the logarithmic function is increasing on $(0, \infty)$ .

Solution

Let

$f (x) = \log x, x > 0$

Differentiate with respect to $x$ :

$f^{'} (x) = \frac{1}{x}$

Now examine the sign of $f^{'} (x)$ over its domain $(0, \infty)$ :
- For all $x > 0$ , $\frac{1}{x} > 0$
Therefore: $f^{'} (x) > 0 for all x \in (0, \infty)$

According to Theorem 1 on Increasing and Decreasing Functions, if

$f^{'} (x) > 0$

for every $x$ in an interval, then $f (x)$ is increasing in that interval.

So:

$The function \log x is increasing on (0, \infty) .$

Question 11

Prove that the function $f (x) = x^{2} - x + 1$ is neither strictly increasing nor strictly decreasing on $(- 1, 1)$ .

Solution

Given:

$f (x) = x^{2} - x + 1$

Differentiate with respect to $x$ :

$f^{'} (x) = 2 x - 1$

Now set $f^{'} (x) = 0$ to locate the critical point:

$2 x - 1 = 0$ $x = \frac{1}{2}$

This point lies inside the interval $(- 1, 1)$ .
So the interval $(- 1, 1)$ is divided into two parts:

$(- 1, \frac{1}{2}) and (\frac{1}{2}, 1)$

Test the sign of $f^{'} (x)$ in these intervals

For $x \in (- 1, \frac{1}{2})$

Choose $x = 0$ :

$f^{'} (0) = 2 (0) - 1 = - 1 < 0$

$\Rightarrow f (x) is decreasing on (- 1, \frac{1}{2})$

For $x \in (\frac{1}{2}, 1)$

Choose $x = \frac{3}{4}$ :

$f^{'} (\frac{3}{4}) = 2 (\frac{3}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2} > 0$ $\Rightarrow f (x) is increasing on (\frac{1}{2}, 1)$

Question 12

Which of the following functions are decreasing on

$(0, \frac{π}{2}) ?$

(A) $\cos x$
(B) $\cos 2 x$
(C) $\cos 3 x$
(D) $\tan x$

Solution

A function is decreasing if its derivative is negative in the interval.

Option (A) $f (x) = \cos x$

$f^{'} (x) = - \sin x$

In $(0, \frac{π}{2})$ , $\sin x > 0$ , so

$f^{'} (x) = - \sin x < 0 \Rightarrow \cos x is decreasing$

Option (B) $f (x) = \cos 2 x$

$f^{'} (x) = - 2 \sin 2 x$

In $(0, \frac{π}{2})$ , $2 x \in (0, π)$ , so $\sin 2 x > 0$ , hence

$f^{'} (x) = - 2 \sin 2 x < 0 \Rightarrow \cos 2 x is decreasing$

Option (C) $f (x) = \cos 3 x$

$f^{'} (x) = - 3 \sin 3 x$

In $(0, \frac{π}{2})$ , $3 x \in (0, \frac{3 π}{2})$ , and in this range $\sin 3 x > 0$ , so

$f^{'} (x) = - 3 \sin 3 x < 0 \Rightarrow \cos 3 x is decreasing$

Option (D) $f (x) = \tan x$

$f^{'} (x) = \sec^{2} x$

In $(0, \frac{π}{2})$ , $\sec^{2} x > 0$

So

$f^{'} (x) > 0 \Rightarrow \tan x is increasing, not decreasing$

Final Answer

$(A) \cos x, (B) \cos 2 x, (C) \cos 3 x$ $\tan x is not decreasing$

Question 13

On which of the following intervals is the function

$f (x) = x^{100} + \sin x - 1$

decreasing?

(A) $(0, 1)$
(B) $(\frac{π}{2}, π)$
(C) $(0, \frac{π}{2})$
(D) None of these

Solution

Differentiate $f (x)$ :

$f^{'} (x) = 100 x^{99} + \cos x$

For $f (x)$ to be decreasing, we need:

$f^{'} (x) < 0$

Analyze sign of each term
- $100 x^{99} > 0$ for all $x > 0$ , because any positive number raised to any power remains positive.
- $\cos x$ in different intervals:
  - In $(0, \frac{π}{2})$ , $\cos x > 0$
  - In $(\frac{π}{2}, π)$ , $\cos x < 0$
So the only interval where $f^{'} (x)$ might be negative is:

$(\frac{π}{2}, π)$

Check sign in this interval:
- $100 x^{99}$ is a very large positive number
- $\cos x$ is negative, but lies between $- 1$ and $0$
Thus:

$f^{'} (x) = 100 x^{99} + \cos x > 100 x^{99} - 1 > 0$

Therefore, $f^{'} (x) > 0$ everywhere on the given intervals.

So there is no interval from the options where the function is decreasing.

Final Answer

$(D) None of these$

Question 14

For what values of $a$ the function

$f (x) = x^{2} + a x + 1$

is increasing on $[1, 2]$ ?

Solution

Differentiate $f (x)$ with respect to $x$ :

$f^{'} (x) = 2 x + a$

For $f (x)$ to be increasing on $[1, 2]$ , we require:

$f^{'} (x) \geq 0 for all x \in [1, 2]$

That is:

$2 x + a \geq 0$

Now check the minimum value of $2 x + a$ in $[1, 2]$ .

Since $2 x + a$ is a linear increasing function of $x$ , the minimum occurs at the left endpoint $x = 1$ :

$2 (1) + a \geq 0$

$2 + a \geq 0$

$a \geq - 2$

Final Answer

$f (x) = x^{2} + a x + 1 is increasing on [1, 2] when a \geq - 2$

Question 15

Let $I$ be any interval disjoint from $[- 1, 1]$ . Prove that the function

$f (x) = x + \frac{1}{x}$ is increasing on $I$ .

Solution

Given: $f (x) = x + \frac{1}{x}$ Differentiate with respect to $x$ :

$f^{'} (x) = 1 - \frac{1}{x^{2}}$ Rewrite: $f^{'} (x) = \frac{x^{2} - 1}{x^{2}}$ $f^{'} (x) = \frac{(x - 1) (x + 1)}{x^{2}}$

We need to determine where $f^{'} (x) > 0$ .

Sign Analysis
- $x^{2} > 0$ for all $x \neq 0$
- So the sign of $f^{'} (x)$ depends on the sign of $(x - 1) (x + 1)$
$(x - 1) (x + 1) > 0$

This product is positive when both factors are positive or both are negative.

Thus:

$x > 1 or x < - 1$

Intervals to Check

Given: The interval $I$ is disjoint from $[- 1, 1]$ .

So $I$ must lie entirely in one of the following:

$(- \infty, - 1) or (1, \infty)$

In these intervals:

Interval Sign of $(x - 1)$ Sign of $(x + 1)$ Sign of $f^{'} (x)$ Nature

$(- \infty, - 1)$ – – + Increasing

$(1, \infty)$ + + + Increasing

So,

$f^{'} (x) > 0 on both intervals$

Conclusion

$f^{'} (x) > 0 everywhere on any interval disjoint from [- 1, 1]$

Therefore,

$f (x) = x + \frac{1}{x} is increasing on any such interval I .$

Question 16

Prove that the function

$f (x) = \log (\sin x)$

is increasing on $(0, \frac{π}{2})$ and decreasing on $(\frac{π}{2}, π)$ .

Solution

Given: $f (x) = \log (\sin x)$

Differentiate using the chain rule:

$f^{'} (x) = \frac{1}{\sin x} \cdot \cos x = \cot x$

So: $f^{'} (x) = \cot x$

We analyze the sign of $f^{'} (x)$ in the given intervals.

1. In the interval $(0, \frac{π}{2})$

In this interval,
- $\sin x > 0$
- $\cos x > 0$
- Therefore, $\cot x = \frac{\cos x}{\sin x} > 0$
So:

$f^{'} (x) > 0 for all x \in (0, \frac{π}{2})$

Hence,

$f (x) = \log (\sin x) is increasing on (0, \frac{π}{2})$

2. In the interval $(\frac{π}{2}, π)$

In this interval,
- $\sin x > 0$
- $\cos x < 0$
- Therefore, $\cot x = \frac{\cos x}{\sin x} < 0$
So: $f^{'} (x) < 0 for all x \in (\frac{π}{2}, π)$

Hence,

$f (x) = \log (\sin x) is decreasing on (\frac{π}{2}, π)$

Final Answer

$\log (\sin x) is increasing on (0, \frac{π}{2}) and decreasing on (\frac{π}{2}, π)$

Question 17

Prove that the function

$f (x) = \log ∣ \cos x ∣$

is decreasing on $(0, \frac{π}{2})$ and increasing on $(\frac{3 π}{2}, 2 π)$ .

Solution

Given:

$f (x) = \log ∣ \cos x ∣$

Differentiate using the chain rule:

$f^{'} (x) = \frac{1}{∣ \cos x ∣} \cdot \frac{d}{d x} (∣ \cos x ∣)$

Derivative of $∣ \cos x ∣$

$\frac{d}{d x} (∣ \cos x ∣) = \frac{\cos x}{∣ \cos x ∣} \cdot (- \sin x)$

Therefore:

$f^{'} (x) = \frac{1}{∣ \cos x ∣} \cdot (\frac{\cos x}{∣ \cos x ∣}) (- \sin x)$

Simplifying:

$f^{'} (x) = \frac{- \sin x}{∣ \cos x ∣}$

So:

$f^{'} (x) = \frac{- \sin x}{∣ \cos x ∣}$

Check sign of $f^{'} (x)$ in the given intervals

1. On $(0, \frac{π}{2})$
- $\sin x > 0$
- $\cos x > 0 \Rightarrow ∣ \cos x ∣ = \cos x > 0$
So: $f^{'} (x) = \frac{- \sin x}{\cos x} = - \tan x < 0$

Therefore:

$f (x) is decreasing on (0, \frac{π}{2})$

2. On $(\frac{3 π}{2}, 2 π)$
- $\sin x < 0$
- $\cos x > 0 \Rightarrow ∣ \cos x ∣ = \cos x > 0$
So: $f^{'} (x) = \frac{- \sin x}{\cos x} = - \tan x > 0$

(negative of a negative becomes positive)

Therefore:

$f (x) is increasing on (\frac{3 π}{2}, 2 π)$

Final Answer

$\log ∣ \cos x ∣ is decreasing on (0, \frac{π}{2}) and increasing on (\frac{3 π}{2}, 2 π)$

Question 18

Prove that the function

$f (x) = x^{3} - 3 x^{2} + 3 x - 100$

is increasing in $R$ (the set of all real numbers).

Solution

Differentiate $f (x)$ with respect to $x$ :

$f^{'} (x) = 3 x^{2} - 6 x + 3$

Factorize:

$f^{'} (x) = 3 (x^{2} - 2 x + 1)$

$f^{'} (x) = 3 (x - 1)^{2}$

Analyze the sign of $f^{'} (x)$
- $(x - 1)^{2} \geq 0$ for all $x \in R$ , since the square of any real number is non-negative.
- Therefore:
$3 (x - 1)^{2} \geq 0 for all x$

Thus:

$f^{'} (x) \geq 0 for all x \in R$

Since the derivative is never negative and is zero only at a single point $x = 1$ , the function does not decrease anywhere.

Conclusion

$f^{'} (x) \geq 0 for all real numbers x$ $Hence, f (x) = x^{3} - 3 x^{2} + 3 x - 100 is increasing on R .$

Question 19

The interval in which

$y = x^{2} e^{- x}$

is increasing is:

(A) $(- \infty, \infty)$
(B) $(- 2, 0)$
(C) $(2, \infty)$
(D) $(0, 2)$

Solution

Given:

$y = x^{2} e^{- x}$

Differentiate using product rule:

$y^{'} = (x^{2})^{'} e^{- x} + x^{2} (e^{- x})^{'}$

$y^{'} = 2 x e^{- x} + x^{2} (- e^{- x})$

$y^{'} = e^{- x} (2 x - x^{2})$

Factor further:

$y^{'} = e^{- x} x (2 - x)$

Analyze the sign of $y^{'}$

$y^{'} = e^{- x} x (2 - x)$
- $e^{- x} > 0$ for all real $x$
- So the sign of $y^{'}$ depends only on $x (2 - x)$
$x (2 - x) > 0$

Solve inequality:
- Product is positive when both factors have the same sign.
Case 1:

$x > 0 and 2 - x > 0 \Rightarrow x < 2$

So:

$0 < x < 2$

Case 2 would be:

$x < 0 and 2 - x < 0 \Rightarrow x > 2$

Impossible.

Therefore, the function is increasing only in:

$(0, 2)$

Final Answer

$The function is increasing on (0, 2)$

$Correct option: (D)$
November 28, 2025
Exercise-6.1, Class 12th, Maths, Chapter 6, NCERT
Question 1

Find the rate of change of the area of a circle with respect to its radius $r$ when
(a) $r = 3$
(b) $r = 4$

Answer

Area of a circle:

$A = π r^{2}$

Differentiate w.r.t. $r$ :

$\frac{d A}{d r} = 2 π r$

(a) When $r = 3$ :

$\frac{d A}{d r} = 2 π (3) = 6 π {cm}^{2} / cm$

(b) When $r = 4$ :

$\frac{d A}{d r} = 2 π (4) = 8 π {cm}^{2} / cm$

Question 2

The volume of a cube is increasing at the rate of $8 {cm}^{3} / s$ .
How fast is the surface area increasing when the edge is $12$ cm?

