NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability - Exercise 5.7

Exercise 5.7 

Find the second order derivatives of the functions given in Exercises 1 to 10. 

1. $\text{y=}{{\text{x}}^{\text{2}}}\text{+3x+2}$.

Ans:  

The given function is $\text{y=}{{\text{x}}^{\text{2}}}\text{+3x+2}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\frac{\text{d}}{\text{dx}}\text{(3x)+}\frac{\text{d}}{\text{dx}}\text{(2)=2x+3+0=2x+3}$

That is,

$\frac{\text{dy}}{\text{dx}}=2\text{x+3}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(2x+3)=}\frac{\text{d}}{\text{dx}}\text{(2x)+}\frac{\text{d}}{\text{dx}}\text{(3)=2+0=2}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2}$.

2. $\mathbf{y=}{{\mathbf{x}}^{\mathbf{20}}}$.

Ans: 

The given function is $\text{y=}{{\text{x}}^{\text{20}}}$.

Then, differentiating both sides with respect to $\text{x}$ gives

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{20}}}\text{)=20}{{\text{x}}^{\text{19}}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(20}{{\text{x}}^{\text{19}}}\text{)=20}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{19}}}\text{)=20}\left( 19 \right){{\text{x}}^{\text{18}}}\text{=380}{{\text{x}}^{\text{18}}}\].

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=380}{{\text{x}}^{\text{18}}}\].

3. $\mathbf{y=x}\cdot \mathbf{cosx}$.

Ans: 

The given function is $\text{y=x}\text{.cosx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(x}\text{.cosx)=cosx}\text{.}\frac{\text{d}}{\text{dx}}\text{(x)+x}\frac{\text{d}}{\text{dx}}\text{(cosx)=cosx}\text{.1+x(-sinx)=cosx-xsinx}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=cosx-xsinx}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(cosx-xsinx)=}\frac{\text{d}}{\text{dx}}\text{(cosx)-}\frac{\text{d}}{\text{dx}}\text{(xsinx)} \\ & \text{=-sinx- }\!\![\!\!\text{ sinx}\cdot \frac{\text{d}}{\text{dx}}\text{(x)+x}\cdot \frac{\text{d}}{\text{dx}}\text{(sinx) }\!\!]\!\!\text{ } \\ & \text{=-sinx-(sinx+xcosx)} \\ \end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-(xcosx+2sinx)}$.

4. $\mathbf{y=logx}$.

Ans: 

The given function is $\text{y=logx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(logx)=}\frac{\text{1}}{\text{x}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{1}}{\text{x}} \right)\text{=}\frac{\text{-1}}{{{\text{x}}^{\text{2}}}}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-}\frac{\text{1}}{{{\text{x}}^{\text{2}}}}\].

5. $\text{y=}{{\text{x}}^{\text{3}}}\text{logx}$.

Ans: 

The given function is $\text{y=}{{\text{x}}^{\text{3}}}\text{logx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{x}}^{\text{3}}}\text{logx} \right]\text{=logx}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+}{{\text{x}}^{\text{3}}}\frac{\text{d}}{\text{dx}}\text{(logx)=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{3}}}\text{.}\frac{\text{1}}{\text{x}}\text{=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{2}}}\text{(1+3logx)}$. 

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{(1+3logx))} \\ & \text{=(1+3logx)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\frac{\text{d}}{\text{dx}}\text{(1+3logx)} \\ & \text{=(1+3logx)}\text{.2x+}{{\text{x}}^{\text{3}}}\text{.}\frac{\text{3}}{\text{x}} \\ & \text{=2x+6logx+3x} \\ & \text{=5x+6xlogx} \end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=x(5+6logx)}$.

6. $\mathbf{y=}{{\mathbf{e}}^{\mathbf{x}}}\mathbf{sin5x}$

Ans: 

The given function is $\text{y=}{{\text{e}}^{\text{x}}}\text{sin5x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align} & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{sin5x} \right]\text{=sinx}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\frac{\text{d}}{\text{dx}}\text{(sin5x)} \\ & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=sin5x}\text{.}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{x}}}\text{.cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)} \\ \end{align}\]

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)}$.

