Tag: NCERT class 12th Chapter 5 Exercise – Miscellaneous Maths solutions

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:

$$\frac{d}{dx}(f(x{)}^{g(x)})=f(x{)}^{g(x)}[g^{\mathrm{\prime }}(x)\ln f(x)+g(x)\frac{f^{\mathrm{\prime }}(x)}{f(x)}]$$

We use logarithmic differentiation.

Solution

Term 1: $x^{x^2-3}$

Let $u=x$, $v=x^2-3$

$$\frac{d}{dx}\left(x^{x^2-3}\right)=x^{x^2-3}[(2x)\ln x+\frac{x^2-3}{x}]$$Term 2: $(x-3{)}^{x^2}$

Let $u=x-3$, $v=x^2$

$$\frac{d}{dx}((x-3{)}^{x^2})=(x-3{)}^{x^2}[(2x)\ln (x-3)+\frac{x^2}{x-3}]$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=x^{x^2-3}(2x\ln x+\frac{x^2-3}{x})+(x-3{)}^{x^2}(2x\ln (x-3)+\frac{x^2}{x-3})}}$$

---

Question 12

Find ${\displaystyle \frac{dy}{dx}}$, if$$y\; =12(1-\cos t),x=10(t-\sin t),-\frac{\pi }{2}<t<\frac{\pi }{2}$$

Solution:

For parametric equations:$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Step 1: Differentiate y w.r.t. t

$$y=12(1-\cos t)$$

$$\frac{dy}{dt}=12(0+\sin t)=12\sin t$$

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

$$x=10(t-\sin t)$$

$$\frac{dx}{dt}=10(1-\cos t)=10(1-\cos t)$$

Step 3: Substitute in formula

$$\frac{dy}{dx}=\frac{12\sin t}{10(1-\cos t)}$$

We have:$$\frac{dy}{dx}=\frac{6\sin t}{5(1-\cos t)}$$

Use trigonometric identities

$$1-\cos t=2{\sin }^2\frac{t}{2}$$
$$\sin t=2\sin \frac{t}{2}\cos \frac{t}{2}$$

Substitute these in:$$\frac{dy}{dx}=\frac{6(2\sin \frac{t}{2}\cos \frac{t}{2})}{5\cdot 2{\sin }^2\frac{t}{2}}$$

Cancel 2 from numerator and denominator:

$$\frac{dy}{dx}=\frac{6(\sin \frac{t}{2}\cos \frac{t}{2})}{5{\sin }^2\frac{t}{2}}$$

Simplify:

$$\frac{dy}{dx}=\frac{6}{5}\cdot \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{6}{5}\cot \frac{t}{2}}}$$

---

Question 13

Find ${\displaystyle \frac{dy}{dx}}$, if

$$y={\sin }^{-1}x+{\sin }^{-1}\sqrt{1-x^2},0<x<1$$

Solution

Differentiate term by term.

Term 1: ${\sin }^{-1}x$

$$\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$

Term 2: ${\sin }^{-1}\sqrt{1-x^2}$

Let $u=\sqrt{1-x^2}=(1-x^2{)}^{1\mathrm{/}2}$

$$\frac{du}{dx}=\frac{1}{2}(1-x^2{)}^{-1\mathrm{/}2}(-2x)=\frac{-x}{\sqrt{1-x^2}}$$

Now,

$$\frac{d}{dx}({\sin }^{-1}u)=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$$

But,

$$u^2=(1-x^2)\Rightarrow 1-u^2=1-(1-x^2)=x^2$$

So,

$$\frac{d}{dx}({\sin }^{-1}\sqrt{1-x^2})=\frac{1}{\sqrt{x^2}}\cdot \frac{-x}{\sqrt{1-x^2}}$$

For $0<x<1$, $\sqrt{x^2}=x$

$$\frac{d}{dx}({\sin }^{-1}\sqrt{1-x^2})=\frac{-x}{x\sqrt{1-x^2}}=\frac{-1}{\sqrt{1-x^2}}$$

Combine both derivatives

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=0}}$$

---

Question 14

If

$$x\sqrt{1+y}+y\sqrt{1+x}=0,-1<x<1$$

prove that$$\frac{dy}{dx}=-\frac{1}{(1+x{)}^2}$$

Solution

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

$$\frac{d}{dx}\left(x\sqrt{1+y}\right)+\frac{d}{dx}\left(y\sqrt{1+x}\right)=0$$

Use product rule for each term

First term:

$$\sqrt{1+y}+x\cdot \frac{1}{2\sqrt{1+y}}\cdot \frac{dy}{dx}$$

Second term:

$$\frac{dy}{dx}\sqrt{1+x}+y\cdot \frac{1}{2\sqrt{1+x}}$$

Put them together:

$$\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}\frac{dy}{dx}+\sqrt{1+x}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}=0$$

