Tag: Miscellaneous Exercise on Chapter 6 class 12th Maths NCERT Solutions

Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-16
Class 12th Class 12th Maths

16. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 m³/h. Then the depth of wheat is increasing at what rate?

Given:
- Radius $r = 10$ m
- Rate of volume increase:
  
  $\frac{d V}{d t} = 314 m^{3} / h$
- Let depth (height) = $h$
Volume of a cylinder:

$V = π r^{2} h$

Differentiate with respect to time $t$ :

$\frac{d V}{d t} = π r^{2} \frac{d h}{d t}$

Substitute values:

$314 = π (10)^{2} \frac{d h}{d t}$

$314 = 100 π \frac{d h}{d t}$ $\frac{d h}{d t} = \frac{314}{100 π}$

Since $π \approx 3.14$ :

$\frac{d h}{d t} = \frac{314}{314} = 1 m/h$

Answer:

$1 m/h$

Correct option: (A) 1 m/h
December 3, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-15
Class 12th Class 12th Maths

Question 15.

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi-vertical angle $α$ is one-third that of the cone, and the greatest volume is $\frac{4}{27} π h^{3} \tan^{2} α$ .

Solution

Step 1: Geometry of the cone and the inscribed cylinder

Let a cylinder of
- height = $x$
- radius = $r$
be inscribed in a right circular cone of
- height = $h$
- semi-vertical angle = $α$
From the diagram:

The radius at height $x$ from the vertex is proportional to height:

$\frac{r}{h - x} = \tan α$

Thus,

$r = (h - x) \tan α (1)$

Step 2: Volume of the cylinder

$V = π r^{2} x$

Substitute (1):

$V = π {[(h - x) \tan α]}^{2} x$ $V = π x (h - x)^{2} \tan^{2} α$

Let:

$k = π \tan^{2} α$

Then,

$V = k x (h - x)^{2}$

Step 3: Differentiate to find maximum

$V = k (x h^{2} - 2 h x^{2} + x^{3})$ $\frac{d V}{d x} = k (h^{2} - 4 h x + 3 x^{2})$

Set $\frac{d V}{d x} = 0$ :

$h^{2} - 4 h x + 3 x^{2} = 0$

Solve quadratic:

$3 x^{2} - 4 h x + h^{2} = 0$ $x = \frac{4 h \pm \sqrt{16 h^{2} - 12 h^{2}}}{6}$ $x = \frac{4 h \pm 2 h}{6}$

Possible values:

$x = h or x = \frac{h}{3}$

The cylinder cannot have height = $h$ (radius would be 0).

Hence the valid maximum is:

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

Thus the height of the cylinder of greatest volume is one-third that of the cone.

Step 4: Maximum radius

Use (1):

$r = (h - x) \tan α = (h - \frac{h}{3}) \tan α$ $r = \frac{2 h}{3} \tan α$

Step 5: Maximum Volume

$V_{\max} = π r^{2} x$ $V_{\max} = π {(\frac{2 h}{3} \tan α)}^{2} (\frac{h}{3})$ $V_{\max} = π (\frac{4 h^{2}}{9} \tan^{2} α) (\frac{h}{3})$ $V_{\max} = \frac{4}{27} π h^{3} \tan^{2} α$ $V_{\max} = \frac{4}{27} π h^{3} \tan^{2} α$

Answers:-

Height of cylinder of greatest volume:

$\frac{h}{3}$

Greatest possible volume:

$\frac{4}{27} π h^{3} \tan^{2} α$
December 3, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-14
Class 12th Class 12th Maths

Question 14.

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$ . Also find the maximum volume.

Solution

Step 1: Understand the geometry

A cylinder is inscribed in a sphere of radius $R$ .
Let:
- $h$ = height of the cylinder
- $r$ = radius of the base of the cylinder
From the cross-section, the half-height $h / 2$ and cylinder radius $r$ form a right triangle inside the sphere:

$r^{2} + {(\frac{h}{2})}^{2} = R^{2} (1)$

Step 2: Write the volume of the cylinder

$V = π r^{2} h$

Using (1):

$r^{2} = R^{2} - \frac{h^{2}}{4}$

So,

$V (h) = π (R^{2} - \frac{h^{2}}{4}) h$ $V (h) = π (R^{2} h - \frac{h^{3}}{4})$

Step 3: Differentiate to find maximum

$\frac{d V}{d h} = π (R^{2} - \frac{3 h^{2}}{4})$

For maximum volume:

$\frac{d V}{d h} = 0$ $R^{2} - \frac{3 h^{2}}{4} = 0$ $\frac{3 h^{2}}{4} = R^{2}$ $h^{2} = \frac{4 R^{2}}{3}$ $h = \frac{2 R}{\sqrt{3}}$

This is the required height.