Again, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)} \right] \\ & \text{=(sin5x+5cos5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(sin5x+5cos5x)} \\ & \text{=(sin5x+5cos5x)(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\left[ \text{cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)+5(-sin5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)} \right] \\ & \text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)+}{{\text{e}}^{\text{x}}}\text{(5cos5x-25sin5x)} \\ \end{align}\]

 $\text{=}{{\text{e}}^{\text{x}}}\text{(10cos5x-24sin5x)}$.

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2}{{\text{e}}^{\text{x}}}\text{(5cox5x-12sin5x)}$.

7. $\text{y=}{{\text{e}}^{\text{6x}}}\text{cos3x}$.

Ans:  

The given function is $\text{y=}{{\text{e}}^{\text{6x}}}\text{cos3x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)=cos3x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(cos3x)} \\ & \Rightarrow \frac{\text{dy}}{\text{dx}}=\text{cos3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(6x)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ (-sin3x) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(3x)} \\ \end{align}$

Therefore,

$\frac{\text{dy}}{\text{dx}}\text{=6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x}$ …… (1)

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x)=6 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)-3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{sin3x)}$

$\text{=6 }\!\!\times\!\!\text{ }\left[ \text{6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x} \right]\text{-3 }\!\!\times\!\!\text{ }\left[ \text{sin3x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(sin3x)} \right]$ [using (1)]

\[\begin{align} & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-3}\left[ \text{sin3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ 6+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ cos3x-3} \right] \\ & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-9}{{\text{e}}^{\text{6x}}}\text{cos3x} \\ \end{align}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=9}{{\text{e}}^{\text{6x}}}\text{(3cos3x-4sin3x)}\].

8. $\mathbf{y=ta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}$.

Ans: 

The given function is $\text{y=ta}{{\text{n}}^{\text{-1}}}\text{x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{1}}{\text{1-}{{\text{x}}^{\text{2}}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\frac{\text{d}}{\text{dx}}{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{-1}}}\text{=(-1) }\!\!\times\!\!\text{ (1+}{{\text{x}}^{\text{2}}}{{\text{)}}^{\mathbf{-2}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(1+}{{\text{x}}^{\text{2}}}\text{)} \\ & \text{=-}\frac{\text{1}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}\text{ }\!\!\times\!\!\text{ 2x} \\ \end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-2x}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}$.

9. $\text{y=log(logx)}$.

Ans: 

The given function is $\text{y=log(logx)}$.

Now, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(logx) }\!\!]\!\!\text{ } \\ & =\frac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \\ & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=(xlogx}{{\text{)}}^{\text{-1}}} \\ \end{align}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{(xlogx)}}^{\text{-1}}} \right]\text{=(-1) }\!\!\times\!\!\text{ (xlogx}{{\text{)}}^{\text{-2}}}\frac{\text{d}}{\text{dx}}\text{(xlogx)} \\ & =\frac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ & =\frac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{x}} \right] \\ \end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-(1+logx)}}{{{\text{(xlogx)}}^{\text{2}}}}$.

10. $\text{y=sin(logx)}$.

Ans: 

 The given function is $\text{y=sin(logx)}$.

Now, differentiating both sides with respect to $\text{x}$ gives

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ =cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)}=\frac{\text{cos(logx)}}{\text{x}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ \frac{\text{cos(logx)}}{\text{x}} \right] \\ & =\frac{\text{x}\left[ \text{cos(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(x)}}{{{\text{x}}^{\text{2}}}} \\ & =\frac{\text{x}\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ 1}}{{{\text{x}}^{\text{2}}}} \\ & =\frac{\text{-xsin(logx) }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{x}}\text{-cos(logx)}}{{{\text{x}}^{\text{2}}}} \\ \end{align}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\left[ \text{-sin(logx)-cos(logx)} \right]}{{{\text{x}}^{\text{2}}}}\].