Now group $\frac{dy}{dx}$ terms:

$$\frac{dy}{dx}(\frac{x}{2\sqrt{1+y}}+\sqrt{1+x})=-\sqrt{1+y}-\frac{y}{2\sqrt{1+x}}$$

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:$$\frac{dy}{dx}=\frac{-\sqrt{1}\frac{2+x}{2(1+x)}}{\frac{2+x}{2(1+x)}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{1}{(1+x{)}^2}}}$$

---

Question 15

For the curve

$$(x-a{)}^2+(y-b{)}^2=c^2,c>0$$

prove that

$$\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3\mathrm{/}2}}{\frac{d^{2}y}{d{x}^2}}$$

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

$$2(x-a)+2(y-b)\frac{dy}{dx}=0$$

$$(x-a)+(y-b)\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x-a}{y-b}$$

Second derivative

Differentiate again w.r.t. $x$ using quotient rule:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=-\frac{(y-b)-(x-a)\frac{dy}{dx}}{(y-b{)}^2}$$

Substitute $\frac{dy}{dx}=-\frac{x-a}{y-b}$:

$$\frac{d^{2}y}{d{x}^2}=-\frac{(y-b)+\frac{(x-a{)}^2}{y-b}}{(y-b{)}^2}$$

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 $1+{\left(\frac{dy}{dx}\right)}^2$

$$1+{\left(\frac{dy}{dx}\right)}^2=1+{\left(\frac{x-a}{y-b}\right)}^2=\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^2}=\frac{c^2}{(y-b{)}^2}$$

Raise to power $3\mathrm{/}2$:

$${[1+{\left(\frac{dy}{dx}\right)}^2]}^{3\mathrm{/}2}={\left(\frac{c^2}{(y-b{)}^2}\right)}^{3\mathrm{/}2}=\frac{c^3}{\mathrm{\mid }y-b{\mathrm{\mid }}^3}$$

Now evaluate the required expression

$$\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3\mathrm{/}2}}{\frac{d^{2}y}{d{x}^2}}=\frac{\frac{c^3}{\mathrm{\mid }y-b{\mathrm{\mid }}^3}}{\frac{-c^2}{(y-b{)}^3}}=\frac{c^3}{c^2}\cdot -1=-c$$

Final Proven Result

$$\boxed{{\displaystyle \frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3\mathrm{/}2}}{\frac{d^{2}y}{d{x}^2}}=-c}}$$

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

$$\cos y=x\cos (a+y),\cos a\mathrm{\ne }\pm 1$$

prove that

$$\frac{dy}{dx}=\frac{{\cos }^2(a+y)}{\sin a}$$

Solution

Given:

$$\cos y=x\cos (a+y)$$Differentiate both sides with respect to $x$:

LHS

$$\frac{d}{dx}(\cos y)=-\sin y\cdot \frac{dy}{dx}$$

RHS

$$\frac{d}{dx}(x\cos (a+y))=1\cdot \cos (a+y)+x\cdot \frac{d}{dx}(\cos (a+y))$$

Now,

$$\frac{d}{dx}[\cos (a+y)]=-\sin (a+y)\cdot \frac{dy}{dx}$$

So RHS becomes:

$$\cos (a+y)-x\sin (a+y)\frac{dy}{dx}$$

Now equate derivatives

$$-\sin y\frac{dy}{dx}=\cos (a+y)-x\sin (a+y)\frac{dy}{dx}$$

Group $\frac{dy}{dx}$ terms:

$$-x\sin (a+y)\frac{dy}{dx}+\sin y\frac{dy}{dx}=\cos (a+y)$$

$$\frac{dy}{dx}(\sin y-x\sin (a+y))=\cos (a+y)$$

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

$$\frac{dy}{dx}=\frac{\cos (a+y)}{-\sin a\mathrm{/}\cos (a+y)}=\frac{{\cos }^2(a+y)}{\sin a}$$