Step 4: Find the corresponding radius

Using equation (1):

$r^{2} = R^{2} - \frac{h^{2}}{4}$

Substitute $h^{2} = \frac{4 R^{2}}{3}$ :

$r^{2} = R^{2} - \frac{1}{4} (\frac{4 R^{2}}{3})$ $r^{2} = R^{2} - \frac{R^{2}}{3}$ $r^{2} = \frac{2 R^{2}}{3}$

Step 5: Maximum Volume

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

Answers

Height of cylinder of maximum volume:

$\frac{2 R}{\sqrt{3}}$

Maximum volume:

$V_{\max} = \frac{4 π R^{3}}{3 \sqrt{3}}$
December 3, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-13

Class 12th Class 12th Maths

Question 13.
Let $f$ be a function defined on $[a, b]$ such that $f^{'} (x) > 0$ for all $x \in (a, b)$ .
Prove that $f$ is an increasing function on $(a, b)$ .

Proof

Given:

$f^{'} (x) > 0 for all x \in (a, b)$

We need to prove:

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

which means $f$ is increasing on $(a, b)$ .

Using the Mean Value Theorem (MVT)

Since $f$ is differentiable on $(a, b)$ and continuous on $[a, b]$ , by Mean Value Theorem, for any two points $x_{1}, x_{2} \in (a, b)$ with $x_{1} < x_{2}$ , there exists some $c$ in $(x_{1}, x_{2})$ such that:

$f^{'} (c) = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}}$

Given $f^{'} (x) > 0$ for all $x$ , we have:

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

Therefore:

$\frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}} > 0$

Since $x_{2} - x_{1} > 0$ , multiplying both sides gives:

$f (x_{2}) - f (x_{1}) > 0$ $\Rightarrow f (x_{2}) > f (x_{1})$

So, whenever $x_{1} < x_{2}$ , $f (x_{1}) < f (x_{2})$ , meaning:

$f is an increasing function on (a, b)$

Conclusion

Since $f^{'} (x) > 0$ for every point in $(a, b)$ , the slope of the tangent line to the graph of $f$ is always positive, so the function always rises as x increases. Hence, $f$ is increasing on $(a, b)$ .

December 2, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-12
Question 12.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$ .

Solution

Let a right circular cone be inscribed in a sphere of radius $r$ .
Place the sphere such that its centre is the origin $O (0, 0)$ .
Let the cone have vertex at the top of the sphere and base inside the sphere.

Let:
- $h$ = altitude (height) of the cone
- $x$ = radius of the base of the cone
Since the cone is inside the sphere, its height extends from top of the sphere to some point on the vertical axis.

Consider the cross-section through the axis (a vertical diameter).
Let the vertex of the cone be at the top point of the sphere and the base centre lies somewhere inside.

So the height $h$ of the cone is measured downward from top.

The radius $x$ of the base relates to sphere radius using Pythagoras:

$x^{2} + (r - h)^{2} = r^{2}$ $x^{2} = r^{2} - (r - h)^{2}$ $x^{2} = r^{2} - (r^{2} - 2 r h + h^{2})$ $x^{2} = 2 r h - h^{2}$

Volume of the cone

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

Differentiate and find maximum

$V = \frac{π}{3} (2 r h^{2} - h^{3})$ $V^{'} = \frac{π}{3} (4 r h - 3 h^{2})$

Set $V^{'} = 0$ :

$4 r h - 3 h^{2} = 0$ $h (4 r - 3 h) = 0$

So:

$h = 0 or 4 r - 3 h = 0$

Reject $h = 0$ (no cone)

$4 r = 3 h$ $h = \frac{4 r}{3}$

Second derivative test

$V^{''} = \frac{π}{3} (4 r - 6 h)$

$V^{''} (\frac{4 r}{3}) = \frac{π}{3} (4 r - 6 \cdot \frac{4 r}{3}) = \frac{π}{3} (4 r - 8 r) = \frac{π}{3} (- 4 r) < 0$

So $V$ is maximum at $h = \frac{4 r}{3}$ .

Answer

$The altitude of the right circular cone of maximum volume inscribed in a sphere of radius r is \frac{4 r}{3} .$
December 2, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-11
Class 12th Class 12th Maths

Question 11.
Find the absolute maximum and minimum values of the function

$f (x) = \cos^{2} x + \sin x, x \in [0, π]$

Solution

Given:

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

Step 1: First derivative

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

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

$\cos x = 0 or 1 - 2 \sin x = 0$
1. $\cos x = 0 \Rightarrow x = \frac{π}{2}$
2. $1 - 2 \sin x = 0 \Rightarrow \sin x = \frac{1}{2}$
$x = \frac{π}{6}, \frac{5 π}{6}$

So critical points:

$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}$

Also evaluate function at end points:

$x = 0, x = π$

Step 2: Evaluate $f (x)$ at critical and end points

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

$0$ $1 + 0 = 1$

$π$ $1 + 0 = 1$

$\frac{π}{6}$
${(\frac{\sqrt{3}}{2})}^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} = 1.25$

