Important Questions for CBSE Class 12 Maths Chapter 5 - Continuity and Differentiability 2024-25

Very Short Questions and Answers (1 Marks Questions)

1. For what value of ${\text{x,}}\,{\text{f}}\left( {\text{x}} \right){\text{ = }}\left| {{\text{2x - 7}}} \right|$. is not derivable.

Ans: We need to find the input value for which the function will not be derivable. In other words, the point where the function has a break (there are two different tangents at that point)

$f\left( x \right) = \left\{ \begin{align} 2x - 7\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \geqslant \dfrac{7}{2} \hfill \\ 7 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < \dfrac{7}{2} \hfill \\ \end{align}  \right.$

Hence, for the value $x = \dfrac{7}{2}$, the function is not derivable.

2. Write the set of points of continuity of ${\text{g}}\left( {\text{x}} \right){\text{ = }}\left| {{\text{x - 1}}} \right|{\text{ + }}\left| {{\text{x + 1}}} \right|$.

Ans: Let us check for the break points of the piecewise function,

$g\left( x \right) = \left\{ \begin{align} \left( {x - 1} \right) + \left( {x + 1} \right)\,\,\,\,\,\,\,\,\,x > 1 \hfill \\ \left( {1 - x} \right) + \left( {x + 1} \right)\,\,\,\,\,\,\,\,\, - 1 \leqslant x \leqslant 1 \hfill \\ \left( {1 - x} \right) - \left( {x + 1} \right)\,\,\,\,\,\,\,\,\,x <  - 1 \hfill \\  \end{align}  \right.$

We see that the function is a polynomial at all the breakpoints, polynomials are continuous for their entire domain and hence the points of continuity of $g\left( x \right)$ is $\mathbb{R}$.

3. What is derivative of $\left| {{\text{x - 3}}} \right|$ at ${\text{x =  - 1}}$

Ans: Writing the function in piecewise form, we get

$f\left( x \right) = \left\{ \begin{align} x - 3\,\,\,\,\,\,\,\,\,\,\,\,x > 3 \hfill \\ 3 - x\,\,\,\,\,\,\,\,\,\,\,\,x \leqslant 3 \hfill \\  \end{align}  \right.$

For $x =  - 1$, we have

$\begin{align} f\left( { - 1} \right) = 3 - \left( { - 1} \right) \hfill \\ = 4 \hfill \\ \end{align} $

$\begin{align} f'\left( x \right) = \dfrac{d}{{dx}}\left( {3 - x} \right) \hfill \\ =  - 1 \hfill \\  \end{align} $

$f'\left( { - 1} \right) =  - 1$

4. What are the points of discontinuity of ${\text{f}}\left( {\text{x}} \right){\text{ = }}\dfrac{{\left( {{\text{x - 1}}} \right){\text{ + }}\left( {{\text{x + 1}}} \right)}}{{\left( {{\text{x - 7}}} \right)\left( {{\text{x - 6}}} \right)}}$.

Ans: From the given function, it is noticeable that it is not defined for the values $x = 6,7$. Considering that the function is a polynomial, we can say that it is continuous for the points mentioned above. Hence, the points of discontinuity will be situated at the inputs $x = 6$ and $x = 7$.

5. Write the number of points of discontinuity of ${\text{f}}\left( {\text{x}} \right){\text{ = }}\left[ {\text{x}} \right]$ in $\left[ {{\text{3,7}}} \right]$.

Ans: We have that step functions are discontinuous at the integer points. At the given domain, the integer inputs are $4,5,6,7$. Hence, the number of points of discontinuity is $4$.

6. The function, ${\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} {\lambda x - 3}\,\,\,\,\,\,\,\,\, x < 2 \hfill \\ {\text{4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x = 2 \hfill \\ {\text{2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x > 2 \hfill \\  \end{align}  \right.$ is a continuous function for all ${\text{x}} \in {\text{R}}$, find ${\lambda}$.

Ans: In piecewise functions, the function is continuous at the break points if the limit of the breakpoints evaluates to the same value. Evaluating the limit at the break points $x < 2$ and $x = 2$.

$\begin{align} \mathop {\lim }\limits_{x \to {2^ - }} \lambda x - 3 = \mathop {\lim }\limits_{x \to 2} \,\,4 \hfill \\ \Rightarrow \lambda \left( 2 \right) - 3 = 4 \hfill \\ \Rightarrow \lambda  = \dfrac{7}{2} \hfill \\  \end{align} $

Therefore, $\lambda  = \dfrac{7}{2}$.

7. For what value of ${\text{K}}$, \[{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} \dfrac{{{\text{tan3x}}}}{{{\text{sin2x}}}}\,\,\,\,\,\,\,{\text{x}} \ne {\text{0}} \hfill \\ {\text{2K}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 0}} \hfill \\  \end{align}  \right.\] is continuous $\forall {\text{x}} \in {\text{R}}$.

Ans: Evaluating the limit at the break points, we get

\[\begin{align} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{\sin 2x}} = 2K \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\tan 3x}}{{3x}} \cdot 3}}{{\dfrac{{\sin 2x}}{{2x}} \cdot 2}} = 2K \\ \Rightarrow \dfrac{3}{2} = 2K \\ \Rightarrow K = \dfrac{3}{4} \\ \end{align} \]

8. Write the derivative of ${\text{sinx}}$ with respect to ${\text{cosx}}$.

Ans: Let $y = \sin x$, and $z = \cos x$

Differentiate both with respect to $x$,

$\dfrac{{dy}}{{dx}} = \cos x$ and $\dfrac{{dz}}{{dx}} =  - \sin x$

Let us divide the first equation with the second equation, as shown below

$\begin{align} \Rightarrow \dfrac{{dy}}{{dx}} \cdot \dfrac{{dx}}{{dz}} = \cos x \cdot \dfrac{1}{{ - \sin x}} \hfill \\ \Rightarrow \dfrac{{dy}}{{dz}} =  - \dfrac{{\cos x}}{{\sin x}} \hfill \\ \Rightarrow \dfrac{{dy}}{{dz}} =  - \cot x \hfill \\  \end{align} $

9. If ${\text{f}}\left( {\text{x}} \right){\text{ = }}{{\text{x}}^{\text{2}}}{\text{g}}\left( {\text{x}} \right)$ and ${\text{g}}\left( {\text{1}} \right){{ = 6,g'}}\left( {\text{1}} \right){\text{ = 3}}$ find the value of ${{f'}}\left( {\text{1}} \right)$.

Ans: Differentiating the given function with respect to $x$,

$\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = 2xg\left( x \right) + g'\left( x \right){x^2}$

Therefore, $f'\left( 1 \right)$ should be,

$\begin{align} f'\left( 1 \right) = 2\left( 1 \right)\left( 6 \right) + \left( 3 \right){\left( 1 \right)^2} \hfill \\ = 12 + 3 \hfill \\   = 15 \hfill \\  \end{align} $

10. Write the derivative of the following functions:

(i) $\mathbf{{\text{lo}}{{\text{g}}_{\text{3}}}\left( {{\text{3x + 5}}} \right)}$

Ans: $y = {\log _3}\left( {3x + 5} \right)$

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\log }_3}\left( {3x + 5} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{3}{{\left( {3x + 5} \right)\ln \left( 3 \right)}} \hfill \\  \end{align} $

(ii)$\mathbf{{{\text{e}}^{{\text{lo}}{{\text{g}}_{\text{2}}}{\text{x}}}}}$

Ans:$y = {e^{{{\log }_2}\left( x \right)}}$

$ \Rightarrow \ln y = {\log _2}\left( x \right)$

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{1}{y}} \right) = \dfrac{d}{{dx}}\left( {{{\log }_2}\left( x \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{1}{y}} \right) = \dfrac{1}{{\left( x \right)\ln \left( 2 \right)}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{{{\log }_2}\left( x \right)}}}}{{x\ln \left( 2 \right)}} \hfill \\ \end{align} $