11. If $\text{y=5cosx-3sinx}$, prove that $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0}$. 

Ans: 

The given equation is $\text{y=5cosx-3sinx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(5cosx)-}\frac{\text{d}}{\text{dx}}\text{(3sinx)=5}\frac{\text{d}}{\text{dx}}\text{(cosx)-3}\frac{\text{d}}{\text{dx}}\text{(sinx)=5(-sinx)-3cosx}$

Therefore, $\frac{\text{dy}}{\text{dx}}\text{=-(5sinx+3cosx)}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ -(5sinx+3cosx) }\!\!]\!\!\text{ }$

$\begin{align} & \text{=-}\left[ \text{5 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(sinx)+3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(cosx)} \right] \\ & \text{= }\!\![\!\!\text{ 5cosx+3(-sinx) }\!\!]\!\!\text{ } \\ & \text{=-y} \\ \end{align}$

That is, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-y}$.

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0}$.

12. If $\mathbf{y=co}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{x}$, Find $\frac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}$ in the terms of $\mathbf{y}$ alone.

Ans: 

The given function is $\text{y=co}{{\text{s}}^{\text{-1}}}\text{x}$.

Now, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{-1}}}\text{x)=}\frac{\text{-1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-1}}{\text{2}}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ \text{-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-1}}{\text{2}}}} \right] \\ & =\left( \frac{\text{-1}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ (1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-3}}{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)} \\ & =\frac{\text{1}}{\sqrt[\text{2}]{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{ (-2x)} \\ \end{align}$

$\Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-x}}{\sqrt{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}}$                                                                       …… (1)

Now, $\text{y=co}{{\text{s}}^{\text{-1}}}\text{x}\Rightarrow \text{x=cosy}$.

Therefore, substituting $\text{x=cosy}$ into equation (1), gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{x}}{\text{d}{{\text{y}}^{\text{2}}}}\text{=}\frac{\text{-cosy}}{\sqrt{{{\text{(1-co}{{\text{s}}^{\text{2}}}\text{y)}}^{\text{3}}}}} \\ & =\frac{\text{-cosy}}{\text{si}{{\text{n}}^{\text{3}}}\text{y}} \\ & =\frac{\text{-cosy}}{\text{siny}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{si}{{\text{n}}^{\text{2}}}\text{y}} \\ \end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{= -coty }\!\!\times\!\!\text{ cose}{{\text{c}}^{\text{2}}}\text{y}$.

13. If $\mathbf{y=3cos(logx)+4sin(logx)}$, show that ${{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{+x}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{+y=0}$.

Ans: 

The given equations are $\text{y=3cos(logx)+4sin(logx)}$                               …… (1)

and ${{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y=0}$                                                                                …… (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

\[\begin{align} & {{\text{y}}_{\text{1}}}\text{=3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ cos(logx) }\!\!]\!\!\text{ +4 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ } \\ & \text{=3 }\!\!\times\!\!\text{ }\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+4 }\!\!\times\!\!\text{ }\left[ \text{cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ \end{align}\]

\[{{\text{y}}_{\text{1}}}\text{=}\frac{\text{-3sin(logx)}}{\text{x}}\text{+}\frac{\text{4cos(logx)}}{\text{x}}\text{=}\frac{\text{4cos(logx)-3sin(logx)}}{\text{x}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[{{\text{y}}_{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{4cos(logx)-3sin(logx)}}{\text{x}} \right)\]

\[\begin{align} & \text{=}\frac{\text{x}\cdot \frac{d}{dx}\left[ \text{4 }\!\!\{\!\!\text{ cos(logx) }\!\!\}\!\!\text{ - }\!\!\{\!\!\text{ 3sin(logx) }\!\!\}\!\!\text{ } \right]\text{- }\!\!\{\!\!\text{ 4cos(logx)-3sin(logx) }\!\!\}\!\!\text{ }\times \text{1}}{{{\text{x}}^{\text{2}}}} \\ & \text{=}\frac{\text{x}\left[ \text{-4sin(logx)}\frac{d}{dx}\text{(logx)-3cos(logx)}\frac{d}{dx}\text{(logx)} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} \\ \end{align}\]