Answer

Let edge = $x$

Volume:

$V = x^{3}$ $\frac{d V}{d t} = 8$

Differentiate:

$\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t}$ $8 = 3 x^{2} \frac{d x}{d t}$ $\frac{d x}{d t} = \frac{8}{3 (12)^{2}} = \frac{8}{432} = \frac{1}{54} cm/s$

Surface Area:

$S = 6 x^{2}$

Differentiate:

$\frac{d S}{d t} = 12 x \frac{d x}{d t}$

Substitute $x = 12$ and $\frac{d x}{d t} = \frac{1}{54}$ :

$\frac{d S}{d t} = 12 (12) (\frac{1}{54}) = \frac{144}{54} = \frac{8}{3} {cm}^{2} / s$

Question 3

The radius of a circle is increasing uniformly at the rate of $3$ cm/s.
Find the rate at which the area is increasing when the radius is $10$ cm.

Answer

$A = π r^{2}$ $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$

Given:

$\frac{d r}{d t} = 3, r = 10$ $\frac{d A}{d t} = 2 π (10) (3) = 60 π {cm}^{2} / s$

Question 4

An edge of a variable cube is increasing at $3$ cm/s.
How fast is the volume increasing when the edge is $10$ cm?

Answer

$V = x^{3}$

Differentiate:

$\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t}$ $\frac{d V}{d t} = 3 (10)^{2} (3) = 900 {cm}^{3} / s$

Question 5

A stone is dropped into a lake and waves move in circles at 5 cm/s.
When radius is 8 cm, how fast is the enclosed area increasing?

Answer

$A = π r^{2}$ $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$

Given:

$\frac{d r}{d t} = 5, r = 8$ $\frac{d A}{d t} = 2 π (8) (5) = 80 π {cm}^{2} / s$

Question 6

The radius of a circle is increasing at the rate of 0.7 cm/s.
What is the rate of increase of its circumference?

Answer

Let radius be $r$

Circumference of a circle:

$C = 2 π r$

Differentiate w.r.t. time $t$ :

$\frac{d C}{d t} = 2 π \frac{d r}{d t}$

Given:

$\frac{d r}{d t} = 0.7 cm/s$

Substitute:

$\frac{d C}{d t} = 2 π (0.7)$ $\frac{d C}{d t} = 1.4 π cm/s$

Question 7

The length $x$ of a rectangle is decreasing at the rate of 5 cm/min and the width $y$ is increasing at the rate of 4 cm/min.
When $x = 8$ and $y = 6$ , find the rates of change of
(a) the perimeter, and (b) the area of the rectangle.

Given

$\frac{d x}{d t} = - 5 cm/min (decreasing)$ $\frac{d y}{d t} = + 4 cm/min (increasing)$

(a) Rate of change of Perimeter

Perimeter of rectangle:

$P = 2 (x + y)$

Differentiate w.r.t. time $t$ :

$\frac{d P}{d t} = 2 (\frac{d x}{d t} + \frac{d y}{d t})$

Substitute values:

$\frac{d P}{d t} = 2 (- 5 + 4) = 2 (- 1) = - 2 cm/min$

Answer (a)

$\frac{d P}{d t} = - 2 cm/min$

So, the perimeter is decreasing at 2 cm/min.

(b) Rate of change of Area

Area: $A = x \cdot y$

Differentiate:

$\frac{d A}{d t} = x \frac{d x}{d t} + y \frac{d y}{d t}$

Substitute values: $x = 8, y = 6$

$\frac{d A}{d t} = 8 (- 5) + 6 (4)$ $\frac{d A}{d t} = - 40 + 24 = - 16 {cm}^{2} / min$

Answer (b)

$\frac{d A}{d t} = - 16 {cm}^{2} / min$

So, the area is decreasing at 16 cm²/min.

Question 8

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer

Let the radius of the balloon = $r$

Volume of a sphere:

$V = \frac{4}{3} π r^{3}$

Differentiate w.r.t. time $t$ :

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

Given:

$\frac{d V}{d t} = 900 {cm}^{3} / s, r = 15 cm$

Substitute in the formula:

$900 = 4 π (15)^{2} \frac{d r}{d t}$ $900 = 4 π \cdot 225 \frac{d r}{d t}$ $900 = 900 π \frac{d r}{d t}$ $\frac{d r}{d t} = \frac{900}{900 π} = \frac{1}{π} cm/s$

Question 9

A balloon, which always remains spherical, has a variable radius.
Find the rate at which its volume is increasing with respect to the radius when the radius is 10 cm.

Answer

Let the radius of the balloon = $r$

Volume of a sphere:

$V = \frac{4}{3} π r^{3}$

We need:

$\frac{d V}{d r}$ Differentiate with respect to $r$ :

$\frac{d V}{d r} = 4 π r^{2}$

Now substitute $r = 10 cm$ :

$\frac{d V}{d r} = 4 π (10)^{2} = 4 π \cdot 100 = 400 π {cm}^{3} / cm$

Question 10

A ladder 5 m long is leaning against a wall.
The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s.
How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer

Let:
- $x$ = distance of the bottom of the ladder from the wall (ground level)
- $y$ = height of the ladder on the wall
- Length of ladder = $5$ m (constant)
From the geometry of the right triangle:

$x^{2} + y^{2} = 5^{2}$

$x^{2} + y^{2} = 25 (1)$

Differentiate w.r.t. time $t$ :

$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$

Divide by 2:

$x \frac{d x}{d t} + y \frac{d y}{d t} = 0$

Given:

$\frac{d x}{d t} = 2 cm/s = 0.02 m/s$

When $x = 4$ , from (1):

$4^{2} + y^{2} = 25$

$16 + y^{2} = 25$ $y^{2} = 9 \Rightarrow y = 3 m$

Now substitute into differentiation equation:

$4 (0.02) + 3 \frac{d y}{d t} = 0$

$0.08 + 3 \frac{d y}{d t} = 0$

$3 \frac{d y}{d t} = - 0.08$ $\frac{d y}{d t} = - \frac{0.08}{3} = - 0.0267 m/s$

Convert to cm/s:

$- 0.0267 \times 100 = - 2.67 cm/s$

Question 11

A particle moves along the curve

$6 y = x^{3} + 2$

Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer

Given:

$6 y = x^{3} + 2$

Differentiate both sides with respect to $t$ (time):

$6 \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t}$

Divide both sides by 3:

$2 \frac{d y}{d t} = x^{2} \frac{d x}{d t}$

Now divide both sides by $\frac{d x}{d t}$ (assuming $\frac{d x}{d t} \neq 0$ ):

$2 \frac{d y}{d t} / \frac{d x}{d t} = x^{2}$

$2 \frac{d y}{d x} = x^{2}$

Given condition:

$\frac{d y}{d x} = 8$

Substitute:

$2 (8) = x^{2}$ $16 = x^{2}$ $x = \pm 4$

Find corresponding $y$ -values

Use original equation:

$6 y = x^{3} + 2$

For $x = 4$ :

$6 y = (4)^{3} + 2 = 64 + 2 = 66$

$y = \frac{66}{6} = 11$

For $x = - 4$ :

$6 y = (- 4)^{3} + 2 = - 64 + 2 = - 62$

$y = \frac{- 62}{6} = - \frac{31}{3}$

Final Answer

$The required points on the curve are (4, 11) and (- 4, - \frac{31}{3}) .$

Question 12

The radius of an air bubble is increasing at the rate of

$\frac{d r}{d t} = \frac{1}{2} cm/s$

At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer

Let $r$ = radius of the spherical bubble.

Volume of a sphere:

$V = \frac{4}{3} π r^{3}$

Differentiate with respect to time $t$ :

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

Given:

$\frac{d r}{d t} = \frac{1}{2}, r = 1 cm$

Substitute values:

$\frac{d V}{d t} = 4 π (1)^{2} (\frac{1}{2})$

$\frac{d V}{d t} = 4 π \cdot \frac{1}{2} = 2 π {cm}^{3} / s$

Question 13

A balloon, which always remains spherical, has a variable diameter

$D = \frac{3}{2} (2 x + 1)$

Find the rate of change of its volume with respect to $x$ .

Answer

Diameter:

$D = \frac{3}{2} (2 x + 1)$

Radius is:

$r = \frac{D}{2} = \frac{1}{2} \cdot \frac{3}{2} (2 x + 1) = \frac{3}{4} (2 x + 1)$

Volume of a sphere:

$V = \frac{4}{3} π r^{3}$

Substitute $r = \frac{3}{4} (2 x + 1)$ :

$V = \frac{4}{3} π {[\frac{3}{4} (2 x + 1)]}^{3}$

$V = \frac{4}{3} π \cdot \frac{27}{64} (2 x + 1)^{3}$

$V = \frac{108 π}{192} (2 x + 1)^{3}$

$V = \frac{9 π}{16} (2 x + 1)^{3}$

Differentiate with respect to $x$

$\frac{d V}{d x} = \frac{9 π}{16} \cdot 3 (2 x + 1)^{2} \cdot \frac{d}{d x} (2 x + 1)$

$\frac{d}{d x} (2 x + 1) = 2$

So,

$\frac{d V}{d x} = \frac{9 π}{16} \cdot 3 \cdot 2 (2 x + 1)^{2}$

$\frac{d V}{d x} = \frac{54 π}{16} (2 x + 1)^{2}$

$\frac{d V}{d x} = \frac{27 π}{8} (2 x + 1)^{2}$

Question 14

Sand is pouring from a pipe at the rate of 12 cm³/s.
The sand forms a cone such that the height is always one-sixth of the radius.
How fast is the height of the cone increasing when the height is 4 cm?

Answer

Let:
- $V$ = volume of the cone
- $r$ = radius of base
- $h$ = height of the cone
Given:

$\frac{d V}{d t} = 12 {cm}^{3} / s$

Also,

$h = \frac{1}{6} r \Rightarrow r = 6 h$

Volume of cone

$V = \frac{1}{3} π r^{2} h$

Substitute $r = 6 h$ :

$V = \frac{1}{3} π (6 h)^{2} h$

$V = \frac{1}{3} π (36 h^{2}) h$

$V = 12 π h^{3}$

Differentiate w.r.t. time $t$

$\frac{d V}{d t} = 36 π h^{2} \frac{d h}{d t}$

Given $\frac{d V}{d t} = 12$ and $h = 4$ :

$12 = 36 π (4)^{2} \frac{d h}{d t}$

$12 = 36 π \cdot 16 \cdot \frac{d h}{d t}$

$12 = 576 π \frac{d h}{d t}$

$\frac{d h}{d t} = \frac{12}{576 π}$

$\frac{d h}{d t} = \frac{1}{48 π} cm/s$

Question 15

The total cost $C (x)$ in Rupees associated with the production of $x$ units of an item is given by:

$C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000$

Find the marginal cost when 17 units are produced.
(Marginal cost means the instantaneous rate of change of total cost, i.e., $\frac{d C}{d x}$ )

Answer

Given:

$C (x) = 0.007 x^{3} - 0.003 x^{2} + 15 x + 4000$

Differentiate with respect to $x$ :

$\frac{d C}{d x} = 0.021 x^{2} - 0.006 x + 15$

We need the marginal cost at $x = 17$ :

$M C = 0.021 (17)^{2} - 0.006 (17) + 15$

Calculate step-by-step:

$17^{2} = 289$ $0.021 \times 289 = 6.069$ $0.006 \times 17 = 0.102$

Now substitute:

$M C = 6.069 - 0.102 + 15$

$M C = 20.967$

Final Answer

$The marginal cost when 17 units are produced is \approx ₹ 20.97$

Question 16

The total revenue in Rupees received from the sale of $x$ units of a product is:

$R (x) = 13 x^{2} + 26 x + 15$

We need to find the marginal revenue when $x = 7$ .
(Marginal revenue = derivative of revenue w.r.t. $x$ )

Solution

Differentiate:

$\frac{d R}{d x} = 26 x + 26$

Substitute $x = 7$ :

$M R = 26 (7) + 26$ $M R = 182 + 26 = 208$

Final Answer

So, the marginal revenue when 7 units are sold is ₹ 208.

Choose the correct answer for questions 17 and 18.

Question 17

The rate of change of the area of a circle with respect to its radius $r$ at $r = 6$ is:

$A = π r^{2}$

Differentiate w.r.t. $r$ :

$\frac{d A}{d r} = 2 π r$

Substitute $r = 6$ :

$\frac{d A}{d r} = 2 π (6) = 12 π$

Correct Option

$(B) 12 π$

Question 18

The total revenue received from the sale of $x$ units of a product is:

$R (x) = 3 x^{2} + 36 x + 5$

We must find the marginal revenue when $x = 15$ .
(Marginal revenue = $\frac{d R}{d x}$ )

Solution

Differentiate:

$\frac{d R}{d x} = 6 x + 36$

Now substitute $x = 15$ :

$M R = 6 (15) + 36$ $M R = 90 + 36 = 126$

Final Answer

$126$

Correct Option

$(D) 126$
November 28, 2025
Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT (11-22)
Question 11.

