Class 12 Important Questions: CBSE 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)

$f\left(x\right)=\{\begin{array}{r}2x-7x⩾{\displaystyle \frac{7}{2}}\\ 7-2xx<{\displaystyle \frac{7}{2}}\end{array}$

Hence, for the value $x={\displaystyle \frac{7}{2}}$, 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,

$g\left(x\right)=\{\begin{array}{r}\left(x-1\right)+\left(x+1\right)x>1\\ \left(1-x\right)+\left(x+1\right)-1⩽x⩽1\\ \left(1-x\right)-\left(x+1\right)x<-1\end{array}$

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

$f\left(x\right)=\{\begin{array}{r}x-3x>3\\ 3-xx⩽3\end{array}$

For $x=-1$, we have

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

$\begin{array}{r}{f}^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}\left(3-x\right)\\ =-1\end{array}$

$f^{\prime }\left(-1\right)=-1$

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

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, $\text{f}\left(\text{x}\right)\text{ = }\{\begin{array}{r}\lambda x-3x<2\\ \text{4}x=2\\ \text{2x}x>2\end{array}$ 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 $x=2$.

$\begin{array}{r}\underset{x\to 2^{-}}{lim}\lambda x-3=\underset{x\to 2}{lim}4\\ \Rightarrow \lambda \left(2\right)-3=4\\ \Rightarrow \lambda ={\displaystyle \frac{7}{2}}\end{array}$

Therefore, $\lambda ={\displaystyle \frac{7}{2}}$.

7. For what value of $\text{K}$, $$\text{f}\left(\text{x}\right)\text{ = }\{\begin{array}{r}{\displaystyle \frac{\text{tan3x}}{\text{sin2x}}}\text{x}\ne \text{0}\\ \text{2K}\text{x = 0}\end{array}$$ is continuous $\mathrm{\forall }\text{x}\in \text{R}$.

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

$$\begin{array}{r}\Rightarrow \underset{x\to 0}{lim}{\displaystyle \frac{\tan 3x}{\sin 2x}}=2K\\ \Rightarrow \underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\tan 3x}{3x}}\cdot 3}{{\displaystyle \frac{\sin 2x}{2x}}\cdot 2}}=2K\\ \Rightarrow {\displaystyle \frac{3}{2}}=2K\\ \Rightarrow K={\displaystyle \frac{3}{4}}\end{array}$$

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 $x$,

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

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

$\begin{array}{r}\Rightarrow {\displaystyle \frac{dy}{dx}}\cdot {\displaystyle \frac{dx}{dz}}=\cos x\cdot {\displaystyle \frac{1}{-\sin x}}\\ \Rightarrow {\displaystyle \frac{dy}{dz}}=-{\displaystyle \frac{\cos x}{\sin x}}\\ \Rightarrow {\displaystyle \frac{dy}{dz}}=-\cot x\end{array}$

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

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

${\displaystyle \frac{d}{dx}}\left(f\left(x\right)\right)=2xg\left(x\right)+g^{\prime }\left(x\right)x^2$

Therefore, $f^{\prime }\left(1\right)$ should be,

$\begin{array}{r}{f}^{\prime }\left(1\right)=2\left(1\right)\left(6\right)+\left(3\right){\left(1\right)}^2\\ =12+3\\ =15\end{array}$

10. Write the derivative of the following functions:

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

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

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{d}{dx}}\left({\log }_3\left(3x+5\right)\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{3}{\left(3x+5\right)\ln \left(3\right)}}\end{array}$

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

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

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

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{d}{dx}}\left({\log }_2\left(x\right)\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{1}{\left(x\right)\ln \left(2\right)}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{e^{{\log }_2\left(x\right)}}{x\ln \left(2\right)}}\end{array}$

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

Ans: 

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

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{d}{dx}}\left({\left(x-1\right)}^6\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=6{\left(x-1\right)}^5\end{array}$

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

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

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{d}{dx}}\left({\sec }^{-1}\sqrt{x}+{\csc }^{-1}\sqrt{x}\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{\left(\sqrt{x}\right)\sqrt{{\left(\sqrt{x}\right)}^2-1}}}+{\displaystyle \frac{-1}{\left(\sqrt{x}\right)\sqrt{{\left(\sqrt{x}\right)}^2-1}}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{\sqrt{x\left(x-1\right)}}}-{\displaystyle \frac{1}{\sqrt{x\left(x-1\right)}}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=0\end{array}$