$=\frac{\text{x}\left[ \text{-4sin(logx)}\cdot \frac{\text{1}}{\text{x}}\text{-3cos(logx)}\cdot \frac{\text{1}}{\text{x}} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}}$

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

Therefore, ${{\text{y}}_{\text{2}}}\text{=}\frac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}}$.

Now, substituting the derivatives ${{\text{y}}_{1}}$ ,${{\text{y}}_{2}}$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{align} & {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y} \\ & \text{=}{{\text{x}}^{\text{2}}}\left( \frac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+x}\left( \frac{\text{4cos(logx)-3sin(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+3cos(logx)+4sin(logx)} \\ & \text{=-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+4sin(logx)} \\ & \text{=0} \\ \end{align}$

Hence, it has been proved that ${{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y}=0$.

14. If $\mathbf{y=A}{{\mathbf{e}}^{\mathbf{mx}}}\mathbf{+B}{{\mathbf{e}}^{\mathbf{nx}}}$, show that $\frac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{-(m+n)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+mny=0}$.

Ans: 

The given equations are $\text{y=A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}}$                                     …… (1)

and \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}\]                                                      ……. (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=A}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+B}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)=A}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+B}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)=Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)=Am}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+Bn}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{nx}}}\text{)} \\ & \text{=Am}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+Bn}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)} \\ \end{align}$

Therefore, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=A}{{\text{m}}^{\text{2}}}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}$.

Thus, substituting the derivatives ${{\text{y}}_{1}}$ ,${{\text{y}}_{2}}$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny} \\ & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-(m+n)}\text{.(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)+mn(A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}}\text{)} \\ & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-Ame}{{\text{x}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}} \\ & \text{=0} \\ \end{align}$

Thus, it has been proved that \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}\].

15. If $\mathbf{y=500}{{\mathbf{e}}^{\mathbf{7x}}}\mathbf{+600}{{\mathbf{e}}^{\mathbf{-7x}}}$, show that $\frac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{=49y}$.

Ans: 

The given equation is $\text{y=500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}}$.                                     …… (1)

Then, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{\text{dy}}{\text{dx}}\text{=500 }\!\!\times\!\!\text{ (}{{\text{e}}^{\text{7x}}}\text{)+600 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(7x)+600 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=3500}{{\text{e}}^{\text{7x}}}\text{-4200}{{\text{e}}^{\text{-7x}}} \\ \end{align}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=3500 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{7x}}}\text{)-4200 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{-7x}}}\text{)} \\ & \text{=3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(7x)-4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=7 }\!\!\times\!\!\text{ 3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{+7 }\!\!\times\!\!\text{ 4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}} \\ & \text{=49 }\!\!\times\!\!\text{ 500}{{\text{e}}^{\text{7x}}}\text{+49 }\!\!\times\!\!\text{ 600}{{\text{e}}^{\text{-7x}}} \\ & \text{=49(500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}}\text{)} \\ \end{align}$

$\text{=49y}$, using the equation (1).

Thus, it has been proved that $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=49y}$.

16. If ${{\mathbf{e}}^{\mathbf{y}}}\mathbf{(x+1)=1}$, show that $\frac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{=}{{\left( \frac{\mathbf{dy}}{\mathbf{dx}} \right)}^{\mathbf{2}}}$.

Ans: 

The given equation is ${{\text{e}}^{\text{y}}}\text{(x+1)=1}$.

Now, ${{\text{e}}^{\text{y}}}\text{(x+1)=1}\Rightarrow {{\text{e}}^{\text{y}}}\text{=}\frac{\text{1}}{\text{x+1}}$.