$y = x^{(x^{2} - 3)} + (x - 3)^{x^{2}}, x > 3$

Formula Used

To differentiate $a^{u (x)}$ , where base is $a (x)$ or exponent is a function:

When both base and exponent are functions of x:

$\frac{d}{d x} (f (x)^{g (x)}) = f (x)^{g (x)} [g^{'} (x) \ln f (x) + g (x) \frac{f^{'} (x)}{f (x)}]$

We use logarithmic differentiation.

Solution

Term 1: $x^{x^{2} - 3}$

Let $u = x$ , $v = x^{2} - 3$

$\frac{d}{d x} (x^{x^{2} - 3}) = x^{x^{2} - 3} [(2 x) \ln x + \frac{x^{2} - 3}{x}]$ Term 2: $(x - 3)^{x^{2}}$

Let $u = x - 3$ , $v = x^{2}$

$\frac{d}{d x} ((x - 3)^{x^{2}}) = (x - 3)^{x^{2}} [(2 x) \ln (x - 3) + \frac{x^{2}}{x - 3}]$

Final Answer

$\frac{d y}{d x} = x^{x^{2} - 3} (2 x \ln x + \frac{x^{2} - 3}{x}) + (x - 3)^{x^{2}} (2 x \ln (x - 3) + \frac{x^{2}}{x - 3})$

Question 12

Find $\frac{d y}{d x}$ , if $y = 12 (1 - \cos t), x = 10 (t - \sin t), - \frac{π}{2} < t < \frac{π}{2}$

Solution:

For parametric equations: $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Step 1: Differentiate y w.r.t. t

$y = 12 (1 - \cos t)$

$\frac{d y}{d t} = 12 (0 + \sin t) = 12 \sin t$

Step 2: Differentiate x w.r.t. t

$x = 10 (t - \sin t)$

$\frac{d x}{d t} = 10 (1 - \cos t) = 10 (1 - \cos t)$

Step 3: Substitute in formula

$\frac{d y}{d x} = \frac{12 \sin t}{10 (1 - \cos t)}$

We have: $\frac{d y}{d x} = \frac{6 \sin t}{5 (1 - \cos t)}$

Use trigonometric identities

$1 - \cos t = 2 \sin^{2} \frac{t}{2}$

Substitute these in: $\frac{d y}{d x} = \frac{6 (2 \sin \frac{t}{2} \cos \frac{t}{2})}{5 \cdot 2 \sin^{2} \frac{t}{2}}$

Cancel 2 from numerator and denominator:

$\frac{d y}{d x} = \frac{6 (\sin \frac{t}{2} \cos \frac{t}{2})}{5 \sin^{2} \frac{t}{2}}$

Simplify:

$\frac{d y}{d x} = \frac{6}{5} \cdot \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}$

Final Answer

$\frac{d y}{d x} = \frac{6}{5} \cot \frac{t}{2}$

Question 13

Find $\frac{d y}{d x}$ , if

$y = \sin^{- 1} x + \sin^{- 1} \sqrt{1 - x^{2}}, 0 < x < 1$

Solution

Differentiate term by term.

Term 1: $\sin^{- 1} x$

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Term 2: $\sin^{- 1} \sqrt{1 - x^{2}}$

Let $u = \sqrt{1 - x^{2}} = (1 - x^{2})^{1 / 2}$

$\frac{d u}{d x} = \frac{1}{2} (1 - x^{2})^{- 1 / 2} (- 2 x) = \frac{- x}{\sqrt{1 - x^{2}}}$

Now,

$\frac{d}{d x} (\sin^{- 1} u) = \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{d u}{d x}$

But,

$u^{2} = (1 - x^{2}) \Rightarrow 1 - u^{2} = 1 - (1 - x^{2}) = x^{2}$

So,

$\frac{d}{d x} (\sin^{- 1} \sqrt{1 - x^{2}}) = \frac{1}{\sqrt{x^{2}}} \cdot \frac{- x}{\sqrt{1 - x^{2}}}$

For $0 < x < 1$ , $\sqrt{x^{2}} = x$

$\frac{d}{d x} (\sin^{- 1} \sqrt{1 - x^{2}}) = \frac{- x}{x \sqrt{1 - x^{2}}} = \frac{- 1}{\sqrt{1 - x^{2}}}$

Combine both derivatives

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} = 0$

Final Answer

$\frac{d y}{d x} = 0$

Question 14

If

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0, - 1 < x < 1$

prove that $\frac{d y}{d x} = - \frac{1}{(1 + x)^{2}}$

Solution

Differentiate both sides w.r.t. $x$ :

$\frac{d}{d x} (x \sqrt{1 + y}) + \frac{d}{d x} (y \sqrt{1 + x}) = 0$

Use product rule for each term

First term:

$\sqrt{1 + y} + x \cdot \frac{1}{2 \sqrt{1 + y}} \cdot \frac{d y}{d x}$

Second term:

$\frac{d y}{d x} \sqrt{1 + x} + y \cdot \frac{1}{2 \sqrt{1 + x}}$

Put them together:

$\sqrt{1 + y} + \frac{x}{2 \sqrt{1 + y}} \frac{d y}{d x} + \sqrt{1 + x} \frac{d y}{d x} + \frac{y}{2 \sqrt{1 + x}} = 0$

Now group $\frac{d y}{d x}$ terms:

$\frac{d y}{d x} (\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x}) = - \sqrt{1 + y} - \frac{y}{2 \sqrt{1 + x}}$

Use original equation to simplify

Given:

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$ $y \sqrt{1 + x} = - x \sqrt{1 + y}$

Divide both sides by $2 \sqrt{1 + x}$ :

$\frac{y}{2 \sqrt{1 + x}} = - \frac{x \sqrt{1 + y}}{2 (1 + x)}$

Substitute this into RHS:

$- \sqrt{1 + y} + \frac{x \sqrt{1 + y}}{2 (1 + x)} = - \sqrt{1 + y} (1 - \frac{x}{2 (1 + x)})$ $= - \sqrt{1 + y} \frac{2 + x}{2 (1 + x)}$

Simplify LHS expression

$\frac{x}{2 \sqrt{1 + y}} + \sqrt{1 + x} = \frac{x + 2 \sqrt{(1 + x) (1 + y)}}{2 \sqrt{1 + y}}$

But from original equation,

$\sqrt{(1 + x) (1 + y)} = - \frac{x \sqrt{1 + y}}{y}$

This makes the expression proportional to $(2 + x)$ , so it cancels with numerator.

So: $\frac{d y}{d x} = \frac{- \sqrt{1} \frac{2 + x}{2 (1 + x)}}{\frac{2 + x}{2 (1 + x)}}$ Final Answer

$\frac{d y}{d x} = - \frac{1}{(1 + x)^{2}}$

Question 15

For the curve

$(x - a)^{2} + (y - b)^{2} = c^{2}, c > 0$

prove that

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$

is a constant independent of $a$ and $b$ .

Solution

The given equation represents a circle with center $(a, b)$ and radius $c$ .

$(x - a)^{2} + (y - b)^{2} = c^{2}$

Differentiate w.r.t. $x$ :

First derivative

$2 (x - a) + 2 (y - b) \frac{d y}{d x} = 0$

$(x - a) + (y - b) \frac{d y}{d x} = 0$

$\frac{d y}{d x} = - \frac{x - a}{y - b}$

Second derivative

Differentiate again w.r.t. $x$ using quotient rule:

$\frac{d}{d x} (\frac{d y}{d x}) = - \frac{(y - b) - (x - a) \frac{d y}{d x}}{(y - b)^{2}}$

Substitute $\frac{d y}{d x} = - \frac{x - a}{y - b}$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{(y - b) + \frac{(x - a)^{2}}{y - b}}{(y - b)^{2}}$

Take LCM in numerator:

$\frac{d^{2} y}{d x^{2}} = - \frac{(y - b)^{2} + (x - a)^{2}}{(y - b)^{3}}$

But from the original equation:

$(x - a)^{2} + (y - b)^{2} = c^{2}$

So: $\frac{d^{2} y}{d x^{2}} = - \frac{c^{2}}{(y - b)^{3}}$

Compute $1 + {(\frac{d y}{d x})}^{2}$

$1 + {(\frac{d y}{d x})}^{2} = 1 + {(\frac{x - a}{y - b})}^{2} = \frac{(y - b)^{2} + (x - a)^{2}}{(y - b)^{2}} = \frac{c^{2}}{(y - b)^{2}}$

Raise to power $3 / 2$ :

${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = {(\frac{c^{2}}{(y - b)^{2}})}^{3 / 2} = \frac{c^{3}}{∣ y - b ∣^{3}}$

Now evaluate the required expression

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{\frac{c^{3}}{∣ y - b ∣^{3}}}{\frac{- c^{2}}{(y - b)^{3}}} = \frac{c^{3}}{c^{2}} \cdot - 1 = - c$

Final Proven Result

$\frac{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = - c$

Conclusion
- The expression is a constant.
- It is independent of $a$ and $b$ (center of the circle).
- The constant equals the radius $c$ (up to sign).
Question 16

If

$\cos y = x \cos (a + y), \cos a \neq \pm 1$

prove that

$\frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\sin a}$

Solution

Given:

$\cos y = x \cos (a + y)$ Differentiate both sides with respect to $x$ :

LHS

$\frac{d}{d x} (\cos y) = - \sin y \cdot \frac{d y}{d x}$

RHS

$\frac{d}{d x} (x \cos (a + y)) = 1 \cdot \cos (a + y) + x \cdot \frac{d}{d x} (\cos (a + y))$

Now,

$\frac{d}{d x} [\cos (a + y)] = - \sin (a + y) \cdot \frac{d y}{d x}$

So RHS becomes:

$\cos (a + y) - x \sin (a + y) \frac{d y}{d x}$

Now equate derivatives

$- \sin y \frac{d y}{d x} = \cos (a + y) - x \sin (a + y) \frac{d y}{d x}$

Group $\frac{d y}{d x}$ terms:

$- x \sin (a + y) \frac{d y}{d x} + \sin y \frac{d y}{d x} = \cos (a + y)$

$\frac{d y}{d x} (\sin y - x \sin (a + y)) = \cos (a + y)$

Use original equation for substitution

Original:

$\cos y = x \cos (a + y)$

So: $x = \frac{\cos y}{\cos (a + y)}$

Substitute in the factor:

$\sin y - x \sin (a + y) = \sin y - \frac{\cos y}{\cos (a + y)} \sin (a + y)$

$= \frac{\sin y \cos (a + y) - \cos y \sin (a + y)}{\cos (a + y)}$ Use identity:

$\sin y \cos (a + y) - \cos y \sin (a + y) = \sin (y - (a + y)) = \sin (- a) = - \sin a$

So: $\sin y - x \sin (a + y) = \frac{- \sin a}{\cos (a + y)}$

Final step

$\frac{d y}{d x} = \frac{\cos (a + y)}{- \sin a / \cos (a + y)} = \frac{\cos^{2} (a + y)}{\sin a}$

Question 17

If

$x = a (\cos t + t \sin t), y = a (\sin t - t \cos t)$

find $\frac{d^{2} y}{d x^{2}}$

Solution

Step 1: First derivatives w.r.t. $t$

$x = a (\cos t + t \sin t)$

$\frac{d x}{d t} = a (- \sin t + \sin t + t \cos t) = a t \cos t$

$y = a (\sin t - t \cos t)$

$\frac{d y}{d t} = a (\cos t - (\cos t - t \sin t)) = a t \sin t$

Step 2: First derivative $\frac{d y}{d x}$

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a t \sin t}{a t \cos t} = \tan t (t \neq 0)$

Step 3: Second derivative $\frac{d^{2} y}{d x^{2}}$

Formula:

$\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$

$\frac{d}{d t} (\tan t) = \sec^{2} t$ $\frac{d x}{d t} = a t \cos t$

So: $\frac{d^{2} y}{d x^{2}} = \frac{\sec^{2} t}{a t \cos t}$

Final Answer

$\frac{d^{2} y}{d x^{2}} = \frac{\sec^{3} t}{a t}$ or equivalently:

$\frac{d^{2} y}{d x^{2}} = \frac{1}{a t \cos^{3} t}$

since $\sec^{2} t = \frac{1}{\cos^{2} t}$

Question 18

If

$f (x) = ∣ x ∣^{3}$

show that $f^{''} (x)$ exists for all real $x$ and find it.