---

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. $t$

$$x=a(\cos t+t\sin t)$$

$$\frac{dx}{dt}=a(-\sin t+\sin t+t\cos t)=at\cos t$$

$$y=a(\sin t-t\cos t)$$

$$\frac{dy}{dt}=a(\cos t-(\cos t-t\sin t))=at\sin t$$

Step 2: First derivative $\frac{dy}{dx}$

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{at\sin t}{at\cos t}=\tan t(t\mathrm{\ne }0)$$

Step 3: Second derivative $\frac{d^{2}y}{d{x}^2}$

Formula:

$$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

$$\frac{d}{dt}(\tan t)={\sec }^{2}t$$
$$\frac{dx}{dt}=at\cos t$$

So:$$\frac{d^{2}y}{d{x}^2}=\frac{{\sec }^{2}t}{at\cos t}$$

Final Answer

$$\boxed{{\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{{\sec }^{3}t}{at}}}$$or equivalently:

$$\boxed{{\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{1}{at{\cos }^{3}t}}}$$

since ${\sec }^{2}t=\frac{1}{{\cos }^{2}t}$

---

Question 18

If

$$f(x)=\mathrm{\mid }x{\mathrm{\mid }}^3$$

show that $f^{\mathrm{\prime }\mathrm{\prime }}(x)$ exists for all real $x$ and find it.

Solution

First, rewrite the function without modulus

$$\mathrm{\mid }x\mathrm{\mid }=\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}$$

So:$$\mathrm{\mid }x{\mathrm{\mid }}^3=\{\begin{array}{ll}{x}^3,& x\ge 0\\ (-x{)}^3=-x^3,& x<0\end{array}$$

Thus$$f(x)=\{\begin{array}{ll}{x}^3,& x\ge 0\\ -x^3,& x<0\end{array}$$

First derivative

$$f^{\mathrm{\prime }}(x)=\{\begin{array}{ll}3x^2,& x>0\\ -3x^2,& x<0\end{array}$$

Check at $x=0$:

$$f^{\mathrm{\prime }}(0)=\underset{x\to 0}{\lim }\frac{f(x)-f(0)}{x-0}=\underset{x\to 0}{\lim }\frac{\mathrm{\mid }x{\mathrm{\mid }}^3}{x}=\underset{x\to 0}{\lim }\mathrm{\mid }x{\mathrm{\mid }}^2=0$$So:$$f^{\mathrm{\prime }}(x)=\{\begin{array}{ll}3x^2,& x>0\\ 0,& x=0\\ -3x^2,& x<0\end{array}$$

Second derivative

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\{\begin{array}{ll}6x,& x>0\\ -6x,& x<0\end{array}$$

Check at $x=0$:

$$f^{\mathrm{\prime }\mathrm{\prime }}(0)=\underset{x\to 0}{\lim }\frac{f^{\mathrm{\prime }}(x)-f^{\mathrm{\prime }}(0)}{x-0}=\underset{x\to 0}{\lim }\frac{f^{\mathrm{\prime }}(x)}{\mathrm{x<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

Right-hand limit:

$$\underset{x\to 0^{+}}{\lim }\frac{3x^2}{x}=\lim 3x=0$$

Left-hand limit:$$\underset{x\to 0^{-}}{\lim }\frac{-3x^2}{x}=\lim -3x=0$$

Both limits exist and equal → 0.

So $f^{\mathrm{\prime }\mathrm{\prime }}(0)=0$

---

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

$$\frac{d}{dB}[\sin (A+B)]=\cos (A+B)$$

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:

$$\frac{d}{dB}[\sin A\cos B+\cos A\sin B]=\sin A(-\sin B)+\cos A(\cos B)$$
$$=\cos A\cos B-\sin A\sin B$$

Equate derivatives of both sides

$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$

Final Answer

$$\boxed{{\displaystyle \cos (A+B)=\cos A\cos B-\sin A\sin B}}$$

---

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

$$f(x)=\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }x-1\mathrm{\mid }$$

Check Continuity

• Both $\mathrm{\mid }x\mathrm{\mid }$ and $\mathrm{\mid }x-1\mathrm{\mid }$ are continuous everywhere.

• Sum of continuous functions is also continuous.

$$\Rightarrow f(x)\text{ is continuous for all real }x\mathrm{.}$$

Check Differentiability

A function involving $\mathrm{\mid }x\mathrm{\mid }$ is not differentiable where the inside part becomes zero.