(iii) $\mathbf{{{\text{e}}^{{\text{6lo}}{{\text{g}}_{\text{e}}}\left( {{\text{x - 1}}} \right)}}{\text{,x > 1}}}$

Ans: 

$\begin{align}  y = {e^{6{{\log }_e}\left( {x - 1} \right)}} \hfill \\   \Rightarrow y = {e^{{{\log }_e}{{\left( {x - 1} \right)}^6}}} \hfill \\   \Rightarrow y = {\left( {x - 1} \right)^6} \hfill \\ \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\left( {x - 1} \right)}^6}} \right) \hfill \\ \Rightarrow \dfrac{{dy}}{{dx}} = 6{\left( {x - 1} \right)^5} \hfill \\ \end{align} $

(iv) $\mathbf{{\text{se}}{{\text{c}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ + cs}}{{\text{c}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{,}}\,\,{\text{x}} \geqslant {\text{1}}}$

Ans: $y = {\sec ^{ - 1}}\sqrt x  + {\csc ^{ - 1}}\sqrt x $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}\sqrt x  + {{\csc }^{ - 1}}\sqrt x } \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {\sqrt x } \right)\sqrt {{{\left( {\sqrt x } \right)}^2} - 1} }} + \dfrac{{ - 1}}{{\left( {\sqrt x } \right)\sqrt {{{\left( {\sqrt x } \right)}^2} - 1} }} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {x\left( {x - 1} \right)} }} - \dfrac{1}{{\sqrt {x\left( {x - 1} \right)} }} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = 0 \hfill \\ \end{align} $

(v) $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{\text{x}}^{{\text{7/2}}}}} \right)}$

Ans: $y = {\sin ^{ - 1}}\left( {{x^{7/2}}} \right)$

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {{x^{7/2}}} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {{x^{7/2}}} \right)}^2}} }} \cdot \left( {\dfrac{7}{2}{{\left( x \right)}^{\dfrac{5}{2}}}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{7}{2} \cdot \dfrac{{\sqrt {{x^5}} }}{{\sqrt {1 - {x^7}} }} \hfill \\ \end{align} $

(vi) $\mathbf{{\text{lo}}{{\text{g}}_{\text{x}}}{\text{5,}}\,\,{\text{x > 0}}}$

Ans: $y = {\log _x}\left( 5 \right)$

$\begin{align}   \Rightarrow y = \dfrac{{\ln 5}}{{\ln x}} \hfill \\   \Rightarrow \dfrac{1}{y} = \dfrac{{\ln x}}{{\ln 5}} \hfill \\ \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{y}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\ln x}}{{\ln 5}}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{1}{y}} \right) = \dfrac{1}{{x\ln 5}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\log }_x}5}}{{x\ln 5}} \hfill \\ \end{align} $

Short Questions and Answers (4 Marks Questions)

11. Discuss the continuity of following functions at the indicated points.

(i) $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} \dfrac{{{\text{x - }}\left| {\text{x}} \right|}}{{\text{x}}}\,\,\,\,\,\,{\text{x}} \ne {\text{0}} \hfill \\ {\text{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 0}} \hfill \\ \end{align}  \right.}$ at $\mathbf{{\text{x = 0}}}$

Ans: Given, $f\left( x \right) = \left\{ \begin{align} \dfrac{{x - \left| x \right|}}{x}\,\,\,\,\,\,x \ne 0 \hfill \\  2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \\ \end{align}  \right.$

Check the limit of the function as it approaches both sides of the point $x = 0$ from either direction,

Left-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{x - \left( { - x} \right)}}{x} \hfill \\   = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{2x}}{x} \hfill \\   = 2 \hfill \\  \end{align} $

Middle Value is,

$f\left( 0 \right) = 2$

Right-Hand limit is,

$\begin{align} \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{x - \left( x \right)}}{x} \hfill \\   = 0 \hfill \\ \end{align} $

As all three limits do not evaluate to the same value, the function is discontinuous at the point $x = 0$.

(ii) $\mathbf{{\text{g}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} \dfrac{{{\text{sin2x}}}}{{{\text{3x}}}}\,\,\,\,\,\,\,{\text{x}} \ne {\text{0}} \hfill \\ \dfrac{{\text{3}}}{{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 0}} \hfill \\ \end{align}  \right.}$ at $\mathbf{{\text{x = 0}}}$

Ans: Given, $g\left( x \right) = \left\{ \begin{align}  \dfrac{{\sin 2x}}{{3x}}\,\,\,\,\,\,x \ne 0 \hfill \\  \dfrac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \\ \end{align}  \right.$

Check the limit of the function as it approaches both sides of the point $x = 0$ from either direction.

Left-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin 2x}}{{3x}} \hfill \\   = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\dfrac{{\sin 2x}}{{2x}} \cdot 2}}{{\dfrac{{3x}}{x}}} \hfill \\   = \dfrac{2}{3} \hfill \\  \end{align} $

Middle Value is,

$f\left( 0 \right) = \dfrac{3}{2}$

Right-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin 2x}}{{3x}} \hfill \\   = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\dfrac{{\sin 2x}}{{2x}} \cdot 2}}{{\dfrac{{3x}}{x}}} \hfill \\   = \dfrac{2}{3} \hfill \\  \end{align} $

As all three limits do not evaluate to the same value, the function is discontinuous at the point $x = 0$.

(iii) $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} {{\text{x}}^{\text{2}}}{\text{cos}}\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right)\,\,\,\,\,\,{\text{x}} \ne {\text{0}} \hfill \\ {\text{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 0}} \hfill \\ \end{align}  \right.}$ at $\mathbf{{\text{x = 0}}}$

Ans: Given, $f\left( x \right) = \left\{ \begin{align} {x^2}\cos \left( {\dfrac{1}{x}} \right)\,\,\,\,\,\,x \ne 0 \hfill \\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \hfill \\  \end{align}  \right.$

Check the limit of the function as it approaches both sides of the point $x = 0$ from either direction.

Left-Hand limit is,

$\begin{align} \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} {x^2}\cos \dfrac{1}{x} \\    = 0 \\  \end{align} $

Middle Value is,

$f\left( 0 \right) = 0$

Right-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\cos \dfrac{1}{x} \\    = 0 \\ \end{align} $

We have that $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$

As all three limits evaluate to the same value, the function is continuous at the point $x = 0$.

(iv) $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left| {\text{x}} \right|{\text{ + }}\left| {{\text{x - 1}}} \right|}$ at $\mathbf{{\text{x = 1}}}$

Ans: Given, $f\left( x \right) = \left| x \right| + \left| {x - 1} \right|$

Check the limit of the function as it approaches both sides of the point $x = 1$ from either direction.

Left-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {1^ - }} \left( {\left| x \right| + \left| {x - 1} \right|} \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {x + \left( {1 - x} \right)} \right) \hfill \\   = 1 \hfill \\  \end{align} $

Middle Value is,

$\begin{align}  f\left( 1 \right) = \left( 1 \right) + \left( {1 - 1} \right) \hfill \\   = 1 \hfill \\  \end{align} $

Right-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {1^ + }} \left( {\left| x \right| + \left| {x - 1} \right|} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + \left( {x - 1} \right)} \right) \hfill \\   = 1 \hfill \\ \end{align} $

We have that $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$

As all three limits evaluate to the same value, the function is continuous at the point $x = 1$.

(v) $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} {\text{x - }}\left[ {\text{x}} \right]\,\,\,\,\,\,{\text{x}} \ne {\text{1}} \hfill \\ {\text{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 1}} \hfill \\  \end{align}  \right.}$ at $\mathbf{{\text{x = 1}}}$

Ans: Given, $f\left( x \right) = \left\{ \begin{align}  x - \left[ x \right]\,\,\,\,\,\,x \ne 1 \hfill \\  0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \hfill \\  \end{align}  \right.$

Check the limit of the function as it approaches both sides of the point $x = 1$ from either direction.