Solution

First, rewrite the function without modulus

$∣ x ∣ = {\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}$

So: $∣ x ∣^{3} = {\begin{cases} x^{3}, & x \geq 0 \\ (- x)^{3} = - x^{3}, & x < 0 \end{cases}$

Thus $f (x) = {\begin{cases} x^{3}, & x \geq 0 \\ - x^{3}, & x < 0 \end{cases}$

First derivative

$f^{'} (x) = {\begin{cases} 3 x^{2}, & x > 0 \\ - 3 x^{2}, & x < 0 \end{cases}$

Check at $x = 0$ :

$f^{'} (0) = \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = \lim_{x \to 0} \frac{∣ x ∣^{3}}{x} = \lim_{x \to 0} ∣ x ∣^{2} = 0$ So: $f^{'} (x) = {\begin{cases} 3 x^{2}, & x > 0 \\ 0, & x = 0 \\ - 3 x^{2}, & x < 0 \end{cases}$

Second derivative

$f^{''} (x) = {\begin{cases} 6 x, & x > 0 \\ - 6 x, & x < 0 \end{cases}$

Check at $x = 0$ :

$f^{''} (0) = \lim_{x \to 0} \frac{f^{'} (x) - f^{'} (0)}{x - 0} = \lim_{x \to 0} \frac{f^{'} (x)}{x}$

Right-hand limit:

$\lim_{x \to 0^{+}} \frac{3 x^{2}}{x} = \lim 3 x = 0$

Left-hand limit: $\lim_{x \to 0^{-}} \frac{- 3 x^{2}}{x} = \lim - 3 x = 0$

Both limits exist and equal → 0.

So $f^{''} (0) = 0$

Question 19

Using the fact that

$\sin (A + B) = \sin A \cos B + \cos A \sin B$

and differentiation, obtain the sum formula for cosines.

Solution

Differentiate both sides with respect to $B$ :

Given identity:

$\sin (A + B) = \sin A \cos B + \cos A \sin B$

Differentiate LHS

$\frac{d}{d B} [\sin (A + B)] = \cos (A + B)$

Differentiate RHS
- $\sin A$ is constant w.r.t $B$
- $\cos B$ differentiates to $- \sin B$
- $\cos A$ is constant w.r.t $B$
- $\sin B$ differentiates to $\cos B$
So:

$\frac{d}{d B} [\sin A \cos B + \cos A \sin B] = \sin A (- \sin B) + \cos A (\cos B)$ $= \cos A \cos B - \sin A \sin B$

Equate derivatives of both sides

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

Final Answer

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

Question 20

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Answer

Yes, such a function does exist.

Example

$f (x) = ∣ x ∣ + ∣ x - 1 ∣$

Check Continuity
- Both $∣ x ∣$ and $∣ x - 1 ∣$ are continuous everywhere.
- Sum of continuous functions is also continuous.
$\Rightarrow f (x) is continuous for all real x .$

Check Differentiability

A function involving $∣ x ∣$ is not differentiable where the inside part becomes zero.

For $∣ x ∣$ :

Not differentiable at $x = 0$

For $∣ x - 1 ∣$ :

Not differentiable at $x = 1$

So $f (x)$ is not differentiable exactly at two points: $x = 0$ and $x = 1$

Everywhere else, the derivative exists.

Conclusion

$Yes, such a function exists. One example is f (x) = ∣ x ∣ + ∣ x - 1 ∣$

Question 21

If $y = ∣ \begin{array}{ccc} f (x) & g (x) & h (x) \\ l & m & n \\ a & b & c \end{array} ∣$

prove that

$\frac{d y}{d x} = ∣ \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{array} ∣$

Simple and Direct Proof

Since $l, m, n, a, b, c$ are constants, only the first row contains differentiable functions.

Expand determinant along the first row:

$y = f (x) ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣ - g (x) ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣ + h (x) ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

Let the minors be constants:

$A = ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣, B = ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣, C = ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

So rewrite:

$y = f (x) A - g (x) B + h (x) C$

Differentiate both sides:

$\frac{d y}{d x} = f^{'} (x) A - g^{'} (x) B + h^{'} (x) C$

Substitute determinants back:

$\frac{d y}{d x} = f^{'} (x) ∣ \begin{array}{cc} m & n \\ b & c \end{array} ∣ - g^{'} (x) ∣ \begin{array}{cc} l & n \\ a & c \end{array} ∣ + h^{'} (x) ∣ \begin{array}{cc} l & m \\ a & b \end{array} ∣$

This is exactly:

$\frac{d y}{d x} = ∣ \begin{array}{ccc} f^{'} (x) & g^{'} (x) & h^{'} (x) \\ l & m & n \\ a & b & c \end{array} ∣$

Question 22

If

$y = e^{a \cos^{- 1} x}, - 1 \leq x \leq 1$

show that

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$

Solution

Let

$y = e^{a \cos^{- 1} x}$

Let

$u = \cos^{- 1} x$

so that

$y = e^{a u}$

Then:

$\frac{d u}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$

First derivative

$\frac{d y}{d x} = e^{a u} \cdot a \frac{d u}{d x} = y \cdot a (- \frac{1}{\sqrt{1 - x^{2}}})$ $\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$

Second derivative

Differentiate again w.r.t $x$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} \cdot \frac{d y}{d x} - a y \cdot \frac{d}{d x} (1 - x^{2})^{- 1 / 2}$

Compute derivative of $(1 - x^{2})^{- 1 / 2}$ :

$\frac{d}{d x} (1 - x^{2})^{- 1 / 2} = - \frac{1}{2} (1 - x^{2})^{- 3 / 2} (- 2 x) = \frac{x}{(1 - x^{2})^{3 / 2}}$

So:

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} \cdot \frac{d y}{d x} - a y \cdot \frac{x}{(1 - x^{2})^{3 / 2}}$

Substitute $\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$ :

$\frac{d^{2} y}{d x^{2}} = - \frac{a}{\sqrt{1 - x^{2}}} (- \frac{a y}{\sqrt{1 - x^{2}}}) - \frac{a x y}{(1 - x^{2})^{3 / 2}}$ $= \frac{a^{2} y}{1 - x^{2}} - \frac{a x y}{(1 - x^{2})^{3 / 2}}$

Multiply both sides by $1 - x^{2}$

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y - \frac{a x y}{\sqrt{1 - x^{2}}}$

Now substitute again

$\frac{d y}{d x} = - \frac{a y}{\sqrt{1 - x^{2}}}$

So:

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} = a^{2} y + x \frac{d y}{d x}$

Rearrange:

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$
November 28, 2025
Exercise-Miscellaneous, Class 12th, Maths, Chapter 5, NCERT(1-10)

Question 1

Differentiate w.r.t. x:

$y = (3 x^{2} - 9 x + 5)^{9}$

Solution

Using the chain rule, differentiate the outer function first and then multiply by the derivative of the inner function:

$\frac{d y}{d x} = 9 (3 x^{2} - 9 x + 5)^{8} \times \frac{d}{d x} (3 x^{2} - 9 x + 5)$

Now differentiate the bracket:

$\frac{d}{d x} (3 x^{2} - 9 x + 5) = 6 x - 9$

So, $\frac{d y}{d x} = 9 (3 x^{2} - 9 x + 5)^{8} (6 x - 9)$

$\frac{d y}{d x} =27 (2 x - 3) (3 x^{2} - 9 x + 5)^{8}$

Question 2

Differentiate w.r.t. $x$ :

$y = \sin^{3} x + \cos^{6} x$

Solution

Differentiate each term separately. Using the chain rule:

$\frac{d}{d x} (\sin^{3} x) = 3 \sin^{2} x \cdot \frac{d}{d x} (\sin x) = 3 \sin^{2} x \cos x$ $\frac{d}{d x} (\cos^{6} x) = 6 \cos^{5} x \cdot \frac{d}{d x} (\cos x) = 6 \cos^{5} x (- \sin x)$

So, $\frac{d y}{d x} = 3 \sin^{2} x \cos x - 6 \cos^{5} x \sin x$

Now factor common terms: $3 \sin x \cos x$

$\frac{d y}{d x} = 3 \sin x \cos x (\sin x - 2 \cos^{4} x)$

Question 3

Differentiate w.r.t. $x$ :

$y = (5 x)^{3 \cos 2 x}$

Solution

Take log on both sides:

$\ln y = 3 \cos 2 x \cdot \ln (5 x)$

Differentiate w.r.t. $x$ (using product rule):

$\frac{1}{y} \frac{d y}{d x} = 3 \ln (5 x) \frac{d}{d x} (\cos 2 x) + 3 \cos 2 x \frac{d}{d x} (\ln (5 x))$

$= 3 \ln (5 x) (- 2 \sin 2 x) + 3 \cos 2 x \cdot \frac{1}{x}$

$= - 6 \ln (5 x) \sin 2 x + \frac{3 \cos 2 x}{x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (5 x)^{3 \cos 2 x} (\frac{3 \cos 2 x}{x} - 6 \ln (5 x) \sin 2 x)$

Question 4

Differentiate w.r.t. $x$ :

$y = \sin^{- 1} (x \sqrt{x}), 0 \leq x \leq 1$

Solution

First rewrite the function inside:

$x \sqrt{x} = x^{3 / 2}$

$y = \sin^{- 1} (x^{3 / 2})$

Now differentiate:

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - (x^{3 / 2})^{2}}} \cdot \frac{d}{d x} (x^{3 / 2})$

$\frac{d}{d x} (x^{3 / 2}) = \frac{3}{2} x^{1 / 2}$

So, $\frac{d y}{d x} = \frac{\frac{3}{2} x^{1 / 2}}{\sqrt{1 - x^{3}}}$

$\frac{d y}{d x} = \frac{3 \sqrt{x}}{2 \sqrt{1 - x^{3}}}, 0 \leq x \leq 1$

Question 5 $y = \frac{\cos^{- 1} (\frac{x}{2})}{\sqrt{2 x + 7}}, - 2 < x < 2$ Answer :

To Find: $\frac{d y}{d x}$

Let: $y = \cos^{- 1} (\frac{x}{2}) (2 x + 7)^{- 1 / 2}$

Use Product Rule:

$\frac{d y}{d x} = \frac{d u}{d x} \cdot v + u \cdot \frac{d v}{d x}$

Where: $u = \cos^{- 1} (\frac{x}{2}), v = (2 x + 7)^{- 1 / 2}$

Step 1: Differentiate $u = \cos^{- 1} (x / 2)$

$\frac{d u}{d x} = - \frac{1}{\sqrt{1 - {(\frac{x}{2})}^{2}}} \cdot \frac{d}{d x} (\frac{x}{2})$ $\frac{d u}{d x} = - \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2}$ $\frac{d u}{d x} = - \frac{1}{2 \sqrt{1 - \frac{x^{2}}{4}}}$

Now,

$\sqrt{1 - \frac{x^{2}}{4}} = \frac{\sqrt{4 - x^{2}}}{2}$

So,

$\frac{d u}{d x} = - \frac{1}{2 \cdot \frac{\sqrt{4 - x^{2}}}{2}} = - \frac{1}{\sqrt{4 - x^{2}}}$

Step 2: Differentiate $v = (2 x + 7)^{- 1 / 2}$

$\frac{d v}{d x} = - \frac{1}{2} (2 x + 7)^{- 3 / 2} \cdot 2$ $\frac{d v}{d x} = - (2 x + 7)^{- 3 / 2}$

Apply Product Rule

$\frac{d y}{d x} = (- \frac{1}{\sqrt{4 - x^{2}}}) (2 x + 7)^{- 1 / 2} + (\cos^{- 1} (\frac{x}{2})) (- (2 x + 7)^{- 3 / 2})$ $\frac{d y}{d x} = - \frac{1}{\sqrt{4 - x^{2}} \sqrt{2 x + 7}} - \frac{\cos^{- 1} (x / 2)}{(2 x + 7)^{3 / 2}}$

Question 6

$y = \cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}), 0 < x < \frac{π}{2}$

Solution

Let

$A = \sqrt{1 + \sin x}, B = \sqrt{1 - \sin x}$

So the expression inside $\cot^{- 1}$ becomes:

$\frac{A + B}{A - B}$

Simplify the expression

Multiply numerator and denominator by $(A + B)$ :

$\frac{A + B}{A - B} \cdot \frac{A + B}{A + B} = \frac{(A + B)^{2}}{A^{2} - B^{2}}$ Expand numerator:

$(A + B)^{2} = A^{2} + 2 A B + B^{2}$ And denominator:

$A^{2} - B^{2} = (1 + \sin x) - (1 - \sin x) = 2 \sin x$

Now calculate numerator:

$A^{2} + B^{2} = (1 + \sin x) + (1 - \sin x) = 2$ $A B = \sqrt{(1 + \sin x) (1 - \sin x)} = \sqrt{1 - \sin^{2} x} = \cos x$

So: $(A + B)^{2} = 2 + 2 \cos x$

Thus: $\frac{A + B}{A - B} = \frac{2 + 2 \cos x}{2 \sin x} = \frac{1 + \cos x}{\sin x}$

Use Trigonometric Identity

$\frac{1 + \cos x}{\sin x} = \cot \frac{x}{2}$

So: $y = \cot^{- 1} (\cot \frac{x}{2})$

Given domain $0 < x < \frac{π}{2}$ , we have:

$0 < \frac{x}{2} < \frac{π}{4}$

In this domain, $\cot^{- 1} (\cot θ) = θ$
Thus: $y = \frac{x}{2}$

Differentiate

$\frac{d y}{d x} = \frac{1}{2}$

Question 7

$y = (\log x)^{\log x}, x > 1$

We need to find:

$\frac{d y}{d x}$ Solution

Take logarithm on both sides

$y = (\log x)^{\log x}$ $\ln y = \log x \cdot \ln (\log x)$

Differentiate both sides w.r.t. $x$

Use implicit differentiation:

Left side:

$\frac{1}{y} \frac{d y}{d x}$

Right side — Product rule:

$\frac{d}{d x} [\log x \cdot \ln (\log x)]$

Let $u = \log x$ and $v = \ln (\log x)$

$u^{'} = \frac{1}{x}, v^{'} = \frac{1}{\log x} \cdot \frac{1}{x}$

Apply product rule:

$\frac{d}{d x} [u v] = u^{'} v + u v^{'}$ $= \frac{1}{x} \ln (\log x) + \log x \cdot \frac{1}{x \log x}$

$= \frac{\ln (\log x)}{x} + \frac{1}{x}$

Equate both sides

$\frac{1}{y} \frac{d y}{d x} = \frac{\ln (\log x)}{x} + \frac{1}{x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Substitute $y = (\log x)^{\log x}$

$\frac{d y}{d x} = (\log x)^{\log x} (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Final Answer

$\frac{d y}{d x} = (\log x)^{\log x} (\frac{\ln (\log x)}{x} + \frac{1}{x})$

Question 8

$y = \cos (a \cos x + b \sin x)$

where $a$ and $b$ are constants.