For $\mathrm{\mid }x\mathrm{\mid }$:

Not differentiable at $x=0$

For $\mathrm{\mid }x-1\mathrm{\mid }$:

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

$$\boxed{{\displaystyle \text{Yes, such a function exists. One example is }f(x)=\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }x-1\mathrm{\mid }}}$$

---

Question 21

If$$y=\mid \begin{array}{ccc}f(x)& g(x)& h(x)\\ l& m& n\\ a& b& c\end{array}\mid$$

prove that

$$\frac{dy}{dx}=\mid \begin{array}{ccc}{f}^{\mathrm{\prime }}(x)& g^{\mathrm{\prime }}(x)& h^{\mathrm{\prime }}(x)\\ l& m& n\\ a& b& c\end{array}\mid$$

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:

$$y=f(x)\mid \begin{array}{cc}m& n\\ b& c\end{array}\mid -g(x)\mid \begin{array}{cc}l& n\\ a& c\end{array}\mid +h(x)\mid \begin{array}{cc}l& m\\ a& b\end{array}\mid$$

Let the minors be constants:

$$A=\mid \begin{array}{cc}m& n\\ b& c\end{array}\mid ,B=\mid \begin{array}{cc}l& n\\ a& c\end{array}\mid ,C=\mid \begin{array}{cc}l& m\\ a& b\end{array}\mid$$

So rewrite:

$$y=f(x)A-g(x)B+h(x)C$$

Differentiate both sides:

$$\frac{dy}{dx}=f^{\mathrm{\prime }}(x)A-g^{\mathrm{\prime }}(x)B+h^{\mathrm{\prime }}(x)C$$

Substitute determinants back:

$$\frac{dy}{dx}=f^{\mathrm{\prime }}(x)\mid \begin{array}{cc}m& n\\ b& c\end{array}\mid -g^{\mathrm{\prime }}(x)\mid \begin{array}{cc}l& n\\ a& c\end{array}\mid +h^{\mathrm{\prime }}(x)\mid \begin{array}{cc}l& m\\ a& b\end{array}\mid$$

This is exactly:

$$\frac{dy}{dx}=\mid \begin{array}{ccc}{f}^{\mathrm{\prime }}(x)& g^{\mathrm{\prime }}(x)& h^{\mathrm{\prime }}(x)\\ l& m& n\\ a& b& c\end{array}\mid$$

---

Question 22

If

$$y=e^{a{\cos }^{-1}x},-1\le x\le 1$$

show that

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$$

Solution

Let

$$y=e^{a{\cos }^{-1}x}$$

Let

$$u={\cos }^{-1}x$$

so that

$$y=e^{au}$$

Then:

$$\frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}$$

First derivative

$$\frac{dy}{dx}=e^{au}\cdot a\frac{du}{dx}=y\cdot a(-\frac{1}{\sqrt{1-x^2}})$$
$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{ay}{\sqrt{1-x^2}}}}$$

Second derivative

Differentiate again w.r.t $x$:

$$\frac{d^{2}y}{d{x}^2}=-\frac{a}{\sqrt{1-x^2}}\cdot \frac{dy}{dx}-ay\cdot \frac{d}{dx}(1-x^2{)}^{-1\mathrm{/}2}$$

Compute derivative of $(1-x^2{)}^{-1\mathrm{/}2}$:

$$\frac{d}{dx}(1-x^2{)}^{-1\mathrm{/}2}=-\frac{1}{2}(1-x^2{)}^{-3\mathrm{/}2}(-2x)=\frac{x}{(1-x^2{)}^{3\mathrm{/}2}}$$

So:

$$\frac{d^{2}y}{d{x}^2}=-\frac{a}{\sqrt{1-x^2}}\cdot \frac{dy}{dx}-ay\cdot \frac{x}{(1-x^2{)}^{3\mathrm{/}2}}$$

Substitute $\frac{dy}{dx}=-\frac{ay}{\sqrt{1-x^2}}$:

$$\frac{d^{2}y}{d{x}^2}=-\frac{a}{\sqrt{1-x^2}}(-\frac{ay}{\sqrt{1-x^2}})-\frac{axy}{(1-x^2{)}^{3\mathrm{/}2}}$$
$$=\frac{a^{2}y}{1-x^2}-\frac{axy}{(1-x^2{)}^{3\mathrm{/}2}}$$

Multiply both sides by $1-x^2$

$$(1-x^2)\frac{d^{2}y}{d{x}^2}=a^{2}y-\frac{axy}{\sqrt{1-x^2}}$$

Now substitute again

$$\frac{dy}{dx}=-\frac{ay}{\sqrt{1-x^2}}$$

So:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}=a^{2}y+x\frac{dy}{dx}$$

Rearrange:

$$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$$

---