NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 - FREE PDF Download
Class 12 Maths Continuity and differentiability class 12 solutions cover fundamental calculus concepts, including the second order derivatives. This exercise focuses on second order derivatives, which involve finding a derivative's derivative. These are crucial for understanding the concavity and points of inflection of functions. In this exercise, consider how second order derivatives are calculated and what they indicate about the function's behavior. Understanding these concepts is important for solving advanced problems in calculus. Vedantu's step-by-step NCERT Solutions for Class 12 Maths will help you understand and apply these ideas effectively, ensuring you are well-prepared for your exams.
Glance on NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.7 | Vedantu
NCERT Solutions Maths Chapter 5 Exercise 5.7 Class 12 covers second order derivatives and finding the second order derivative of a differential function.
The second derivative of a function, is the derivative of the first derivative.
The second order derivative of a function f(x) is denoted by $ \frac{\mathrm{d^2} y }{\mathrm{d} x^2} $ or f{}''(x).
There are 17 questions in class 12 maths ex 5.7 which are fully solved by experts at Vedantu.
NCERT Solutions Maths Chapter 5 Exercise 5.7 Class 12 covers second order derivatives and finding the second order derivative of a differential function.
The second derivative of a function, is the derivative of the first derivative.
The second order derivative of a function f(x) is denoted by $ \frac{\mathrm{d^2} y }{\mathrm{d} x^2} $ or f{}''(x).
There are 17 questions in class 12 maths ex 5.7 which are fully solved by experts at Vedantu.
Access Class 12 Maths NCERT Chapter 5 Continuity and Differentiability 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
\[{{\text{y}}_{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{4cos(logx)-3sin(logx)}}{\text{x}} \right)\]
\[\begin{align}
& \text{=}\frac{\text{x}\cdot \frac{d}{dx}\left[ \text{4 }\!\!\{\!\!\text{ cos(logx) }\!\!\}\!\!\text{ - }\!\!\{\!\!\text{ 3sin(logx) }\!\!\}\!\!\text{ } \right]\text{- }\!\!\{\!\!\text{ 4cos(logx)-3sin(logx) }\!\!\}\!\!\text{ }\times \text{1}}{{{\text{x}}^{\text{2}}}} \\
& \text{=}\frac{\text{x}\left[ \text{-4sin(logx)}\frac{d}{dx}\text{(logx)-3cos(logx)}\frac{d}{dx}\text{(logx)} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} \\
\end{align}\]
$=\dfrac{\text{x}\left[ \text{-4sin(logx)}\cdot \dfrac{\text{1}}{\text{x}}\text{-3cos(logx)}\cdot \dfrac{\text{1}}{\text{x}} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}}$
$=\dfrac{\text{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}}$
Therefore, ${{\text{y}}_{\text{2}}}\text{=}\dfrac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}}$.
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 $\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2}$.
Conclusion
Exercise 5.7 of Chapter 5 in Class 12 Maths focuses on second order derivatives, a crucial concept for understanding the behavior of functions. It is important to focus on accurately calculating second order derivatives. Regular practice of these problems will enhance your ability to analyze functions effectively. Vedantu's solutions provide detailed, step-by-step explanations to help you master these concepts, ensuring you are well-prepared for your exams. By understanding and practicing these key ideas, you'll build a solid foundation in calculus, which is essential for success in higher mathematics and various applications.
Class 12 Maths Chapter 5: Exercises Breakdown
S.No. | Chapter 5 - Continuity and Differentiability Exercises in PDF Format | |
1 | Class 12 Maths Chapter 5 Exercise 5.1 - 34 Questions & Solutions (10 Short Answers, 24 Long Answers) | |
2 | Class 12 Maths Chapter 5 Exercise 5.2 - 10 Questions & Solutions (2 Short Answers, 8 Long Answers) | |
3 | Class 12 Maths Chapter 5 Exercise 5.3 - 15 Questions & Solutions (9 Short Answers, 6 Long Answers) | |
4 | Class 12 Maths Chapter 5 Exercise 5.4 - 10 Questions & Solutions (5 Short Answers, 5 Long Answers) | |
5 | Class 12 Maths Chapter 5 Exercise 5.5 - 18 Questions & Solutions (4 Short Answers, 14 Long Answers) | |
6 | Class 12 Maths Chapter 5 Exercise 5.6 - 11 Questions & Solutions (7 Short Answers, 4 Long Answers) | |
7 | Class 12 Maths Chapter 5 Miscellaneous Exercise - 22 Questions & Solutions |
CBSE Class 12 Maths Chapter 5 Other Study Materials
S.No. | Important Links for Chapter 5 Continuity and Differentiability |
1 | Class 12 Continuity and Differentiability Important Questions |
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4 | Class 12 Continuity and Differentiability NCERT Exemplar Solutions |
5 | Class 12 Continuity and Differentiability RS Aggarwal Solutions |
NCERT Solutions for Class 12 Maths | Chapter-wise List
Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.
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Related Links for NCERT Class 12 Maths in Hindi
Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.
S.No. | Related NCERT Solutions for Class 12 Maths |
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Important Related Links for NCERT Class 12 Maths
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FAQs on NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability Exercise 5.7
1. What is ex 5.7 class 12 about?
Exercise 5.7 focuses on second order derivatives, which involve taking the derivative of the first derivative of a function. This is crucial for understanding the rate at which a function's slope changes.
2. Why are second order derivatives important in ex 5.7 class 12?
Second order derivatives help determine the concavity of functions (whether they curve upwards or downwards) and identify points of inflection where the function changes its concavity. These are important for graphing functions and solving real-world problems.
3. What should I focus on in continuity and differentiability class 12 solutions?
Focus on accurately calculating second order derivatives and understanding their significance. Practice problems that involve interpreting the concavity of functions and finding points of inflection to strengthen your grasp of these concepts.
4. How many questions on this topic appeared in previous exams from continuity and differentiability class 12 pdf?
Typically, around 2-3 questions related to second order derivatives appear in board exams, indicating their importance in the syllabus.
5. How do you calculate a second order derivative in class 12 maths ex 5.7?
To calculate a second order derivative, first find the derivative of the function (the first derivative), then differentiate the first derivative. This gives you the second order derivative.
6. What is the significance of the second order derivative test in exercise 5.7 class 12 maths?
The second order derivative test helps determine whether a critical point is a local maximum, local minimum, or a point of inflection. It is a crucial tool in calculus for analyzing the behavior of functions.
7. How can Vedantu's solutions help me with class 12 maths chapter 5 exercise 5.7?
Vedantu's solutions provide clear, step-by-step explanations that simplify complex concepts related to second order derivatives. By following these solutions, you can gain a better understanding and improve your problem-solving skills, which are essential for performing well in exams.