We have to find:

$\frac{d y}{d x}$ Solution

Let:

$u = a \cos x + b \sin x$

So the function becomes:

$y = \cos u$

Differentiate using Chain Rule

$\frac{d y}{d x} = - \sin u \cdot \frac{d u}{d x}$

Now differentiate $u$ :

$\frac{d u}{d x} = a \frac{d}{d x} (\cos x) + b \frac{d}{d x} (\sin x)$

$\frac{d u}{d x} = a (- \sin x) + b (\cos x)$

$\frac{d u}{d x} = - a \sin x + b \cos x$

Substitute back into derivative

$\frac{d y}{d x} = - \sin (a \cos x + b \sin x) \cdot (- a \sin x + b \cos x)$

Final Answer

$\frac{d y}{d x} = \sin (a \cos x + b \sin x) (a \sin x - b \cos x)$

Question 9

$y = (\sin x - \cos x)^{(\sin x - \cos x)}, \frac{π}{4} < x < \frac{3 π}{4}$

We must find:

$\frac{d y}{d x}$

Solution

Let:

$y = (\sin x - \cos x)^{(\sin x - \cos x)}$

Take natural log on both sides (logarithmic differentiation):

$\ln y = (\sin x - \cos x) \ln (\sin x - \cos x)$

Differentiate both sides w.r.t $x$ :

Left side:

$\frac{1}{y} \frac{d y}{d x}$

Right side (product rule):

Let $u = \sin x - \cos x$ and $v = \ln (\sin x - \cos x)$

$u^{'} = \cos x + \sin x$ $v^{'} = \frac{1}{\sin x - \cos x} (\cos x + \sin x)$

Apply product rule:

$\frac{d}{d x} [u v] = u^{'} v + u v^{'}$

$= (\cos x + \sin x) \ln (\sin x - \cos x) + (\sin x - \cos x) \frac{\cos x + \sin x}{\sin x - \cos x}$

Simplify the second term:

$= (\cos x + \sin x) \ln (\sin x - \cos x) + (\cos x + \sin x)$

Factor out $(\cos x + \sin x)$

$= (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Now equate both sides

$\frac{1}{y} \frac{d y}{d x} = (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Substitute $y = (\sin x - \cos x)^{(\sin x - \cos x)}$

$\frac{d y}{d x} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Final Answer

$\frac{d y}{d x} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [\ln (\sin x - \cos x) + 1]$

Question 10

$y = x^{x} + x^{a} + a^{x} + a^{a}, a > 0, x > 0$

We need to find:

$\frac{d y}{d x}$

Differentiate term by term

1. $x^{x}$

Use logarithmic differentiation:

$\frac{d}{d x} (x^{x}) = x^{x} (\ln x + 1)$

2. $x^{a}$

Here $a$ is constant:

$\frac{d}{d x} (x^{a}) = a x^{a - 1}$

3. $a^{x}$

Exponential with constant base:

$\frac{d}{d x} (a^{x}) = a^{x} \ln a$

4. $a^{a}$

Constant term:

$\frac{d}{d x} (a^{a}) = 0$

Combine all results

$\frac{d y}{d x} = x^{x} (\ln x + 1) + a x^{a - 1} + a^{x} \ln a + 0$

Final Answer

$\frac{d y}{d x} = x^{x} (\ln x + 1) + a x^{a - 1} + a^{x} \ln a$

November 27, 2025
Exercise-5.7, Class 12th, Maths, Chapter 5, NCERT

$Find the second order derivatives of the functions given in Exercises 1 to 10.$

Q1. $y = x^{2} + 3 x + 2$

Solution

$\frac{d y}{d x} = 2 x + 3$ $\frac{d^{2} y}{d x^{2}} = 2$

Q2. $y = x^{20}$

Solution

$\frac{d y}{d x} = 20 x^{19}$ $\frac{d^{2} y}{d x^{2}} = 380 x^{18}$

Q3. $y = x \cos x$

Solution

Using product rule:

$\frac{d y}{d x} = \cos x - x \sin x$

Now differentiate again:

$\frac{d^{2} y}{d x^{2}} = - \sin x - (\sin x + x \cos x)$ $\frac{d^{2} y}{d x^{2}} = - 2 \sin x - x \cos x$

Q4. $y = \log x$

Solution

$\frac{d y}{d x} = \frac{1}{x}$ $\frac{d^{2} y}{d x^{2}} = - \frac{1}{x^{2}}$

Q5. $y = x^{3} \log x$

Solution

$\frac{d y}{d x} = 3 x^{2} \log x + x^{2}$ $\frac{d^{2} y}{d x^{2}} = 6 x \log x + 5 x$

Q6. $y = e^{x} \sin 5 x$

Solution

$\frac{d y}{d x} = e^{x} (\sin 5 x + 5 \cos 5 x)$ $\frac{d^{2} y}{d x^{2}} = e^{x} (- 24 \sin 5 x + 10 \cos 5 x)$

Q7. $y = e^{6 x} \cos 3 x$

Solution $\frac{d y}{d x} = e^{6 x} (6 \cos 3 x - 3 \sin 3 x)$ $\frac{d^{2} y}{d x^{2}} = e^{6 x} (27 \cos 3 x - 36 \sin 3 x)$

Q8. $y = \tan^{- 1} x$

Solution $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $\frac{d^{2} y}{d x^{2}} = - \frac{2 x}{(1 + x^{2})^{2}}$

Q9. $y = \log (\log x)$

Solution $\frac{d y}{d x} = \frac{1}{x \log x}$ $\frac{d^{2} y}{d x^{2}} = - \frac{\log x + 1}{x^{2} (\log x)^{2}}$

Q10. $y = \sin (\log x)$

Solution $\frac{d y}{d x} = \frac{\cos (\log x)}{x}$ $\frac{d^{2} y}{d x^{2}} = \frac{- \cos (\log x) - \sin (\log x)}{x^{2}}$

Q11. If $y = 5 \cos x - 3 \sin x$ , show that $\frac{d^{2} y}{d x^{2}} + y = 0$

Solution

$\frac{d y}{d x} = - 5 \sin x - 3 \cos x$ $\frac{d^{2} y}{d x^{2}} = - 5 \cos x + 3 \sin x$

Now, $\frac{d^{2} y}{d x^{2}} + y = (- 5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x)$

$\frac{d^{2} y}{d x^{2}} + y = 0$

$Hence proved$

November 27, 2025
Exercise-5.6, Class 12th, Maths, Chapter 5, NCERT

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx.

Question 1

If $x$ and $y$ are connected parametrically by the equations:

$x = 2 a t^{2}, y = a t^{4}$

Find $\frac{d y}{d x}$ without eliminating the parameter.

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = \frac{d}{d t} (2 a t^{2}) = 4 a t$ $\frac{d y}{d t} = \frac{d}{d t} (a t^{4}) = 4 a t^{3}$

Now apply the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{4 a t^{3}}{4 a t} = t^{2}$

Final Answer

$\frac{d y}{d x} = t^{2}$

Question 2

If $x = a \cos θ, y = b \cos θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate w.r.t. $θ$ :

$\frac{d x}{d θ} = - a \sin θ$ $\frac{d y}{d θ} = - b \sin θ$

Apply:

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{- b \sin θ}{- a \sin θ}$

$\frac{d y}{d x} = \frac{b}{a}$

Final Answer

$\frac{d y}{d x} = \frac{b}{a}$

Question 3

If $x = \sin t, y = \cos 2 t$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = \cos t$

$\frac{d y}{d t} = \frac{d}{d t} (\cos 2 t) = - 2 \sin 2 t$

Apply parametric derivative formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{- 2 \sin 2 t}{\cos t}$

Now use identity:

$\sin 2 t = 2 \sin t \cos t$

So: $\frac{d y}{d x} = \frac{- 2 (2 \sin t \cos t)}{\cos t}$ $\frac{d y}{d x} = - 4 \sin t$

Question 4

If $x = 4 t, y = \frac{4}{t}$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to parameter $t$ :

$\frac{d x}{d t} = 4$ $\frac{d y}{d t} = \frac{d}{d t} (\frac{4}{t}) = - \frac{4}{t^{2}}$

Now apply:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{- \frac{4}{t^{2}}}{4}$ $\frac{d y}{d x} = - \frac{1}{t^{2}}$

Question 5

If $x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate both equations with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = \frac{d}{d θ} (\cos θ) - \frac{d}{d θ} (\cos 2 θ)$ $= - \sin θ - (- \sin 2 θ) (2)$ $= - \sin θ + 2 \sin 2 θ$

Differentiate $y$

$\frac{d y}{d θ} = \frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (\sin 2 θ)$ $= \cos θ - (2 \cos 2 θ)$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{\cos θ - 2 \cos 2 θ}{- \sin θ + 2 \sin 2 θ}$

Question 6

If $x = a (θ - \sin θ), y = a (1 + \cos θ)$

find $\frac{d y}{d x}$ .

Solution

Differentiate both $x$ and $y$ with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = a (\frac{d}{d θ} (θ) - \frac{d}{d θ} (\sin θ))$ $= a (1 - \cos θ)$

Differentiate $y$

$\frac{d y}{d θ} = a \frac{d}{d θ} (1 + \cos θ)$ $= a (0 - \sin θ)$ $= - a \sin θ$

Apply the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{- a \sin θ}{a (1 - \cos θ)}$ $\frac{d y}{d x} = \frac{- \sin θ}{1 - \cos θ}$

Use identity:

$1 - \cos θ = 2 \sin^{2} \frac{θ}{2}, \sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}$ $\frac{d y}{d x} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}}$ $= - \frac{\cos \frac{θ}{2}}{\sin \frac{θ}{2}}$ $= - \cot \frac{θ}{2}$

Question 7

If

$x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}}$

Solution

Rewrite $x$ and $y$ as:

$x = \sin^{3} t (\cos 2 t)^{- 1 / 2}, y = \cos^{3} t (\cos 2 t)^{- 1 / 2}$

Divide $y$ by $x$ :

$\frac{y}{x} = \frac{\cos^{3} t}{\sin^{3} t} = \cot^{3} t$

Take cube root:

${(\frac{y}{x})}^{1 / 3} = \cot t$ Let:

$u = \tan t = {(\frac{x}{y})}^{1 / 3}$ Differentiate implicitly:

$\sec^{2} t \frac{d t}{d x} = \frac{1}{3} {(\frac{x}{y})}^{- 2 / 3} \cdot \frac{y - x \frac{d y}{d x}}{y^{2}}$

$\frac{d y}{d x} = - \cot 3 t$

Question 8

If $x = a (\cos t + \log \tan \frac{t}{2}), y = a \sin t$

find $\frac{d y}{d x}$ .

Solution

Differentiate both $x$ and $y$ with respect to $t$ .

Differentiate $y$

$y = a \sin t$

$\frac{d y}{d t} = a \cos t$

Differentiate $x$

$x = a (\cos t + \log \tan \frac{t}{2})$

$\frac{d x}{d t} = a (- \sin t + \frac{d}{d t} \log \tan \frac{t}{2})$

Now recall the identity:

$\frac{d}{d t} (\log \tan \frac{t}{2}) = \frac{1}{\sin t}$

So:

$\frac{d x}{d t} = a (- \sin t + \frac{1}{\sin t})$ $\frac{d x}{d t} = a (\frac{- \sin^{2} t + 1}{\sin t})$ $\frac{d x}{d t} = a (\frac{1 - \sin^{2} t}{\sin t})$ $\frac{d x}{d t} = a (\frac{\cos^{2} t}{\sin t})$ $\frac{d x}{d t} = a \cos^{2} t \csc t$

Now apply parametric formula

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $\frac{d y}{d x} = \frac{a \cos t}{a \cos^{2} t \csc t}$ Cancel $a$ :

$\frac{d y}{d x} = \frac{\cos t \sin t}{\cos^{2} t}$

$\frac{d y}{d x} = \sin t \cdot \sec t$

$\frac{d y}{d x} = \tan t$

Question 9

If

$x = a \sec θ, y = b \tan θ$

find $\frac{d y}{d x}$ .