Left-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {x - \left[ x \right]} \right) \hfill \\   = 1 - \left( 0 \right) \hfill \\   = 1 \hfill \\  \end{align} $

Middle Value is,

$\begin{align}  f\left( 1 \right) = \left( 1 \right) - \left( 1 \right) \hfill \\   = 0 \hfill \\  \end{align} $

Right-Hand limit is,

$\begin{align}  \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x - \left[ x \right]} \right) \hfill \\   = 1 - \left( 1 \right) \hfill \\   = 0 \hfill \\ \end{align} $

As all three limits do not evaluate to the same value, the function is discontinuous at the point $x = 1$.

12. For what value of k, \[\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} {\text{3}}{{\text{x}}^{\text{2}}}{\text{ - kx + 5}}\,\,\,\,\,\,\,\,\,{\text{0}} \leqslant {\text{x < 2}} \hfill \\ {\text{1 - 3x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{2}} \leqslant {\text{x}} \leqslant {\text{3}} \hfill \\  \end{align}  \right.}\] is continuous $\mathbf{\forall {\text{x}} \in \left[ {{\text{0,3}}} \right]}$.

Ans: If the function is continuous, the limit at the break points must evaluate to the same value.

Therefore, the following condition must be met,

$\begin{align}  \mathop {\lim }\limits_{x \to {2^ - }} 3{x^2} - kx + 5 = \mathop {\lim }\limits_{x \to {2^ + }} 1 - 3x \hfill \\   \Rightarrow 3{\left( 2 \right)^2} - k\left( 2 \right) + 5 = 1 - 3\left( 2 \right) \hfill \\   \Rightarrow 12 - 2k + 5 =  - 5 \hfill \\   \Rightarrow 2 = 2k \hfill \\   \Rightarrow k = 1 \hfill \\  \end{align} $

Secondly, since the piecewise function consists of polynomials, we know that the polynomials are continuous for their respective domains. Hence, the value of $k$ must be $k = 1$ if the function is continuous $\forall x \in \left[ {0,3} \right]$.

13. For what values of $\mathbf{{\text{a}}}$ and $\mathbf{{\text{b}}}$, $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} \dfrac{{{\text{x + 2}}}}{{\left| {{\text{x + 2}}} \right|}}{\text{ + a}}\,\,\,\,\,{\text{if x <  - 2}} \hfill \\ {\text{a + b}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{if x =  - 2}} \hfill \\ \dfrac{{{\text{x + 2}}}}{{\left| {{\text{x + 2}}} \right|}}{\text{ + 2b}}\,\,{\text{if x >  - 2}} \hfill \\ \end{align}  \right.}$ is continuous at $\mathbf{{\text{x = 2}}}$

Ans: For the function to be continuous, the following condition has to be necessarily met,

$\mathop {\lim }\limits_{x \to  - {2^ - }} \left( {\dfrac{{x + 2}}{{\left| {x + 2} \right|}} + a} \right) = a + b = \mathop {\lim }\limits_{x \to  - {2^ + }} \left( {\dfrac{{x + 2}}{{\left| {x + 2} \right|}} + 2b} \right)$

Left-Hand limit is,

$\begin{align}   \Rightarrow \mathop {\lim }\limits_{x \to  - {2^ - }} \left( {\dfrac{{x + 2}}{{\left| {x + 2} \right|}} + a} \right) = a + b \hfill \\   \Rightarrow \mathop {\lim }\limits_{x \to  - {2^ - }} \left( {\dfrac{{x + 2}}{{ - \left( {x + 2} \right)}} + a} \right) = a + b \hfill \\   \Rightarrow  - 1 + a = a + b \hfill \\   \Rightarrow b =  - 1 \hfill \\  \end{align} $

Right-Hand limit is,

$\begin{align}   \Rightarrow a + b = \mathop {\lim }\limits_{x \to  - {2^ + }} \left( {\dfrac{{x + 2}}{{\left| {x + 2} \right|}} + 2b} \right) \hfill \\   \Rightarrow a + b = \mathop {\lim }\limits_{x \to  - {2^ + }} \left( {\dfrac{{x + 2}}{{x + 2}} + 2b} \right) \hfill \\   \Rightarrow a + b = 1 + 2b \hfill \\   \Rightarrow a - 1 = 1 - 2 \hfill \\   \Rightarrow a = 0 \hfill \\  \end{align} $

The values of $a$ and $b$ should be $a = 0$ and $b =  - 1$.

14. Prove that $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left| {{\text{x + 1}}} \right|}$ is continuous at $\mathbf{{\text{x =  - 1}}}$ but not derivable at $\mathbf{{\text{x =  - 1}}}$.

Ans: To prove that the function is continuous at $x =  - 1$, take the limit of the breaking points,

$\begin{align} \mathop {\lim }\limits_{x \to  - {1^ - }} \left| {x + 1} \right| = \left| { - 1 + 1} \right| = \mathop {\lim }\limits_{x \to  - {1^ + }} \left| {x + 1} \right| \hfill \\ \Rightarrow \mathop {\lim }\limits_{x \to  - {1^ - }}  - \left( {x + 1} \right) = \left| 0 \right| = \mathop {\lim }\limits_{x \to  - {1^ + }} \left( {x + 1} \right) \hfill \\ \Rightarrow 0 = 0 = 0 \hfill \\  \end{align} $

Since the above condition is satisfied, we conclude that the function is continuous.

We have proved continuity, now we need to check for the left-hand derivative and the right-hand derivative.

$\begin{align}  \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f\left( { - 1 + h} \right) - f\left( { - 1} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( { - 1 + h} \right) - f\left( { - 1} \right)}}{h} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{\left| {\left( { - 1 + h} \right) + 1} \right| - \left| { - 1 + 1} \right|}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{\left| {\left( { - 1 + h} \right) + 1} \right| - \left| { - 1 + 1} \right|}}{h} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{\left( { - 1 + h + 1} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - \left( { - 1 + h + 1} \right)}}{h} \hfill \\   \Rightarrow 1 \ne  - 1 \hfill \\  \end{align} $

Since the above condition could not be met, the function is not differentiable or derivable at $x =  - 1$.

15. For what value of ${\text{p}}$, $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}\left\{ \begin{align} {{\text{x}}^{\text{p}}}{\text{sin}}\left( {{\text{1/x}}} \right)\,\,\,{\text{x}} \ne {\text{0}} \hfill \\ {\text{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x = 0}} \hfill \\ \end{align}  \right.}$ is derivable at $\mathbf{{\text{x = 0}}}$

Ans: To ensure derivability, we must ensure the following condition

$\begin{align}  \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{{\left( {0 + h} \right)}^p}\sin \left( {\dfrac{1}{{0 + h}}} \right) - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{{{\left( {0 + h} \right)}^p}\sin \left( {\dfrac{1}{{0 + h}}} \right) - 0}}{h} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^p}\sin \left( {\dfrac{1}{h}} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{{h^p}\sin \left( {\dfrac{1}{h}} \right)}}{h} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^{p - 1}}\sin \left( {\dfrac{1}{h}} \right)}}{{h \cdot \dfrac{1}{h}}} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{{h^{p - 1}}\sin \left( {\dfrac{1}{h}} \right)}}{{h \cdot \dfrac{1}{h}}} \hfill \\   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^{p - 1}}}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{{h^{p - 1}}}}{h} \hfill \\  \end{align} $

Therefore, the condition that we have obtained for $p$ is as shown below,

$p - 1 > 0 \Rightarrow p > 1$ which means $p$ must be greater than \[1\]. 

16. If $\mathbf{{\text{y = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} \right){\text{ + 2ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{\text{x}}}} \right)} \right]{\text{,}}\,\,{\text{0 < x < 1}}}$, find $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$.

Ans: Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) + \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{1}{x}} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\left( {1 + {{\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)}^2}} \right)}} \cdot \dfrac{{2\left( {1 - {x^2}} \right) - 2x\left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} + \dfrac{1}{{1 + {{\left( {\dfrac{1}{x}} \right)}^2}}} \cdot \left( { - \dfrac{1}{{{x^2}}}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2\left( {1 - {x^2}} \right) + 4{x^2}}}{{2\left( {{{\left( {1 - {x^2}} \right)}^2} + 4{x^2}} \right)}} - \dfrac{1}{{{x^2} + 1}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} - \dfrac{1}{{{x^2} + 1}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = 0 \hfill \\ \end{align} $

17. If $\mathbf{{\text{y = sin}}\left[ {{\text{2ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\dfrac{{{\text{1 - x}}}}{{{\text{1 + x}}}}} } \right]}$ then $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}}$?