So, taking logarithm bth sides of the equation gives

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

Therefore, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=(x+1)}\frac{\text{d}}{\text{dx}}\left( \frac{\text{1}}{\text{x+1}} \right)\text{=(x+1) }\!\!\times\!\!\text{ }\frac{\text{-1}}{{{\text{(x+1)}}^{\text{2}}}}\text{=}\frac{\text{-1}}{\text{x+1}}$

That is,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{-1}}{\text{x+1}}$                                                     …… (1)

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{=}\left( \frac{\text{1}}{\text{x+1}} \right)\text{=-}\left( \frac{\text{-1}}{{{\text{(x+1)}}^{\text{2}}}} \right)\text{=}\frac{\text{1}}{{{\text{(x+1)}}^{\text{2}}}} \\ & \Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{-1}}{\text{x+1}} \right)}^{\text{2}}} \\ \end{align}$

$\Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}$, using the equation (1).

Thus, it is proved that $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}$.

17. If $\mathbf{y=(ta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}{{\mathbf{)}}^{\mathbf{2}}}$, show that ${{\mathbf{(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{+2x(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{=2}$.

Ans:  

The given equations are $\text{y=(ta}{{\text{n}}^{\text{-1}}}\text{x}{{\text{)}}^{\text{2}}}$.

Then, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align} & {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}\frac{\text{d}}{\text{dx}}\text{(ta}{{\text{n}}^{\text{-1}}}\text{x)} \\ & \Rightarrow {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \\ & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x} \\ \end{align}\]

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align} & \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x}{{\text{y}}_{\text{1}}}\text{=2}\left( \frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) \\ & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2} \\ \end{align}$

Thus, it has been proved that $\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2}$.

What Do You Learn from NCERT Maths 5.7 Exercise Class 12?

• What Can You Expect From the Chapter in CBSE Exams?

The chapter will also introduce you to one of the most important laws in Class 12 Physics, Coulomb’s law, which provides you the formula for finding the magnitude of the force between two point charges. Numerical problems based on the Coulomb’s Laws are common from this chapter. You will also encounter numerical problems that require you to calculate the electric fields of charge distribution over one, two, and/or three-dimensional mediums. You can expect problems from this section in Electric Charges and Fields Class 12 based on different scenarios. Students who are familiar with the chapter on Differentiation which they learned in Maths, may find it easy to attempt numerical problems from this section.

• Summary

‘Electric Charges and Fields’ acquaints you with electric charges. So far, students have mostly received an elementary understanding of electric charges. However, Class 12 Physics delves into the depth of the subject, to inform students about the types of charges, attraction, and repulsion between charges, etc. Students may have come across electrons and protons in ‘Molecular Chemistry’ as well, but Class 12 Physics acclimates them to numerical problems related to the positive charge and negative charge carried by protons and electrons, respectively; students will also learn about conductors, insulators, and induction charging, on the theoretical side, when they study Physics Class 12 chapter 1. 

The chapter also introduces a fair number of important diagrams to help students visualize the field of a source charge. Students may be asked to draw such electric field diagrams based on variable scenarios. Lastly, this chapter will teach you about electric dipoles. Small numerical problems that require you to find the strength of polarization or net field, can be found in this section. Most of the problems described here manifest in the CBSE Class 12 Physics paper. Class 12 Physics is a subject in which you cannot afford to be complacent. It is a subject with a vast syllabus, that goes on to play a very important role for those taking up science, architecture or engineering fields, further down their career path. Chapter 1, ‘Electric Charges and Fields’ is one of the most critically important chapters in the Physics syllabus

• Exam strategy 

The best way to master any doubts you may have in this chapter is to try and attempt the NCERT exercise questions after finishing the lesson on your own. Try to solve as many of the problems and theoretical questions without consulting any solution for guidance. If you notice that you still require assistance in understanding the chapter, just head over to Vedantu’s website and download the chapter 1 Physics Class 12 solution PDF. 