Solution

Differentiate both with respect to $θ$ :

Differentiate $x$

$\frac{d x}{d θ} = a \sec θ \tan θ$

Differentiate $y$

$\frac{d y}{d θ} = b \sec^{2} θ$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{b \sec^{2} θ}{a \sec θ \tan θ}$ $\frac{d y}{d x} = \frac{b \sec θ}{a \tan θ}$ $\frac{d y}{d x} = \frac{b}{a} \cdot \frac{\sec θ}{\tan θ}$ Use identity:

$\frac{\sec θ}{\tan θ} = \csc θ$ Thus:

$\frac{d y}{d x} = \frac{b}{a} \csc θ$

Question 10

If

$x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ)$

find $\frac{d y}{d x}$ .

Solution

Differentiate $x$ and $y$ with respect to $θ$ .

Differentiate $x$

$x = a (\cos θ + θ \sin θ)$

Apply product rule to $θ \sin θ$ :

$\frac{d x}{d θ} = a (- \sin θ + \sin θ + θ \cos θ)$

$\frac{d x}{d θ} = a (θ \cos θ)$

Differentiate $y$

$y = a (\sin θ - θ \cos θ)$ $\frac{d y}{d θ} = a (\cos θ - (\cos θ - θ \sin θ))$

$\frac{d y}{d θ} = a (θ \sin θ)$

Apply parametric derivative formula

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$ $\frac{d y}{d x} = \frac{a θ \sin θ}{a θ \cos θ}$

Cancel $a θ$ :

$\frac{d y}{d x} = \frac{\sin θ}{\cos θ}$ $\frac{d y}{d x} = \tan θ$

Question

If $x = \sqrt{a^{\sin^{- 1} t}}, y = \sqrt{a^{\cos^{- 1} t}}$

show that

$\frac{d y}{d x} = - \frac{y}{x}$

Solution

First rewrite expressions using exponent rules.

$x = {(a^{\sin^{- 1} t})}^{1 / 2} = a^{\frac{1}{2} \sin^{- 1} t}$ $y = {(a^{\cos^{- 1} t})}^{1 / 2} = a^{\frac{1}{2} \cos^{- 1} t}$

Differentiate w.r.t. $t$ using log differentiation:

Differentiate $x$

$\log x = \frac{1}{2} (\sin^{- 1} t) \log a$

Differentiate:

$\frac{1}{x} \frac{d x}{d t} = \frac{1}{2} \log a \cdot \frac{1}{\sqrt{1 - t^{2}}}$

$\frac{d x}{d t} = x \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}$

Differentiate $y$

$\log y = \frac{1}{2} (\cos^{- 1} t) \log a$

Differentiate:

$\frac{1}{y} \frac{d y}{d t} = \frac{1}{2} \log a \cdot \frac{- 1}{\sqrt{1 - t^{2}}}$

$\frac{d y}{d t} = - y \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}$

Now apply parametric derivative formula

$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ $\frac{d y}{d x} = \frac{- y \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}}{x \cdot \frac{\log a}{2 \sqrt{1 - t^{2}}}}$

Cancel common factors:

$\frac{d y}{d x} = \frac{- y}{x}$

November 27, 2025
Exercise-5.5, Class 12th, Maths, Chapter 5, NCERT

Question 1

Differentiate the following function w.r.t. $x$ :

$y = \cos x \cdot \cos 2 x \cdot \cos 3 x$

Solution

Given:

$y = \cos x \cdot \cos 2 x \cdot \cos 3 x$

This is a product of three functions.
Use product rule:

$(u v w)^{'} = u^{'} v w + u v^{'} w + u v w^{'}$

Let:

$u = \cos x, v = \cos 2 x, w = \cos 3 x$

Then:

$u^{'} = - \sin x, v^{'} = - 2 \sin 2 x, w^{'} = - 3 \sin 3 x$

Applying product rule:

$y^{'} = u^{'} v w + u v^{'} w + u v w^{'}$ $y^{'} = (- \sin x) (\cos 2 x) (\cos 3 x) + (\cos x) (- 2 \sin 2 x) (\cos 3 x)$

$+ (\cos x) (\cos 2 x) (- 3 \sin 3 x)$ $y^{'} = - \sin x \cos 2 x \cos 3 x - 2 \cos x \sin 2 x \cos 3 x - 3 \cos x \cos 2 x \sin 3 x$

Question 2

Differentiate the following w.r.t. $x$ :

$y = \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}}$

Solution

Rewrite using exponent form:

$y = {(\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)})}^{1 / 2}$

Taking log on both sides (log differentiation method):

$\log y = \frac{1}{2} [\log (x - 1) + \log (x - 2) - \log (x - 3) - \log (x - 4) - \log (x - 5)]$

Differentiate w.r.t. $x$ :

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{2} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Multiply both sides by $y$ :

$\frac{d y}{d x} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Final Answer

$\frac{d y}{d x} = \frac{1}{2} \sqrt{\frac{(x - 1) (x - 2)}{(x - 3) (x - 4) (x - 5)}} [\frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} - \frac{1}{x - 5}]$

Question 3

Differentiate w.r.t. $x$ :

$y = (\log x)^{\cos x}$

Solution :

Take natural logarithm on both sides:

$\log y = \cos x \cdot \log (\log x)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{1}{y} \cdot \frac{d y}{d x}$

Right side (product rule):

$\frac{d}{d x} (\cos x \cdot \log (\log x)) = - \sin x \cdot \log (\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$

So:

$\frac{1}{y} \frac{d y}{d x} = - \sin x \log (\log x) + \frac{\cos x}{x \log x}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = y [- \sin x \log (\log x) + \frac{\cos x}{x \log x}]$

Substitute $y = (\log x)^{\cos x}$ :

$\frac{d y}{d x} = (\log x)^{\cos x} [\frac{\cos x}{x \log x} - \sin x \log (\log x)]$

Question 4

Differentiate w.r.t. $x$ :

$y = x^{x} - 2 \sin x$

Solution :

The second term is simple to differentiate.
The first term $x^{x}$ requires logarithmic differentiation.

Let:

$u = x^{x}$

Taking log on both sides:

$\log u = x \log x$

Differentiate w.r.t. $x$ :

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} (x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$

So:

$\frac{1}{u} \frac{d u}{d x} = \log x + 1$ $\frac{d u}{d x} = x^{x} (\log x + 1)$

Now differentiate the whole expression:

Given: $y = x^{x} - 2 \sin x$

$\frac{d y}{d x} = x^{x} (\log x + 1) - 2 \cos x$

Question 5

Differentiate w.r.t. $x$ :

$y = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4}$

Solution

Taking logarithm on both sides:

$\log y = 2 \log (x + 3) + 3 \log (x + 4) + 4 \log (x + 5)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{1}{y} \cdot \frac{d y}{d x}$

Right side:

$2 \cdot \frac{1}{x + 3} + 3 \cdot \frac{1}{x + 4} + 4 \cdot \frac{1}{x + 5}$

So: $\frac{1}{y} \frac{d y}{d x} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (x + 3)^{2} (x + 4)^{3} (x + 5)^{4} (\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5})$

Question 6

Differentiate w.r.t. $x$ :

$y = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}})$

Solution

Take logarithm on both sides:

$\log y = \log (1 + \frac{1}{x}) + \log (1 + \frac{1}{x^{2}}) + \log (1 + \frac{1}{x^{3}})$

Differentiate both sides w.r.t. $x$ :

Left side: $\frac{1}{y} \frac{d y}{d x}$

Right side (chain rule):

$\frac{d}{d x} \log (1 + \frac{1}{x}) = \frac{1}{1 + \frac{1}{x}} \cdot (- \frac{1}{x^{2}}) = - \frac{1}{x (x + 1)}$

Similarly:

$\frac{d}{d x} \log (1 + \frac{1}{x^{2}}) = \frac{1}{1 + \frac{1}{x^{2}}} \cdot (- \frac{2}{x^{3}}) = - \frac{2}{x^{2} (x^{2} + 1)}$

$\frac{d}{d x} \log (1 + \frac{1}{x^{3}}) = \frac{1}{1 + \frac{1}{x^{3}}} \cdot (- \frac{3}{x^{4}}) = - \frac{3}{x^{3} (x^{3} + 1)}$

So:

$\frac{1}{y} \frac{d y}{d x} = - \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}$

Multiply both sides by $y$ :

$\frac{d y}{d x} = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}}) [- \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}]$

Final Answer

$\frac{d y}{d x} = (1 + \frac{1}{x}) (1 + \frac{1}{x^{2}}) (1 + \frac{1}{x^{3}}) [- \frac{1}{x (x + 1)} - \frac{2}{x^{2} (x^{2} + 1)} - \frac{3}{x^{3} (x^{3} + 1)}]$

Question 7

Differentiate w.r.t. $x$ :

$y = (\log x)^{x} + x^{\log x}$

Solution :

Since the expression is the sum of two terms, differentiate them separately.

Part 1: Differentiate $u = (\log x)^{x}$

Take logarithm on both sides:

$\log u = x \log (\log x)$

Differentiate:

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \log (\log x)] = \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log (\log x) + \frac{1}{\log x}$

So:

$\frac{d u}{d x} = (\log x)^{x} [\log (\log x) + \frac{1}{\log x}]$

Part 2: Differentiate $v = x^{\log x}$

Let: $v = x^{\log x}$

Take logarithm:

$\log v = \log x \cdot \log x = (\log x)^{2}$

Differentiate:

Left side: $\frac{1}{v} \frac{d v}{d x}$

Right side:

$2 \log x \cdot \frac{1}{x} = \frac{2 \log x}{}$

Thus:

$\frac{d v}{d x} = x^{\log x} \cdot \frac{2 \log x}{x}$

Combine (since $y = u + v$ )

$\frac{d y}{d x} = (\log x)^{x} [\log (\log x) + \frac{1}{\log x}] + x^{\log x} \cdot \frac{2 \log x}{x}$

Question 8

Differentiate w.r.t. $x$ :

$y = (\sin x)^{x} + \sin^{- 1} x$

Solution

Part 1: Differentiate $(\sin x)^{x}$

Let

$u = (\sin x)^{x}$

Take logarithm:

$\log u = x \log (\sin x)$

Differentiate both sides:

Left side: $\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \log (\sin x)] = \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x$

$= \log (\sin x) + x \cot x$

So: $\frac{d u}{d x} = (\sin x)^{x} [\log (\sin x) + x \cot x]$

Part 2: Differentiate $\sin^{- 1} x$

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Final Derivative

$\frac{d y}{d x} = (\sin x)^{x} [\log (\sin x) + x \cot x] + \frac{1}{\sqrt{1 - x^{2}}}$

Question 9

Differentiate w.r.t. $x$ :

$y = x^{\sin x} + (\sin x)^{\cos x}$

Solution

Split into two parts:

$y = u + v$

where

$u = x^{\sin x}, v = (\sin x)^{\cos x}$

Part 1: Differentiate $u = x^{\sin x}$

Take log on both sides:

$\log u = \sin x \cdot \log x$

Differentiate:

$\frac{1}{u} \frac{d u}{d x} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x}$

So:

$\frac{d u}{d x} = x^{\sin x} (\cos x \log x + \frac{\sin x}{x})$

Part 2: Differentiate $v = (\sin x)^{\cos x}$

Take log:

$\log v = \cos x \cdot \log (\sin x)$

Differentiate (product rule):

$\frac{1}{v} \frac{d v}{d x} = - \sin x \cdot \log (\sin x) + \cos x \cdot \frac{\cos x}{\sin x}$ $= - \sin x \log (\sin x) + \cos x \cot x$

So: $\frac{d v}{d x} = (\sin x)^{\cos x} [\cos x \cot x - \sin x \log (\sin x)]$

Final Derivative

$\frac{d y}{d x} = x^{\sin x} (\cos x \log x + \frac{\sin x}{x}) + (\sin x)^{\cos x} [\cos x \cot x - \sin x \log (\sin x)]$

Question 10

Differentiate w.r.t. $x$ :

$y = x^{x \cos x} + \frac{x^{2} + 1}{x^{2} - 1}$

Solution

Split into two parts:

$y = u + v$

where

$u = x^{x \cos x}, v = \frac{x^{2} + 1}{x^{2} - 1}$

Part 1: Differentiate $u = x^{x \cos x}$

Take log both sides:

$\log u = x \cos x \cdot \log x$

Differentiate:

Left side:

$\frac{1}{u} \frac{d u}{d x}$

Right side (product rule):

$\frac{d}{d x} [x \cos x \cdot \log x]$ $= \cos x \cdot \log x + x (- \sin x) \log x + x \cos x \cdot \frac{1}{x}$

$= \cos x \log x - x \sin x \log x + \cos x$

$= \cos x (\log x + 1) - x \sin x \log x$

So:

$\frac{d u}{d x} = x^{x \cos x} [\cos x (\log x + 1) - x \sin x \log x]$

Part 2: Differentiate $v = \frac{x^{2} + 1}{x^{2} - 1}$

Use quotient rule:

$v^{'} = \frac{(x^{2} - 1) (2 x) - (x^{2} + 1) (2 x)}{(x^{2} - 1)^{2}}$

$= \frac{2 x (x^{2} - 1 - x^{2} - 1)}{(x^{2} - 1)^{2}}$

$= \frac{2 x (- 2)}{(x^{2} - 1)^{2}}$

$= \frac{- 4 x}{(x^{2} - 1)^{2}}$

Final Derivative

$\frac{d y}{d x} = x^{x \cos x} [\cos x (\log x + 1) - x \sin x \log x] + \frac{- 4 x}{(x^{2} - 1)^{2}}$

Question 11

Differentiate w.r.t. $x$ :

$y = (x \cos x)^{x} + (x \sin x)^{1 / x}$

Solution

Part 1: Let

$u = (x \cos x)^{x}$

Taking log:

$\log u = x \log (x \cos x)$

Differentiate:

$\frac{1}{u} \frac{d u}{d x} = \log (x \cos x) + x \cdot \frac{1}{x \cos x} (\cos x - x \sin x)$

$= \log (x \cos x) + \frac{\cos x - x \sin x}{x \cos x}$

$= \log (x \cos x) + \frac{1}{x} - \tan x$

Thus:

$\frac{d u}{d x} = (x \cos x)^{x} [\log (x \cos x) + \frac{1}{x} - \tan x]$

Part 2: Let

$v = (x \sin x)^{1 / x}$

Taking log:

$\log v = \frac{1}{x} \log (x \sin x)$

Differentiate:

$\frac{1}{v} \frac{d v}{d x} = - \frac{1}{x^{2}} \log (x \sin x) + \frac{1}{x} \cdot \frac{\sin x + x \cos x}{x \sin x}$ $= - \frac{\log (x \sin x)}{x^{2}} + \frac{\sin x + x \cos x}{x^{2} \sin x}$

$= - \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}$

Thus:

$\frac{d v}{d x} = (x \sin x)^{1 / x} [- \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}]$

Answer

$\frac{d y}{d x} = (x \cos x)^{x} [\log (x \cos x) + \frac{1}{x} - \tan x] + (x \sin x)^{1 / x} [- \frac{\log (x \sin x)}{x^{2}} + \frac{1}{x^{2}} + \frac{\cos x}{x \sin x}]$

Question 12

Differentiate with respect to $x$ :

$x^{y} + y^{x} = 1$

This is an implicit function (y also depends on x), so we will use implicit differentiation + logarithmic differentiation.

Solution :

Differentiate both sides w.r.t.

$x$ :

Step 1: Differentiate $x^{y}$

Write:

$x^{y} = e^{y \log x}$

Differentiate:

$\frac{d}{d x} (x^{y}) = x^{y} \cdot \frac{d}{d x} (y \log x)$

$= x^{y} (\log x \cdot \frac{d y}{d x} + \frac{y}{x})$

Step 2: Differentiate $y^{x}$

Write:

$y^{x} = e^{x \log y}$

Differentiate:

$\frac{d}{d x} (y^{x}) = y^{x} \cdot \frac{d}{d x} (x \log y)$ $= y^{x} (\log y + x \cdot \frac{1}{y} \cdot \frac{d y}{d x})$

$= y^{x} (\log y + \frac{x}{y} \frac{d y}{d x})$

Step 3: Differentiate RHS

$\frac{d}{d x} (1) = 0$

Step 4: Combine all terms

$x^{y} (\log x \cdot \frac{d y}{d x} + \frac{y}{x}) + y^{x} (\log y + \frac{x}{y} \frac{d y}{d x}) = 0$

Now collect $\frac{d y}{d x}$ terms together:

$x^{y} \log x \cdot \frac{d y}{d x} + y^{x} \frac{x}{y} \frac{d y}{d x} = - x^{y} \frac{y}{x} - y^{x} \log y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x^{y} \log x + y^{x} \frac{x}{y}) = - (\frac{y x^{y}}{x} + y^{x} \log y)$

Final Answer

$\frac{d y}{d x} = \frac{- (\frac{y x^{y}}{x} + y^{x} \log y)}{x^{y} \log x + \frac{x}{y} y^{x}}$

Or more neatly:

$\frac{d y}{d x} = \frac{- (\frac{y}{x} x^{y} + y^{x} \log y)}{x^{y} \log x + \frac{x}{y} y^{x}}$

Question 13

Differentiate w.r.t. $x$ :

$y^{x} = x^{y}$

This is an implicit relation involving both $x$ and $y$ , so we will use logarithmic implicit differentiation.

Solution

Given:

$y^{x} = x^{y}$

Take natural log on both sides:

$\log (y^{x}) = \log (x^{y})$

Apply log rule: $\log (a^{b}) = b \log a$

$x \log y = y \log x$

Differentiate both sides w.r.t. $x$

Left side:

$\frac{d}{d x} (x \log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{d y}{d x}$

Right side:

$\frac{d}{d x} (y \log x) = \frac{d y}{d x} \log x + y \cdot \frac{1}{x}$

Arrange terms

$\log y + \frac{x}{y} \frac{d y}{d x} = \log x \cdot \frac{d y}{d x} + \frac{y}{x}$

Bring all $\frac{d y}{d x}$ terms to one side:

$\frac{x}{y} \frac{d y}{d x} - \log x \cdot \frac{d y}{d x} = \frac{y}{x} - \log y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (\frac{x}{y} - \log x) = \frac{y}{x} - \log y$

Final Answer

$\frac{d y}{d x} = \frac{\frac{y}{x} - \log y}{\frac{x}{y} - \log x}$

QUESTION 14

Differentiate with respect to $x$ :

$(\cos x)^{y} = (\cos y)^{x}$

Solution:

Take logarithm on both sides:

$\log ((\cos x)^{y}) = \log ((\cos y)^{x})$

Using the rule $\log (a^{b}) = b \log a$ :

$y \log (\cos x) = x \log (\cos y)$

Differentiate both sides w.r.t. $x$ :

Left side:

$\frac{d}{d x} [y \log (\cos x)] = \frac{d y}{d x} \log (\cos x) + y \cdot \frac{1}{\cos x} (- \sin x)$

$= \frac{d y}{d x} \log (\cos x) - y \tan x$

Right side:

$\frac{d}{d x} [x \log (\cos y)] = 1 \cdot \log (\cos y) + x \cdot \frac{1}{\cos y} (- \sin y) \frac{d y}{d x}$

$= \log (\cos y) - x \tan y \frac{d y}{d x}$

Rearranging terms:

$\frac{d y}{d x} \log (\cos x) - (- x \tan y \frac{d y}{d x}) = \log (\cos y) + y \tan x$

$\frac{d y}{d x} (\log (\cos x) + x \tan y) = \log (\cos y) + y \tan x$

$\frac{d y}{d x} = \frac{\log (\cos y) + y \tan x}{\log (\cos x) + x \tan y}$

QUESTION 15

Differentiate w.r.t. $x$ :

$x y = e^{x - y}$

This is an implicit function, so we differentiate both sides w.r.t. $x$ .

Solution :

Differentiate both sides:

Left side

$\frac{d}{d x} (x y) = x \frac{d y}{d x} + y$

(using product rule)

Right side

$\frac{d}{d x} (e^{x - y}) = e^{x - y} \cdot \frac{d}{d x} (x - y) = e^{x - y} (1 - \frac{d y}{d x})$

Form the equation

$x \frac{d y}{d x} + y = e^{x - y} (1 - \frac{d y}{d x})$

Expand RHS:

$x \frac{d y}{d x} + y = e^{x - y} - e^{x - y} \frac{d y}{d x}$

Bring $\frac{d y}{d x}$ terms together:

$x \frac{d y}{d x} + e^{x - y} \frac{d y}{d x} = e^{x - y} - y$

Factor out $\frac{d y}{d x}$ :

$\frac{d y}{d x} (x + e^{x - y}) = e^{x - y} - y$

$\frac{d y}{d x} = \frac{e^{x - y} - y}{x + e^{x - y}}$

Since from the given equation:

$x y = e^{x - y}$

Replace $e^{x - y}$ in the derivative:

$\frac{d y}{d x} = \frac{x y - y}{x + x y}$

Factor numerator and denominator:

$= \frac{y (x - 1)}{x (1 + y)}$

QUESTION 16

Find the derivative of the function

$f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})$

and hence find $f^{'} (1)$ .

SOLUTION

Take logarithm on both sides

$\log f (x) = \log (1 + x) + \log (1 + x^{2}) + \log (1 + x^{4}) + \log (1 + x^{8})$

Differentiate w.r.t. $x$ :

$\frac{f^{'} (x)}{f (x)} = \frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}$

Multiply both sides by $f (x)$ :

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$

Final derivative

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$

Now find $f^{'} (1)$

Substitute $x = 1$ :

First compute $f (1)$ :

$f (1) = (1 + 1) (1 + 1^{2}) (1 + 1^{4}) (1 + 1^{8}) = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

Now compute the bracket part:

$\frac{1}{1 + 1} + \frac{2 \cdot 1}{1 + 1} + \frac{4 \cdot 1^{3}}{1 + 1} + \frac{8 \cdot 1^{7}}{1 + 1}$ $= \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}$

$= \frac{1}{2} + 1 + 2 + 4 = \frac{1}{2} + 7 = \frac{15}{2}$

Therefore:

$f^{'} (1) = f (1) \cdot \frac{15}{2}$

$f^{'} (1) = 16 \cdot \frac{15}{2} = 8 \cdot 15 = 120$

$f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) [\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}]$ $f^{'} (1) = 120$

QUESTION 17

Differentiate the function:

$y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$ by:

(i) Product rule

(ii) Expanding the product

(iii) Logarithmic differentiation

and check that all answers are the same.

(i) DIFFERENTIATION BY PRODUCT RULE

Let:

$u = x^{2} - 5 x + 8, v = x^{3} + 7 x + 9$

$u^{'} = 2 x - 5, v^{'} = 3 x^{2} + 7$

Using product rule:

$y^{'} = u^{'} v + u v^{'}$

$y^{'} = (2 x - 5) (x^{3} + 7 x + 9) + (x^{2} - 5 x + 8) (3 x^{2} + 7)$

(ii) DIFFERENTIATION BY EXPANDING FIRST

Expand: $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$

Multiply:

$= x^{2} (x^{3} + 7 x + 9) - 5 x (x^{3} + 7 x + 9) + 8 (x^{3} + 7 x + 9)$

$= x^{5} + 7 x^{3} + 9 x^{2} - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72$

Combine like terms:

$y = x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72$

Differentiate term-wise:

$y^{'} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11$

(iii) BY LOGARITHMIC DIFFERENTIATION

$y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$

Take log:

$\log y = \log (x^{2} - 5 x + 8) + \log (x^{3} + 7 x + 9)$

Differentiate:

$\frac{y^{'}}{y} = \frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}$

Multiply by $y$ :

$y^{'} = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9) [\frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}]$

Simplify (cancel terms):

$y^{'} = (2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)$

QUESTION 18

If $u, v, w$ are functions of $x$ , prove that:

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

(i) By repeated application of the product rule

Let:

$y = u v w$

First consider:

$y = (u v) w$

Differentiate using product rule:

$\frac{d y}{d x} = \frac{d (u v)}{d x} \cdot w + (u v) \frac{d w}{d x}$

Now apply product rule again to $\frac{d (u v)}{d x}$ :

$\frac{d (u v)}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}$

Substitute back:

$\frac{d y}{d x} = (\frac{d u}{d x} v + u \frac{d v}{d x}) w + u v \frac{d w}{d x}$

Distribute $w$ :

$\frac{d y}{d x} = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

Result (Method 1 final statement)

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

(ii) By logarithmic differentiation

Given:

$y = u v w$

Take logarithm on both sides:

$\log y = \log u + \log v + \log w$

Differentiate:

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x}$

Multiply both sides by $y = u v w$ :

$\frac{d y}{d x} = u v w (\frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x})$

Distribute:

$\frac{d y}{d x} = v w \frac{d u}{d x} + u w \frac{d v}{d x} + u v \frac{d w}{d x}$

Final Answer

$\frac{d}{d x} (u v w) = \frac{d u}{d x} v w + u \frac{d v}{d x} w + u v \frac{d w}{d x}$

This verifies the required identity by both methods.