Ans: Let $x = \cos t$

We have,

$\begin{align}   \Rightarrow y = \sin \left[ {2{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos t}}{{1 + \cos t}}} } \right] \hfill \\   \Rightarrow y = \sin \left[ {2{{\tan }^{ - 1}}\sqrt {\tan \left( {\dfrac{t}{2}} \right)} } \right] \hfill \\   \Rightarrow y = \sin \left[ {2\left( {\dfrac{t}{2}} \right)} \right] \hfill \\   \Rightarrow y = \sin t \hfill \\   \Rightarrow y = \sqrt {1 - {{\cos }^2}t}  \hfill \\   \Rightarrow y = \sqrt {1 - {x^2}}  \hfill \\  \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}{\left( {1 - {x^2}} \right)^{ - \dfrac{1}{2}}}\left( { - 2x} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{x}{{\sqrt {1 - {x^2}} }} \hfill \\  \end{align} $

18. If $\mathbf{{{\text{5}}^{\text{x}}}{\text{ + }}{{\text{5}}^{\text{y}}}{\text{ = }}{{\text{5}}^{{\text{x + y}}}}}$ then prove that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ + }}{{\text{5}}^{{\text{y - x}}}}{\text{ = 0}}}$.

Ans: Differentiating with respect to $x$, we get

$\begin{align}  {5^x}\ln 5 + {5^y}\ln 5\dfrac{{dy}}{{dx}} = {5^{x + y}}\ln 5\left( {1 + \dfrac{{dy}}{{dx}}} \right) \hfill \\   \Rightarrow {5^x} + \dfrac{{dy}}{{dx}}{5^y} = \left( {1 + \dfrac{{dy}}{{dx}}} \right){5^{x + y}} \hfill \\   \Rightarrow {5^x} + \dfrac{{dy}}{{dx}}{5^y} = {5^{x + y}} + \dfrac{{dy}}{{dx}}{5^{x + y}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {{5^{x + y}} - {5^y}} \right) + \left( {{5^{x + y}} - {5^x}} \right) = 0 \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} + \dfrac{{{5^{x + y}} - {5^x}}}{{{5^{x + y}} - {5^y}}} = 0 \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} + \dfrac{{{5^y}}}{{{5^x}}} = 0 \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} + {5^{y - x}} = 0 \hfill \\  \end{align} $

19. If $\mathbf{{\text{x}}\sqrt {{\text{1 - }}{{\text{y}}^{\text{2}}}} {\text{ + y}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ = a}}}$ then show that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ =  - }}\sqrt {\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}}} $.

Ans: Differentiating with respect to $x$

$\begin{align}   \Rightarrow  - \dfrac{{2xy}}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} + \sqrt {1 - {y^2}}  - \dfrac{{2xy}}{{\sqrt {1 - {x^2}} }} + \dfrac{{dy}}{{dx}}\sqrt {1 - {x^2}}  = 0 \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\sqrt {1 - {x^2}}  - \dfrac{{2xy}}{{\sqrt {1 - {y^2}} }}} \right) = \dfrac{{2xy}}{{\sqrt {1 - {x^2}} }} - \sqrt {1 - {y^2}}  \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{{\dfrac{{2xy - \left( {\sqrt {1 - {y^2}} \sqrt {1 - {x^2}} } \right)}}{{\sqrt {1 - {x^2}} }}}}{{\dfrac{{2xy - \left( {\sqrt {1 - {y^2}} \sqrt {1 - {x^2}} } \right)}}{{\sqrt {1 - {y^2}} }}}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} =  - \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}}  \hfill \\  \end{align} $

20. If $\mathbf{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\sqrt {{\text{1 - }}{{\text{y}}^{\text{2}}}} {\text{ = a}}\left( {{\text{x - y}}} \right)}$ then show that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\sqrt {\dfrac{{{\text{1 - }}{{\text{y}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} }$.

Ans: Let,

$\begin{align}  x = \sin m \hfill \\  m = {\sin ^{ - 1}}x \hfill \\  y = \sin n \hfill \\  n = {\sin ^{ - 1}}y \hfill \\  \end{align} $

We have,

$\begin{align}   \Rightarrow \sqrt {1 - {{\sin }^2}m}  + \sqrt {1 - {{\sin }^2}n}  = a\left( {\sin m - \sin n} \right) \hfill \\   \Rightarrow \cos m + \cos n = a\left( {2\cos \dfrac{{m + n}}{2}\sin \dfrac{{m - n}}{2}} \right) \hfill \\   \Rightarrow 2\cos \dfrac{{m + n}}{2}\cos \dfrac{{m - n}}{2} = 2a\cos \dfrac{{m + n}}{2}\cos \dfrac{{m - n}}{2} \hfill \\   \Rightarrow \dfrac{1}{a} = \tan \dfrac{{m - n}}{2} \hfill \\   \Rightarrow {\tan ^{ - 1}}\dfrac{1}{a} = \dfrac{{m - n}}{2} \hfill \\   \Rightarrow 2{\tan ^{ - 1}}\dfrac{1}{a} = {\sin ^{ - 1}}x - {\sin ^{ - 1}}y \hfill \\ \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow 0 = \dfrac{1}{{\sqrt {1 - {x^2}} }} - \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}}  \hfill \\  \end{align} $

21. If $\mathbf{{\left( {{\text{x + y}}} \right)^{{\text{m + n}}}}{\text{ = }}{{\text{x}}^{\text{m}}}{{\text{y}}^{\text{n}}}}$ then prove that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}}$.

Ans: Take the natural log on both sides as shown below,

$\begin{align}  \ln \left( {{{\left( {x + y} \right)}^{m + n}}} \right) = \ln \left( {{x^m}{y^n}} \right) \hfill \\   \Rightarrow \left( {m + n} \right)\ln \left( {x + y} \right) = m\ln x + n\ln y \hfill \\  \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{m}{x} + \dfrac{n}{y}\dfrac{{dy}}{{dx}} = \dfrac{{m + n}}{{x + y}}\left( {1 + \dfrac{{dy}}{{dx}}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{n}{y} - \dfrac{{m + n}}{{x + y}}} \right) = \dfrac{{m + n}}{{x + y}} - \dfrac{m}{x} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{{nx + ny - my - ny}}{{y\left( {x + y} \right)}}} \right) = \dfrac{{mx + nx - mx - my}}{{x\left( {x + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x} \hfill \\  \end{align} $

22. Find the derivative of $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x}}}}{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}} \right)}$ w.r.t.$\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right)}$.

Ans:

$\begin{align}  y = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)}^2}}} \cdot \dfrac{{2\left( {1 - {x^2}} \right) - 2x\left( { - 2x} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{1 + {x^2}}} \hfill \\  \end{align} $

Let $x = \tan t$.

$\begin{align}  z = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)  \\ \Rightarrow z = {\sin ^{ - 1}}\left( {\sin 2t} \right)  \\  \Rightarrow z = 2{\tan ^{ - 1}}x \end{align} $

$\dfrac{{dz}}{{dx}} = \left\{ \begin{align} \dfrac{2}{{1 + {x^2}}}\,\,\,\,\,\,\,\left| x \right| < 1  \\   - \dfrac{2}{{1 + {x^2}}}\,\,\,\,\left| x \right| \geqslant 1  \\  \end{align}  \right. $

Hence, we get,

$\dfrac{{dy}}{{dz}} = \left\{ \begin{align}  1\,\,\,\,\,\left| x \right| < 1 \hfill \\    - 1\,\,\left| x \right| \geqslant 1 \hfill \\  \end{align}  \right.$

23. Find the derivative of $\mathbf{{\text{ln}}\left( {{\text{sinx}}} \right)}$ w.r.t.$\mathbf{{\text{lo}}{{\text{g}}_{\text{a}}}\left( {{\text{cosx}}} \right)}$.