It is immensely important for students to learn this chapter well as it goes on to play an important role in the future, in competitive exams, and in higher studies. The exercises in the NCERT exercise book, which is a respectable template for the actual CBSE test papers, illustrate what kind of questions you may have to face in the exams. 

Vedantu’s Class 12 Physics Chapter 1 NCERT Solutions

Students who are struggling to master the very fast chapter of Class 12 Physics can depend on Vedantu as a helping hand. We all know that the NCERT exercises give us a fair idea of the kind of questions students may face in the corresponding CBSE papers. So, if you wish to test your depth of understanding in the subject, then you must try to solve the NCERT exercises after finishing the chapter. That a lot of students find it impossible to answer the questions in the NCERT exercise correctly or confidently. If you are unable to answer the NCERT exercise questions, then do not fret. Vedantu’s NCERT Solutions for Class 12 Physics Chapter 1, provides you the answers to all of the questions in the exercise.

Vedantu’s exercise solutions for the Class 12 NCERT exercises are guaranteed to be a difference-maker in your Class 12 Physics grades. Students can download an assimilated or individual exercise solutions, based on the scope of their doubt. All NCERT Solutions reference material, including that of chapter 1, is available for free download as PDF documents, on the Vedantu website. The solution guide exists to filter out all doubts you may have related to the lesson on ‘Electric Charges and Fields’. 

The solution PDF for CBSE Class 12 Physics, chapter 1, comprises of solutions prepared by Vedantu’s in-house tutors. Vedantu’s veteran in-house academicians have years of individual and collective experience of teaching students; they enjoy a familiarity with the CBSE question paper model that allows them to prepare answers best suited for scoring marks. They are familiar with the question patterns, so you can expect to-the-point, precise solutions in the Vedantu PDF that clarify your doubts and strengthens your base understanding of the topic.

Benefits of Vedantu’s Chapter 1 Physics Class 12 NCERT Solutions

So, students, if you want to build your confidence and breeze through your Class 12 Physics paper, then you must refer to Vedantu’s ‘Electric Charges and Fields’ for assistance. Vedantu’s solutions are guaranteed to build your foundation in advanced Physics stronger so you can excel at competitive exams and grad-level Physics later should you pursue them.

If you are still not convinced that Vedantu’s Class 12 Physics NCERT Solutions can help you clarify your doubts in Class 12 Physics, then read on; here’s what you stand to benefit when you refer to Vedantu’s solutions:

• Precise theoretical answers and numerical solutions with broadened, simplified steps for better understanding.

• Solution papers prepared by veteran tutors at Vedantu.

• Solutions to the latest updated NCERT exercise curriculum.

Advantages of Vedantu’s Digital Learning Platform 

Hence, Vedantu’s Class 12 Physics solutions to NCERT Chapter 1 exercises, can help you gain a competitive edge over the rest of your batch. Plus, Vedantu’s digital learning platform provides you other unforeseen advantages over traditional learning systems:

• Acquiring Vedantu’s NCERT Class 12 Physics Chapter 1 pdf solutions is simple; all you need to do is head over to Vedantu’s website where you will find the solution categorized class and subject-wise for your easy download.

• Download the free PDF solution guide to your computer, smartphone or tablet. Consult the solution guide to clarify any doubts you have regarding the NCERT exercise. You can consult the guide on-the-go as well since it is portable.

• Try solving the exercise without assistance from the solution. Continue to revise the NCERT sums, and past ten year questions based on what you learned from the Vedantu PDF guide.

Thus, learning with Vedantu is not only free but more convenient; and, if you still need help clarifying doubts then just head over to the Vedantu website to register for one-on-one classes.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Students can also download the following additional study material -

• Chapter 5: Important Questions

• Chapter 5: Revision Notes

• Chapter 5: Formulas

• Chapter 5: NCERT Exemplar Solutions

• Chapter 5: RD Sharma Solutions

• Chapter 5: RS Aggarwal Solutions

NCERT Solutions for Class 12 Mathematics Chapter 5 Exercises