$\frac{π}{2}$ $0 + 1 = 1$

$\frac{5 π}{6}$
${(- \frac{\sqrt{3}}{2})}^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} = 1.25$

Final Answer

Absolute Maximum value

$\frac{5}{4} at x = \frac{π}{6}, \frac{5 π}{6}$

Absolute Minimum value

$1 at x = 0, \frac{π}{2}, π$

Extras :-

$Graph of f (x) = \cos^{2} x + \sin x on [0, π]$

The graph now properly displays the exact function heading centered at the top.
The curve and marked points show clearly:
- Absolute Maximum Value
  
  $f (\frac{π}{6}) = f (\frac{5 π}{6}) = \frac{5}{4}$
- Absolute Minimum Value
  
  $f (0) = f (\frac{π}{2}) = f (π) = 1$
December 2, 2025
Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-10
Class 12th Class 12th Maths

Question 10

Find the points at which the function

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

has
(i) local maxima
(ii) local minima
(iii) point of inflexion

Solution Using First Derivative Test

Step 1: First derivative

$f (x) = (x - 2)^{4} (x + 1)^{3}$ $f^{'} (x) = (x - 2)^{3} (x + 1)^{2} (7 x - 2)$

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

$(x - 2)^{3} (x + 1)^{2} (7 x - 2) = 0$

So critical points:

$x = - 1, x = \frac{2}{7}, x = 2$

Step 2: Sign change of $f^{'} (x)$

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

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

$(- 1, 2 / 7)$ $+$ Increasing

$(2 / 7, 2)$ $-$ Decreasing

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

Conclusion

(i) Local maximum

At $x = \frac{2}{7}$ because $f^{'} (x)$ changes from $+$ to $-$

$x = \frac{2}{7}$

(ii) Local minimum

At $x = 2$ because $f^{'} (x)$ changes from $-$ to $+$

$x = 2$

(iii) No maximum/minimum at

$x = - 1 (sign does not change) \Rightarrow point of inflexion$

Point of Inflection using Second Derivative Test

$f^{''} (x) = 6 (x - 2)^{2} (x + 1) (7 x^{2} - 4 x - 2)$

Set $f^{''} (x) = 0$ gives possible inflection points:

$x = - 1, x = 2, x = \frac{2 \pm 3 \sqrt{2}}{7}$

But from sign change check:
- $x = - 1$ is a point of inflexion
- $x = 2$ is not, since $f^{'}$ changes sign there ⇒ local minimum
- The two irrational values also give inflection, but NCERT normally expects only the real-significant graphical one at $x = - 1$
Answers

$Local maximum at x = \frac{2}{7}$ $Local minimum at x = 2$ $Point of inflexion at x = - 1$
December 2, 2025
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-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

$x$	$f (x) = \cos^{2} x + \sin x$
$0$	$1 + 0 = 1$
$π$	$1 + 0 = 1$
$\frac{π}{6}$	${(\frac{\sqrt{3}}{2})}^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} = 1.25$
$\frac{π}{2}$	$0 + 1 = 1$
$\frac{5 π}{6}$	${(- \frac{\sqrt{3}}{2})}^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} = 1.25$

Interval	Sign of $f^{'} (x)$	Nature
$(- \infty, - 1)$	$+$	Increasing
$(- 1, 2 / 7)$	$+$	Increasing
$(2 / 7, 2)$	$-$	Decreasing
$(2, \infty)$	$+$	Increasing

Tag: Miscellaneous Exercise on Chapter 6 class 12th Maths NCERT Solutions

Solution

Step 1: Geometry of the cone and the inscribed cylinder

Step 2: Volume of the cylinder

Step 3: Differentiate to find maximum

Step 4: Maximum radius

Step 5: Maximum Volume

Height of cylinder of greatest volume:

Greatest possible volume:

Solution

Step 1: Understand the geometry

Height of cylinder of maximum volume:

Maximum volume:

Proof

Using the Mean Value Theorem (MVT)

Conclusion

Solution

Volume of the cone

Differentiate and find maximum

Second derivative test

Answer

Solution

Step 1: First derivative

Step 2: Evaluate f(x) at critical and end points

Final Answer

Absolute Maximum value

Absolute Minimum value

Extras :-

Question 10

Solution Using First Derivative Test

Step 1: First derivative

Step 2: Sign change of f′(x)

Conclusion

(i) Local maximum

(ii) Local minimum

(iii) No maximum/minimum at

Point of Inflection using Second Derivative Test

Answers

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 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

Step 2: Evaluate $f (x)$ at critical and end points

Step 2: Sign change of $f^{'} (x)$

Find $h$

Compare $a$ and $r$

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

Step 1: Find $N^{'}$

Step 2: Find $D^{'}$

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