Ans: Let,

$\begin{align}  y = \ln \left( {\sin x} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin x}}\cos x \hfill \\  z = {\log _a}\cos x \hfill \\    \Rightarrow z = \dfrac{{\ln \cos x}}{{\ln a}} \hfill \\   \Rightarrow \dfrac{{dz}}{{dx}} = \dfrac{{1\left( { - \sin x} \right)}}{{\cos x}} \cdot \dfrac{1}{{\ln a}} \hfill \\  \end{align} $

Hence, we get

$\begin{align}  \dfrac{{dy}}{{dx}} \cdot \dfrac{{dx}}{{dz}} = \dfrac{{\cos x}}{{\sin x}} \cdot \dfrac{{\ln a\cos x}}{{ - \sin x}} \\    =  - \ln a{\cot ^2}x \\  \end{align} $

24. If $\mathbf{{{\text{x}}^{\text{y}}}{\text{ + }}{{\text{y}}^{\text{x}}}{\text{ + }}{{\text{x}}^{\text{x}}}{\text{ = }}{{\text{m}}^{\text{n}}}}$, then find the value of $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$.

Ans: On differentiating with respect to $x$, we get 

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( {{x^y}} \right) + \dfrac{d}{{dx}}\left( {{y^x}} \right) + \dfrac{d}{{dx}}\left( {{x^x}} \right) = 0 \hfill \\   \Rightarrow {x^y}\left( {\dfrac{{dy}}{{dx}}\ln x + \dfrac{y}{x}} \right) + {y^x}\left( {\ln y + \dfrac{x}{y}\dfrac{{dy}}{{dx}}} \right) + {x^x}\left( {\ln x + 1} \right) = 0 \hfill \\  \end{align} $

Now, further simplify the expression obtained in the equation above as shown below,

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}}{x^y}\ln x + {y^x}\dfrac{x}{y}\dfrac{{dy}}{{dx}} =  - {x^x}\left( {\ln x + 1} \right) - {y^x}\ln y - {x^y}\left( {\dfrac{y}{x}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - {x^x}\left( {\ln  + 1} \right) - {y^x}\ln y - {x^{y - 1}}y}}{{{y^{x - 1}}x + {x^y}\ln x}} \hfill \\  \end{align} $

25. If $\mathbf{{\text{x = aco}}{{\text{s}}^{\text{3}}}{\theta}\text{,}\,\,{\text{y = asi}}{{\text{n}}^{\text{3}}}{\theta}}$ then find $\mathbf{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}}$.

Ans: Differentiate with respect to $\theta $,

$\begin{align}   \Rightarrow \dfrac{d}{{d\theta }}\left( x \right) = \dfrac{d}{{d\theta }}\left( {a{{\cos }^3}\theta } \right) \hfill \\   \Rightarrow \dfrac{{dx}}{{d\theta }} = 3a{\cos ^2}\theta \left( { - \sin \theta } \right) \hfill \\ \end{align} $

Also,

$\begin{align}   \Rightarrow \dfrac{d}{{d\theta }}\left( y \right) = \dfrac{d}{{d\theta }}\left( {a{{\sin }^3}\theta } \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \left( {\cos \theta } \right) \hfill \\ \end{align} $

Therefore, we have

$\begin{align}  \dfrac{{dy}}{{d\theta }} \cdot \dfrac{{d\theta }}{{dx}} =  - \dfrac{{3a{{\sin }^2}\theta \cos \theta }}{{3a{{\cos }^2}\theta \sin \theta }} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} =  - \tan \theta  \hfill \\  \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( { - \tan \theta } \right) \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{d\theta }}\left( { - \tan \theta } \right)\dfrac{{d\theta }}{{dx}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - {\sec ^2}\theta \left( { - \dfrac{1}{{3a{{\cos }^2}\theta \sin \theta }}} \right) \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{3a\sin \theta {{\cos }^4}\theta }} \hfill \\  \end{align} $

26. If $\mathbf{{\text{x = a}}{{\text{e}}^{\text{t}}}\left( {{\text{sint - cost}}} \right),\,{\text{y = a}}{{\text{e}}^{\text{t}}}\left( {{\text{sint + cost}}} \right)}$ then show that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$ at $\mathbf{{\text{x = }}\dfrac{{\pi }}{{\text{4}}}}$ is $\mathbf{{\text{1}}}$.

Ans: Differentiate the first equation with respect to $t$

$\begin{align}   \Rightarrow \dfrac{d}{{dt}}\left( x \right) = \dfrac{d}{{dt}}\left( {a{e^t}\left( {\sin t - \cos t} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dx}}{{dt}} = a{e^t}\left( {\sin t - \cos t} \right) + a{e^t}\left( {\cos t + \sin t} \right) \hfill \\   \Rightarrow \dfrac{{dx}}{{dt}} = 2a{e^t}\sin t \hfill \\ \end{align} $

Differentiate the second equation with respect to $t$

$\begin{align}   \Rightarrow \dfrac{d}{{dt}}\left( y \right) = \dfrac{d}{{dt}}\left( {a{e^t}\left( {\sin t + \cos t} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dt}} = a{e^t}\left( {\sin t + \cos t} \right) + a{e^t}\left( {\cos t - \sin t} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dt}} = 2a{e^t}\cos t \hfill \\ \end{align} $

Therefore, we have

$\begin{align}  \dfrac{{dy}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \dfrac{{2a{e^t}\left( {\cos t} \right)}}{{2a{e^t}\left( {\sin t} \right)}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \cos t \hfill \\  \end{align} $

Therefore at $t = \dfrac{\pi }{4}$, we have

$\begin{align}  \dfrac{{dy}}{{dx}} = \cot \left( {\dfrac{\pi }{4}} \right) \\     = 1 \\ \end{align} $

27. If $\mathbf{{\text{y = si}}{{\text{n}}^{{\text{ - 1}}}}\left[ {{\text{x}}\sqrt {{\text{1 - x}}} {\text{ - }}\sqrt {\text{x}} \sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } \right]}$ then find $\mathbf{{\text{ - }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$.

Ans: We have,

$y = {\sin ^{ - 1}}\left[ {x\sqrt {1 - {{\left( {\sqrt x } \right)}^2}}  - \sqrt x \sqrt {1 - {x^2}} } \right]$

Using the inverse sine addition trigonometric identity 

$[{\sin ^{ - 1}}x - {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left[ {x\sqrt {1 - {y^2}}  - y\sqrt {1 - {x^2}} } \right]$, we have

$y = {\sin ^{ - 1}}x - {\sin ^{ - 1}}\sqrt x $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }} - \dfrac{1}{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}\dfrac{1}{{2\sqrt x }} \hfill \\   \Rightarrow  - \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x\left( {1 - x} \right)} }} - \dfrac{1}{{\sqrt {1 - {x^2}} }} \hfill \\  \end{align} $

28. If $\mathbf{{\text{y = }}{{\text{x}}^{{\text{lnx}}}}{\text{ + }}{\left( {{\text{lnx}}} \right)^{\text{x}}}}$ then find $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$.