Conclusion

Yes, both methods give the same derivative result:

$\frac{d}{d x} (u v w) = u^{'} v w + u v^{'} w + u v w^{'}$

November 27, 2025
Exercise-5.4, Class 12th, Maths, Chapter 5, NCERT

Differentiate the following w.r.t. x:

Question 1

Differentiate:

$y = \frac{e^{x}}{\sin x}$ Solution

This is a quotient, so use the Quotient Rule:

$\frac{d}{d x} (\frac{u}{v}) = \frac{v u^{'} - u v^{'}}{v^{2}}$

Let:

$u = e^{x}, v = \sin x$

Then:

$u^{'} = e^{x}, v^{'} = \cos x$

Apply the rule:

$\frac{d y}{d x} = \frac{\sin x \cdot e^{x} - e^{x} \cdot \cos x}{(\sin x)^{2}}$

Factor out $e^{x}$ :

$\frac{d y}{d x} = \frac{e^{x} (\sin x - \cos x)}{\sin^{2} x}$ Final Answer

$\frac{d y}{d x} = \frac{e^{x} (\sin x - \cos x)}{\sin^{2} x}$

Question 2

Differentiate w.r.t. $x$ :

$y = e^{\sin^{- 1} x}$ Solution

Let:

$y = e^{\sin^{- 1} x}$

Use the chain rule:

$\frac{d y}{d x} = e^{\sin^{- 1} x} \cdot \frac{d}{d x} (\sin^{- 1} x)$

We know:

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

Therefore:

$\frac{d y}{d x} = e^{\sin^{- 1} x} \cdot \frac{1}{\sqrt{1 - x^{2}}}$ Final Answer

$\frac{d y}{d x} = \frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}}$

Question 3

Differentiate:

$y = e^{x^{3}}$ Solution

Use the chain rule:

Let:

$y = e^{u}, u = x^{3}$

Then:

$\frac{d y}{d u} = e^{u}, \frac{d u}{d x} = 3 x^{2}$

Apply chain rule:

$\frac{d y}{d x} = e^{u} \cdot 3 x^{2}$

Substitute $u = x^{3}$ :

$\frac{d y}{d x} = 3 x^{2} e^{x^{3}}$ Final Answer

$\frac{d y}{d x} = 3 x^{2} e^{x^{3}}$

Question 4

Differentiate with respect to $x$ :

$y = \sin (\tan^{- 1} (e^{- x}))$ Solution

Let:

$u = \tan^{- 1} (e^{- x})$

So:

$y = \sin (u)$

Step 1: Differentiate outer function

$\frac{d y}{d u} = \cos u$

Step 2: Differentiate inner function

$u = \tan^{- 1} (e^{- x})$

Derivative of $\tan^{- 1} t$ is $\frac{1}{1 + t^{2}} \cdot t^{'}$

So,

$\frac{d u}{d x} = \frac{1}{1 + (e^{- x})^{2}} \cdot \frac{d}{d x} (e^{- x})$

$\frac{d u}{d x} = \frac{1}{1 + e^{- 2 x}} \cdot (- e^{- x})$

Step 3: Apply chain rule

$\frac{d y}{d x} = \cos (u) \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = \cos (\tan^{- 1} (e^{- x})) \cdot (- \frac{e^{- x}}{1 + e^{- 2 x}})$

We know the identity:

$\cos (\tan^{- 1} t) = \frac{1}{\sqrt{1 + t^{2}}}$

So:

$\cos (\tan^{- 1} (e^{- x})) = \frac{1}{\sqrt{1 + e^{- 2 x}}}$

Final Simplified Answer

$\frac{d y}{d x} = - \frac{e^{- x}}{(1 + e^{- 2 x})^{3 / 2}}$ Final Answer

$\frac{d y}{d x} = - \frac{e^{- x}}{(1 + e^{- 2 x})^{3 / 2}}$

Question 5

Differentiate w.r.t. $x$ :

$y = \log (\cos (e^{x}))$ Solution

Use the chain rule multiple times.

Let:

$y = \log (\cos (e^{x}))$

Step 1: Differentiate outer logarithm

$\frac{d y}{d x} = \frac{1}{\cos (e^{x})} \cdot \frac{d}{d x} (\cos (e^{x}))$

Step 2: Differentiate $\cos (e^{x})$

$\frac{d}{d x} (\cos (e^{x})) = - \sin (e^{x}) \cdot \frac{d}{d x} (e^{x})$ $= - \sin (e^{x}) \cdot e^{x}$

Combine the results

$\frac{d y}{d x} = \frac{1}{\cos (e^{x})} \cdot (- e^{x} \sin (e^{x}))$ $\frac{d y}{d x} = - e^{x} \frac{\sin (e^{x})}{\cos (e^{x})}$ Final Answer

$\frac{d y}{d x} = - e^{x} \tan (e^{x})$

Question 6

Differentiate w.r.t. $x$ :

$y = e^{x} + e^{x^{2}} + e^{x^{3}} + e^{x^{4}} + e^{x^{5}}$

Solution

Differentiate term-by-term:

1. $\frac{d}{d x} (e^{x}) = e^{x}$

2. $\frac{d}{d x} (e^{x^{2}}) = e^{x^{2}} \cdot \frac{d}{d x} (x^{2}) = e^{x^{2}} \cdot 2 x$

3. $\frac{d}{d x} (e^{x^{3}}) = e^{x^{3}} \cdot \frac{d}{d x} (x^{3}) = e^{x^{3}} \cdot 3 x^{2}$

4. $\frac{d}{d x} (e^{x^{4}}) = e^{x^{4}} \cdot \frac{d}{d x} (x^{4}) = e^{x^{4}} \cdot 4 x^{3}$

5. $\frac{d}{d x} (e^{x^{5}}) = e^{x^{5}} \cdot \frac{d}{d x} (x^{5}) = e^{x^{5}} \cdot 5 x^{4}$

Final Answer

$\frac{d y}{d x} = e^{x} + 2 x e^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}}$

Question 7

Differentiate w.r.t. $x$ :

$y = \sqrt{e^{\sqrt{x}}}, x > 0$ Solution

Rewrite the function:

$y = {(e^{\sqrt{x}})}^{1 / 2} = e^{\frac{1}{2} \sqrt{x}}$

Let: $u = \frac{1}{2} \sqrt{x} = \frac{1}{2} x^{1 / 2}$

So:

$y = e^{u}$

Differentiate

$\frac{d y}{d u} = e^{u}$ $\frac{d u}{d x} = \frac{1}{2} \cdot \frac{1}{2} x^{- 1 / 2} = \frac{1}{4 \sqrt{x}}$

Apply chain rule:

$\frac{d y}{d x} = e^{u} \cdot \frac{d u}{d x}$

$\frac{d y}{d x} = e^{\frac{1}{2} \sqrt{x}} \cdot \frac{1}{4 \sqrt{x}}$ Final Answer

$\frac{d y}{d x} = \frac{e^{\frac{1}{2} \sqrt{x}}}{4 \sqrt{x}}, x > 0$

Question 8

Differentiate w.r.t. $x$ :

$y = \log (\log x), x > 1$ Solution

This is a composition of two logarithmic functions, so apply the chain rule.

Let:

$u = \log x$ $y = \log u$

Differentiate step-by-step

$\frac{d y}{d u} = \frac{1}{u}$ $\frac{d u}{d x} = \frac{1}{x}$

Apply chain rule

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $\frac{d y}{d x} = \frac{1}{\log x} \cdot \frac{1}{x}$ Final Answer

$\frac{d y}{d x} = \frac{1}{x \log x}, x > 1$

Question 9

Differentiate w.r.t. $x$ :

$y = \frac{\cos x}{\log x}, x > 0$ Solution

This is a quotient, so apply the Quotient Rule:

$\frac{d}{d x} (\frac{u}{v}) = \frac{v u^{'} - u v^{'}}{v^{2}}$ Let:

$u = \cos x, v = \log x$ Then:

$u^{'} = - \sin x$

$v^{'} = \frac{1}{x}$

Apply quotient rule

$\frac{d y}{d x} = \frac{(\log x) (- \sin x) - (\cos x) (\frac{1}{x})}{(\log x)^{2}}$ $\frac{d y}{d x} = \frac{- \sin x \log x - \frac{\cos x}{x}}{(\log x)^{2}}$ Final Answer

$\frac{d y}{d x} = \frac{- \sin x \log x - \frac{\cos x}{x}}{(\log x)^{2}}, x > 0$

Question 10

Differentiate w.r.t. $x$ :

$y = \cos (\log x + e^{x}), x > 0$ Solution

Use the chain rule.

Let:

$u = \log x + e^{x}$ So:

$y = \cos (u)$

Differentiate outer function

$\frac{d y}{d u} = - \sin (u)$

Differentiate inner function

$\frac{d u}{d x} = \frac{d}{d x} (\log x) + \frac{d}{d x} (e^{x}) = \frac{1}{x} + e^{x}$

Apply chain rule

$\frac{d y}{d x} = - \sin (u) (\frac{1}{x} + e^{x})$

Substitute back $u = \log x + e^{x}$ :

Final Answer

$\frac{d y}{d x} = - \sin (\log x + e^{x}) (\frac{1}{x} + e^{x}), x > 0$

November 25, 2025

Interval	Sign of (x-3)	Sign of (x+2)	Sign of $f^{'} (x)$	Nature of $f (x)$
$(- \infty, - 2)$	–	–	$(+) (+) = +$	Increasing
$(- 2, 3)$	–	+	$(-) (+) = -$	Decreasing
$(3, \infty)$	+	+	$(+) (+) = +$	Increasing

Interval	(x+1)	(x+2)	$f^{'} (x)$ sign	Nature
$(- \infty, - 2)$	–	–	–6(+)= –	Decreasing
$(- 2, - 1)$	–	+	–6(−)= +	Increasing
$(- 1, \infty)$	+	+	–6(+)= –	Decreasing

Interval	x	x–1	x–2	Sign of $y^{'}$	Nature
$(- \infty, 0)$	–	–	–	–	Decreasing
$(0, 1)$	+	–	–	+	Increasing
$(1, 2)$	+	+	–	–	Decreasing
$(2, \infty)$	+	+	+	+	Increasing

Interval	Sign of $(x - 1)$	Sign of $(x + 1)$	Sign of $f^{'} (x)$	Nature
$(- \infty, - 1)$	–	–	+	Increasing
$(1, \infty)$	+	+	+	Increasing

x	f(x)
1	85
2	75
3	89

x	f(x)
-3	125
-2	139
-1	129

Tag: Best NCERT Class 12th Solutions Maths

Key Idea

Let

Using AM ≥ GM:

Question 8

Find h

Conclusion

Question 7

Given:

Compare a and r

Conclusion

Question 6

Given: Volume

Base area

Area of 4 rectangular side walls

Question 5

Area of the triangle

Maximize A(θ)

Final Answer

Question 4

Solution

Step 1: Differentiate the function

Step 2: Determine where f′(x)>0 or f′(x)<0

Solve:

(i) Increasing intervals

(ii) Decreasing intervals

Summary

Question 3

Solution

Differentiate using Quotient Rule

Step 1: Find N′

Step 2: Find D′

Step 3: Substitute into quotient rule

Sign of f′(x)

Increasing

Decreasing

Question 2

Solution

Step 1: Find the height of the triangle

Step 2: Write the area formula

Step 3: Differentiate with respect to time t

Step 4: Substitute x=b

Miscellaneous Exercise on Chapter 6

Question 1

Solution

To find critical points, set f′(x)=0

Check whether it is maximum (Second Derivative Test)

Final Answer

Question 21

Solution

Question 22

For the Square

For the Circle

Total AreaA=As+Ac=x216+(28−x)24π

Final Calculated Values

Question 23

Using the figure

Volume of cone

Differentiate to find maxima

Substitute back

Volume of the largest cone

Volume of sphere

Required ratio

Question 24

Express l in terms of r and h

Using the volume constraint

Differentiate to find minima

Now find relation between h and r

Question 25

Volume of the cone

Question 26

Given: Total Surface Area is constant

Volume of cone

Question 27

Solution

D2=(x−0)2+(x22−5)2

D2=x2+(x22−5)2d(D2)dx=2x+2(x22−5)x=0

2x+x(x2−10)=0

x(x2−8)=0

y=x22=82=4

Find $h$

Compare $a$ and $r$

Maximize $A (θ)$

Step 2: Determine where $f^{'} (x) > 0$ or $f^{'} (x) < 0$

Step 1: Find $N^{'}$

Step 2: Find $D^{'}$

Sign of $f^{'} (x)$

Step 3: Differentiate with respect to time $t$

Step 4: Substitute $x = b$

To find critical points, set $f^{'} (x) = 0$

Total Area $A = A_{s} + A_{c} = \frac{x^{2}}{16} + \frac{(28 - x)^{2}}{4 π}$

Express $l$ in terms of $r$ and $h$

Now find relation between $h$ and $r$

$D^{2} = (x - 0)^{2} + {(\frac{x^{2}}{2} - 5)}^{2}$

$D^{2} = x^{2} + {(\frac{x^{2}}{2} - 5)}^{2}$ $\frac{d (D^{2})}{d x} = 2 x + 2 (\frac{x^{2}}{2} - 5) x = 0$

$2 x + x (x^{2} - 10) = 0$

$x (x^{2} - 8) = 0$

$y = \frac{x^{2}}{2} = \frac{8}{2} = 4$

Correct option: (D) $\frac{1}{3}$

Step 2: Put $f^{'} (x) = 0$ to find critical points

Step 3: Solve for $x$

Step 2: Set $P^{'} (x) = 0$

Step 1: Write $S$ in terms of $x$

Step 2: Differentiate w.r.t. $x$

Find $h$

(i) $f (x) = (2 x – 1)^{2} + 3$

(ii) $f (x) = 9 x^{2} + 12 x + 2$

(iii) $f (x) = – (x – 1)^{2} + 10$

(iv) $g (x) = x^{3} + 1$

(i) $f (x) = ∣ x + 2 ∣ - 1$

(ii) $g (x) = - ∣ x + 1 ∣ + 3$

(iii) $h (x) = \sin (2 x) + 5$