Ans: Let,

$y = u + v$ where $u = {x^{\ln x}}$ and $v = {\left( {\ln x} \right)^x}$

Differentiating with respect to $x$

$\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}$

Now, for $u$

$u = {x^{\ln x}}$

Take the natural logarithm of both sides

$\begin{align}   \Rightarrow \ln u = \ln x\ln x \hfill \\   \Rightarrow \ln u = {\left( {\ln x} \right)^2} \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{du}}{{dx}}\dfrac{1}{u} = 2\ln x\left( {\dfrac{1}{x}} \right) \hfill \\   \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{2{x^{\ln x}}\ln x}}{x} \hfill \\  \end{align} $

Now, for $v$

$v = {\left( {\ln x} \right)^x}$

Take the natural logarithm of both sides

$\ln v = x\ln \left( {\ln x} \right)$

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{dv}}{{dx}}\dfrac{1}{v} = x\left[ {\dfrac{1}{{\ln x}}\dfrac{1}{x}} \right] + \ln \left( {\ln x} \right) \hfill \\   \Rightarrow \dfrac{{dv}}{{dx}} = {\left( {\ln x} \right)^x}\left( {\dfrac{1}{{\ln x}} + \ln \left( {\ln x} \right)} \right) \hfill \\  \end{align} $

Putting both back in the equation, we get

$\dfrac{{dy}}{{dx}} = \dfrac{{2{x^{\ln x}}\ln x}}{x} + {\left( {\ln x} \right)^x}\left( {\dfrac{1}{{\ln x}} + \ln \left( {\ln x} \right)} \right)$

29. Differentiate $\mathbf{{{\text{x}}^{{{\text{x}}^{\text{x}}}}}}$ w.r.t. $\mathbf{{\text{x}}}$.

Ans: Taking the natural logarithm on both sides and then simplifying further, it can be obtained as show below,

$\begin{align}  y = {x^{{x^x}}} \hfill \\   \Rightarrow \ln y = \ln \left( {{x^{{x^x}}}} \right) \hfill \\   \Rightarrow \ln y = {x^x}\ln x \hfill \\  \end{align} $

Differentiating with respect to $x$,

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}}\dfrac{1}{y} = \dfrac{{{x^x}}}{x} + \ln x\left( {{x^x}\left( {1 + \ln x} \right)} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = {x^{{x^x}}}\left( {{x^{x - 1}} + \ln x\left( {{x^x}\left( {1 + \ln x} \right)} \right)} \right) \hfill \\  \end{align} $

30. Find $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$, if $\mathbf{{\left( {{\text{cosx}}} \right)^{\text{y}}}{\text{ = }}{\left( {{\text{cosy}}} \right)^{\text{x}}}}$.

Ans: Taking the natural logarithm on both sides

$\begin{align}   \Rightarrow \ln \left( {{{\left( {\cos x} \right)}^y}} \right) = \ln \left( {{{\left( {\cos y} \right)}^x}} \right) \hfill \\   \Rightarrow y\ln \cos x = x\ln \cos y \hfill \\  \end{align} $

Differentiating both sides with respect to $x$

$\begin{align}   \Rightarrow \dfrac{d}{{dx}}\left( {y\ln \left( {\cos x} \right)} \right) = \dfrac{d}{{dx}}\left( {x\ln \left( {\cos y} \right)} \right) \hfill \\   \Rightarrow y\left( {\dfrac{1}{{\cos x}}\left( { - \sin x} \right)} \right) + \dfrac{{dy}}{{dx}}\ln \cos x = x\left( {\dfrac{1}{{\cos y}}\left( { - \sin y} \right)\left( {\dfrac{{dy}}{{dx}}} \right)} \right) + \ln \cos y \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\ln \cos x} \right) - y\tan x = \dfrac{{dy}}{{dx}}\left( { - x\tan y} \right) + \ln \cos y \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}}\left( {\ln \cos x + x\tan y} \right) = y\tan x + \ln \cos y \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\tan x + \ln \cos y}}{{x\tan y + \ln \cos x}} \hfill \\ \end{align} $

31. If $\mathbf{{\text{y = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\sqrt {{\text{1 + sinx}}} {\text{ - }}\sqrt {{\text{1 - sinx}}} }}{{\sqrt {{\text{1 + sinx}}} {\text{ + }}\sqrt {{\text{1 - sinx}}} }}} \right)}$ where $\mathbf{\dfrac{{\pi }}{{\text{2}}}{\text{ < x <} \pi }}$, find $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}$.

Ans: We have

$\begin{align}  y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x}  + \sqrt {1 + \sin x} }}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x}  + \sqrt {1 + \sin x} }} \times \dfrac{{\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} }}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{{{\left( {\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} } \right)}^2}}}{{\left( {1 - \sin x} \right) - \left( {1 + \sin x} \right)}}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{{{\left( {\sqrt {1 - \sin x} } \right)}^2} + {{\left( {\sqrt {1 + \sin x} } \right)}^2} - 2\sqrt {1 - \sin x} \sqrt {1 + \sin x} }}{{ - 2\sin x}}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{2 - 2\sqrt {{{\cos }^2}x} }}{{ - 2\sin x}}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos x}}{{ - \sin x}}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{ - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} \right) \hfill \\   \Rightarrow y = {\tan ^{ - 1}}\left( { - \tan \dfrac{x}{2}} \right) \hfill \\   \Rightarrow y =  - {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right) \hfill \\   \Rightarrow y =  - \dfrac{x}{2} \hfill \\  \end{align} $

Differentiating with respect to $x$

$\dfrac{{dy}}{{dx}} =  - \dfrac{1}{2}$

32. If $\mathbf{{\text{x = sin}}\left( {\dfrac{{\text{1}}}{{\text{a}}}{\text{lny}}} \right)}$, then show that $\mathbf{\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{y'' - xy' - }}{{\text{a}}^{\text{2}}}{\text{y = 0}}}$.

Ans: We have

$\begin{align}  x = \sin \left( {\dfrac{1}{a}\ln y} \right) \hfill \\   \Rightarrow {\sin ^{ - 1}}x = \dfrac{1}{a}\ln y \hfill \\   \Rightarrow a{\sin ^{ - 1}}x = \ln y \hfill \\   \Rightarrow {e^{a{{\sin }^{ - 1}}x}} = y \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}} = {e^{a{{\sin }^{ - 1}}x}}\left( {\dfrac{a}{{\sqrt {1 - {x^2}} }}} \right) \hfill \\   \Rightarrow y' = \dfrac{{ay}}{{\sqrt {1 - {x^2}} }} \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{a\dfrac{{dy}}{{dx}}\sqrt {1 - {x^2}}  + \dfrac{{xay}}{{\sqrt {1 - {x^2}} }}}}{{{{\left( {1 - {x^2}} \right)}^2}}} \hfill \\   \Rightarrow y'' = \dfrac{{ay'\left( {1 - {x^2}} \right) + xay}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }} \hfill \\   \Rightarrow y'' = \dfrac{{ay'}}{{\sqrt {1 - {x^2}} }} + \dfrac{{xay}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }} \hfill \\   \Rightarrow y'' = \dfrac{{{a^2}y}}{{1 - {x^2}}} + \dfrac{{xy'}}{{\left( {1 - {x^2}} \right)}} \hfill \\   \Rightarrow \left( {1 - {x^2}} \right)y'' - xy' - {a^2}y = 0 \hfill \\  \end{align} $

33. Differentiate $\mathbf{{\left( {{\text{lnx}}} \right)^{{\text{lnx}}}}}$, $\mathbf{{\text{x > 1}}}$ w.r.t.$\mathbf{x}$.

Ans: Taking the natural log on both sides

$\ln y = \ln x\left( {\ln \left( {\ln x} \right)} \right)$

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}}\dfrac{1}{y} = \ln x\left( {\dfrac{1}{{x\ln x}}} \right) + \dfrac{{\ln \left( {\ln x} \right)}}{x} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\ln x} \right)^{\ln x}}\left( {\dfrac{{1 + \ln \left( {\ln x} \right)}}{x}} \right) \hfill \\  \end{align} $

34. If $\mathbf{{\text{siny = xsin}}\left( {{\text{a + y}}} \right)}$ then show that $\mathbf{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}\left( {{\text{a + y}}} \right)}}{{{\text{sina}}}}}$.

Ans: We have

$\begin{align}  \sin y = x\sin \left( {a + y} \right) \hfill \\   \Rightarrow x = \dfrac{{\sin y}}{{\sin \left( {a + y} \right)}} \hfill \\  \end{align} $

Differentiating with respect to $y$

\[\begin{align}   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{\cos y\left( {\sin \left( {a + y} \right)} \right) - \sin y\left( {\cos \left( {a + y} \right)} \right)}}{{{{\sin }^2}\left( {a + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{\cos y\left( {\sin a\cos y + \cos a\sin y} \right) - \sin y\left( {\cos a\cos y - \sin a\sin y} \right)}}{{{{\sin }^2}\left( {a + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{{{\cos }^2}y\sin a + \cos a\sin y\cos y - \sin y\cos a\cos y + \sin a{{\sin }^2}y}}{{{{\sin }^2}\left( {a + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{{{\cos }^2}y\sin a + {{\sin }^2}y\sin a}}{{{{\sin }^2}a + y}} \hfill \\   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{\sin a\left( {{{\cos }^2}y + {{\sin }^2}y} \right)}}{{{{\sin }^2}\left( {a + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{\sin a}}{{{{\sin }^2}\left( {a + y} \right)}} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\sin }^2}\left( {a + y} \right)}}{{\sin a}} \hfill \\  \end{align} \]

35. If $\mathbf{{\text{y = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$, find $\mathbf{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}}$ in terms of $\mathbf{{\text{y}}}$.

Ans: Differentiating with respect to $x$

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

Differentiating with respect to $x$

$\begin{align}  \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{1}{2}{\left( {1 - {x^2}} \right)^{ - \dfrac{3}{2}}}\left( { - 2x} \right) \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{x}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}}} \hfill \\  \end{align} $

Now, we have

$\begin{align}  y = {\sin ^{ - 1}}x \hfill \\   \Rightarrow x = \sin y \hfill \\ \end{align} $

Replacing in our equation

$\begin{align}   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\sin y}}{{{{\left( {\sqrt {1 - {{\sin }^2}y} } \right)}^3}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\sin y}}{{{{\left( {\cos y} \right)}^3}}} \hfill \\    \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \tan y{\sec ^2}y \hfill \\  \end{align} $

36. If $\mathbf{\dfrac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ + }}\dfrac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = 1}}}$, then show that $\mathbf{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{ - }}{{\text{b}}^{\text{4}}}}}{{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{3}}}}}}$.

Ans: Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{2x}}{{{a^2}}} + \dfrac{{dy}}{{dx}}\left( {\dfrac{{2y}}{{{b^2}}}} \right) = 0 \hfill \\   \Rightarrow \dfrac{x}{a} =  - \dfrac{{dy}}{{dx}}\left( {\dfrac{y}{{{b^2}}}} \right) \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{1}{{{a^2}}} =  - \left( {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}\dfrac{1}{{{b^2}}} + \dfrac{{{d^2}y}}{{d{x^2}}}\dfrac{y}{{{b^2}}}} \right) \hfill \\   \Rightarrow \dfrac{1}{{{a^2}}} + \dfrac{{{{\left( {y'} \right)}^2}}}{{{b^2}}} + \dfrac{{yy''}}{{{b^2}}} = 0 \hfill \\  \end{align} $

Now, we know

$\begin{align}  \dfrac{x}{{{a^2}}} =  - \dfrac{{yy'}}{{{b^2}}} \hfill \\   \Rightarrow \dfrac{{y'}}{{{b^2}}} =  - \dfrac{x}{{{a^2}y}} \hfill \\   \Rightarrow {\left( {\dfrac{{y'}}{{{b^2}}}} \right)^2} = {\left( { - \dfrac{x}{{{a^2}y}}} \right)^2} \hfill \\   \Rightarrow \dfrac{{{{\left( {y'} \right)}^2}}}{{{b^4}}} = \dfrac{{{x^2}}}{{{a^4}{y^2}}} \hfill \\   \Rightarrow \dfrac{{{{\left( {y'} \right)}^2}}}{{{b^2}}} = \dfrac{{{x^2}{b^2}}}{{{a^4}{y^2}}} \hfill \\  \end{align} $

Substituting in our previous equation

$\begin{align}  \dfrac{1}{{{a^2}}} + \dfrac{{{x^2}{b^2}}}{{{a^4}{y^2}}} + \dfrac{{yy''}}{{{b^2}}} = 0 \hfill \\   \Rightarrow \dfrac{1}{{{a^2}}}\left[ {1 + \dfrac{{{x^2}}}{{{a^2}}} \cdot \dfrac{{{b^2}}}{{{y^2}}}} \right] + \dfrac{{yy''}}{{{b^2}}} = 0 \hfill \\   \Rightarrow \dfrac{1}{{{a^2}}}\left[ {1 + \left( {1 - \dfrac{{{y^2}}}{{{b^2}}}} \right)\dfrac{{{b^2}}}{{{y^2}}}} \right] + \dfrac{{yy''}}{{{b^2}}} = 0 \hfill \\   \Rightarrow \dfrac{1}{{{a^2}}}\left[ {\dfrac{{{b^2}}}{{{y^2}}}} \right] + \dfrac{{yy''}}{{{b^2}}} = 0 \hfill \\   \Rightarrow \dfrac{{yy''}}{{{b^2}}} =  - \dfrac{{{b^2}}}{{{a^2}{y^2}}} \hfill \\   \Rightarrow y'' =  - \dfrac{{{b^4}}}{{{a^2}{y^3}}} \hfill \\  \end{align} $

37. If $\mathbf{{\text{y = }}{{\text{e}}^{{\text{aco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}}$, \[\mathbf{{\text{ - 1}} \leqslant {\text{x}} \leqslant {\text{1}}}\], show that $\mathbf{\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ - x}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ - }}{{\text{a}}^{\text{2}}}{\text{y = 0}}}$.

Ans: Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{dy}}{{dx}} = {e^{a{{\cos }^{ - 1}}x}}\left( { - \dfrac{a}{{\sqrt {1 - {x^2}} }}} \right) \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{{ay}}{{\sqrt {1 - {x^2}} }} \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{{a\dfrac{{dy}}{{dx}}\sqrt {1 - {x^2}}  - \dfrac{{xay}}{{\sqrt {1 - {x^2}} }}}}{{\left( {1 - {x^2}} \right)}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{{a\dfrac{{dy}}{{dx}}\left( {1 - {x^2}} \right) - xay}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{{a\dfrac{{dy}}{{dx}}}}{{\sqrt {1 - {x^2}} }} + \dfrac{{xay}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{{{a^2}y}}{{1 - {x^2}}} + \dfrac{{x\dfrac{{dy}}{{dx}}}}{{\left( {1 - {x^2}} \right)}} \hfill \\   \Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + {a^2}y - x\dfrac{{dy}}{{dx}} = 0 \hfill \\  \end{align} $

38. If $\mathbf{{{\text{y}}^{\text{3}}}{\text{ = 3a}}{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{3}}}}$ then prove that $\mathbf{\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{ - 2}}{{\text{a}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}{{{{\text{y}}^{\text{5}}}}}}$.

Ans: Differentiating with respect to $x$

$\begin{align}   \Rightarrow 3{y^2}\dfrac{{dy}}{{dx}} = 6ax - 3{x^2} \hfill \\   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2ax}}{{{y^2}}} - \dfrac{{{x^2}}}{{{y^2}}} \hfill \\  \end{align} $

Differentiating with respect to $x$

$\begin{align}   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {2a - 2x} \right){y^2} - \left( {2ax - {x^2}} \right)2y\left( {\dfrac{{dy}}{{dx}}} \right)}}{{{y^4}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {2a - 2x} \right){y^2} - \left( {2ax - {x^2}} \right)2y\left( {\dfrac{{2ax - {x^2}}}{{{y^2}}}} \right)}}{{{y^4}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {2a - 2x} \right){y^3} - 2{{\left( {2ax - {x^2}} \right)}^2}}}{{{y^5}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\left( {2a - 2x} \right)\left( {3a{x^2} - {x^3}} \right) - 2\left( {4{a^2}{x^2} + {x^4} - 4a{x^3}} \right)}}{{{y^5}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{6{a^2}{x^2} - 6a{x^3} - 2a{x^3} + 2{x^4} - 8a{x^2}{x^2} - 2{x^4} + 8a{x^3}}}{{{y^5}}} \hfill \\   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} =  - \dfrac{{2{a^2}{x^2}}}{{{y^5}}} \hfill \\  \end{align} $

39. Verify Rolle’s theorem for the function, $\mathbf{{\text{y = }}{{\text{x}}^{\text{2}}}{\text{ + 2}}}$ in the interval $\mathbf{\left[ {{\text{a,b}}} \right]}$ where $\mathbf{{\text{a =  - 2,b = 2}}}$.

Ans: We have been given a continuous function which is also differentiable in the given domain [since it is a polynomial].

We also have

$f\left( a \right) = f\left( { - 2} \right) = {\left( { - 2} \right)^2} + 2 = 6$

$f\left( b \right) = f\left( 2 \right) = {\left( 2 \right)^2} + 2 = 6$

Therefore, we have $f\left( a \right) = f\left( b \right)$

So, according to Rolle’s theorem, there must exist a point $c$, such that $f'\left( c \right) = 0$

Let us find $c$

$f'\left( x \right) = 2x$

Replacing $f'\left( x \right) = 0$

$\begin{align}  0 = 2x \hfill \\  x = 0 = c \hfill \\  \end{align} $

We know that $c \in \left[ { - 2,2} \right]$

Hence, Rolle’s theorem is verified for the given function.

40. Verify Mean Value Theorem for the function, $\mathbf{{\text{f}}\left( {\text{x}} \right){\text{ = }}{{\text{x}}^{\text{2}}}}$ in $\mathbf{\left[ {{\text{2,4}}} \right]}$

Ans: We have been given a continuous function which is also differentiable in the given domain (since it is a polynomial).

The Mean Value Theorem states that there must exist a point $c$ such that $f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 2 \right)}}{{4 - 2}}$

We know that $f'\left( x \right) = 2x$, Let us find the point $c$

$\begin{align}  f'\left( c \right) = \dfrac{{{{\left( 4 \right)}^2} - {{\left( 2 \right)}^2}}}{{4 - 2}} \hfill \\   \Rightarrow 2c = \dfrac{{12}}{2} \hfill \\   \Rightarrow 2c = 6 \hfill \\   \Rightarrow c = 3 \hfill \\  \end{align} $

We know that $c \in \left[ {2,4} \right]$

Hence, Mean Value Theorem is verified for the given function.

Download Class 12 Maths Important Questions Chapter 5 With Solutions

Class 12 Maths Important Questions Chapter 5 with solutions are designed and prepared by Vedantu's subject experienced teachers. All the important topics of Continuity and Differentiability like Rolle's theorem, Derivative functions, Mean Value theorem, etc are considered by the experts while preparing important questions of Continuity and Differentiability Class 12.

The solutions to each question are also provided in a stepwise manner along with the marking scheme. Through this, students will get an idea of representing answers in the best possible way so that they do not lose any marks in exams. To have a thorough understanding of the chapter, you can download important questions of Continuity and Differentiability Class 12 free pdf through the link given above.

A Brief Overview of Class 12 Maths Chapter 5

Class 12 Maths Chapter 5 discusses the important concepts of Continuity and Differentiability. Students will also learn differentiation of inverse trigonometric functions along with the new class of functions known as exponential and logarithmic functions. These functions provide strong techniques of differentiation. Certain geometric obvious conditions are also introduced in the chapter which enables students to learn fundamental theorems like Mean value theorem, Rolle’s theorem, etc.

Let us discuss some important terms covered in Class 12 Maths Chapter 5

Continuous Functions

A continuous function in calculus is a real-values function whose graphs do not have any holes or breaks.

A function f(x) is said to be continuous at x = k, if the following first three conditions satisfy.

1. f (k) exists.

2. \[\lim_{x\rightarrow k} f(x)\] exists.

3. \[\lim_{x\rightarrow k} f(x) = f(k)\].

Algebra of Composite Function

Sum, difference, product and quotient of continuous functions are continuous. It implies that if f and gbe two real-valued functions continuous at real number c, then 

• f + g is continuous at x = C.

• f - g is continuous at x = C.

• f . g is continuous at x = C.

• f/g is continuous at x = C.

Differentiable Function

The derivative function is a continuous function, a continuous function whose derivative exists at each point on its domain. This implies that, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain,

A function is said to differentiable at x₀ if the following condition satisfies.

• The function f is continuous at x₀, and 

• The slope of the tangent at x₀ is well defined.

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be determined using the inverse function theorem. For example, the sine function x = φ(y) is equal to sin y is the inverse function for y = f(x) = arcsin x. Hence, the derivative of function y = arcsin x is derived by:

(arc sinx) = \[f'(x) = \frac{1}{\psi'(y)} = \frac{1}{Sin'(y)} = \frac{1}{Cos(y)}\]

\[= \frac{1}{\sqrt{1 - sin^{2}y}} = \frac{1}{\sqrt{1 - sin^{2} (\text{arc sinx})}} = \frac{1}{\sqrt{1 - x^{2}}} (-1 < x < 1)\]

Using the same method, we can find the derivative of other inverse trigonometric functions like arccos x, arctan x, arccot x, arcsec x, and arccosec x.

Logarithmic Differentiation

The process of differentiating functions by first determining logarithms and then differentiation is called logarithmic differentiation.

The derivative of the logarithmic functions is known as the logarithmic derivative of first function y = f(x). These differentiable functions enable us to calculate derivatives of power- exponential functions effectively, that is the function in the form of y = u(x)ⱽ⁽ˣ⁾, where u (x)and v(x)are differentiable functions of x.

Derivative of Functions in Parametric Form

There are certain cases when the relationship between two variables is not explicit or implicit, but some linkage of a third variable with each of the two variables establishes a relation between each of the two variables. In this case, we say that the relation between is expressed through a third variable. In other words, a relation expressed between variables x and y in the form of x = f(t), y = f(t) is said to be in parametric form with t as a parameter.

We can determine the derivative of such functions through chain rule:

dy/dx = dy/dx . dt/dx

Or

dy/dx = dy/dt ÷ dx/dt (Whenever dx/dt ≠ 0)

Rolle's Theorem

Rolle’s theorem states that if f: [x,y] → R is continuous on [x, y], and differentiable on (x, y) such that f(x) = f(y), where x and y are some real numbers, then there exists some c in (x, y) such that f'(c) =0.

Mean Value Theorem

Mean Value Theorem states that if f: [x, y] → R is continuous on [x, y], and differentiable on (x, y). Then there exists some c in (x, y) such that:

\[f'(c) = \frac{f(x) - f(y)}{y - a}\]

List of the Topics Covered in Class 12 Maths Chapter 5

5.1: Introduction

5.2: Continuity

5.2.1: Algebra of Continuous Function

5.3: Differentiability

5.3.1: Derivatives of Composite Function

5.3.2: Derivatives of Implicit Function

5.3.3: Derivatives of Inverse Trigonometric Function

5.4: Exponential And Logarithmic Function

5.5: Logarithmic Differentiation

5.6: Derivatives of Function in Parametric Form

5.7: Second Order Derivative

5.8: Mean Value Theorem

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Conclusion

Continuity and Differentiability Class 12 important questions is a great way to prepare the chapter concepts. The important questions are formulated considering the latest NCERT textbook. Class 12 Maths Chapter 5 important questions pdf given on this page also includes questions which are repeatedly asked in last 10-year previous year question papers. If students find any doubt considering any topic of the chapter, they can easily clear it by practising these important questions. These questions will help students to prepare for upcoming Class 12 board exams.

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Conclusion

The important questions for Class 12 Maths Chapter 5 - Continuity and Differentiability of Vedantu cover a range of concepts and problem types related to the continuity and differentiability of functions. These questions are designed to help students build a strong foundation in these topics and develop problem-solving skills that will be useful in future studies and careers in mathematics and related fields. By practicing these questions, students will gain a deeper understanding of the definitions and properties of continuous and differentiable functions, as well as techniques for finding derivatives and solving related problems. Overall, the important questions for this chapter provide a valuable resource for students looking to master the concepts of continuity and different