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

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

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{d}{dx}}\left({\sin }^{-1}\left(x^{7/2}\right)\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{\sqrt{1-{\left(x^{7/2}\right)}^2}}}\cdot \left({\displaystyle \frac{7}{2}}{\left(x\right)}^{{\displaystyle \frac{5}{2}}}\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{7}{2}}\cdot {\displaystyle \frac{\sqrt{x^5}}{\sqrt{1-x^7}}}\end{array}$

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

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

$\begin{array}{r}\Rightarrow y={\displaystyle \frac{\ln 5}{\ln x}}\\ \Rightarrow {\displaystyle \frac{1}{y}}={\displaystyle \frac{\ln x}{\ln 5}}\end{array}$

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{\ln x}{\ln 5}}\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{1}{y}}\right)={\displaystyle \frac{1}{x\ln 5}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{{\log }_{x}5}{x\ln 5}}\end{array}$

Short Questions and Answers (4 Marks Questions)

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

(i) $\mathbf{\text{f}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\{\begin{array}{r}{\displaystyle \frac{\text{x - }\left|\text{x}\right|}{\text{x}}}\text{x}\ne \text{0}\\ \text{2}\text{x = 0}\end{array}$ at $\mathbf{\text{x = 0}}$

Ans: Given, $f\left(x\right)=\{\begin{array}{r}{\displaystyle \frac{x-\left|x\right|}{x}}x\ne 0\\ 2x=0\end{array}$

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

Left-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}{\displaystyle \frac{x-\left(-x\right)}{x}}\\ =\underset{x\to 0^{-}}{lim}{\displaystyle \frac{2x}{x}}\\ =2\end{array}$

Middle Value is,

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

Right-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}{\displaystyle \frac{x-\left(x\right)}{x}}\\ =0\end{array}$

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

(ii) $\mathbf{\text{g}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\{\begin{array}{r}{\displaystyle \frac{\text{sin2x}}{\text{3x}}}\text{x}\ne \text{0}\\ {\displaystyle \frac{\text{3}}{\text{2}}}\text{x = 0}\end{array}$ at $\mathbf{\text{x = 0}}$

Ans: Given, $g\left(x\right)=\{\begin{array}{r}{\displaystyle \frac{\sin 2x}{3x}}x\ne 0\\ {\displaystyle \frac{3}{2}}x=0\end{array}$

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

Left-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}{\displaystyle \frac{\sin 2x}{3x}}\\ =\underset{x\to 0^{-}}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 2x}{2x}}\cdot 2}{{\displaystyle \frac{3x}{x}}}}\\ ={\displaystyle \frac{2}{3}}\end{array}$

Middle Value is,

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

Right-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}{\displaystyle \frac{\sin 2x}{3x}}\\ =\underset{x\to 0^{+}}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 2x}{2x}}\cdot 2}{{\displaystyle \frac{3x}{x}}}}\\ ={\displaystyle \frac{2}{3}}\end{array}$

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

(iii) $\mathbf{\text{f}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\{\begin{array}{r}{\text{x}}^{\text{2}}\text{cos}\left({\displaystyle \frac{\text{1}}{\text{x}}}\right)\text{x}\ne \text{0}\\ \text{0}\text{x = 0}\end{array}$ at $\mathbf{\text{x = 0}}$

Ans: Given, $f\left(x\right)=\{\begin{array}{r}{x}^2\cos \left({\displaystyle \frac{1}{x}}\right)x\ne 0\\ 0x=0\end{array}$

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

Left-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}{x}^2\cos {\displaystyle \frac{1}{x}}\\ =0\end{array}$

Middle Value is,

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

Right-Hand limit is,

$\begin{array}{r}\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}{x}^2\cos {\displaystyle \frac{1}{x}}\\ =0\end{array}$

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

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

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

$\begin{array}{r}\underset{x\to 1^{-}}{lim}\left(\left|x\right|+\left|x-1\right|\right)=\underset{x\to 1^{-}}{lim}\left(x+\left(1-x\right)\right)\\ =1\end{array}$

Middle Value is,

$\begin{array}{r}f\left(1\right)=\left(1\right)+\left(1-1\right)\\ =1\end{array}$

Right-Hand limit is,

$\begin{array}{r}\underset{x\to 1^{+}}{lim}\left(\left|x\right|+\left|x-1\right|\right)=\underset{x\to 1^{+}}{lim}\left(x+\left(x-1\right)\right)\\ =1\end{array}$

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

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

(v) $\mathbf{\text{f}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\{\begin{array}{r}\text{x - }\left[\text{x}\right]\text{x}\ne \text{1}\\ \text{0}\text{x = 1}\end{array}$ at $\mathbf{\text{x = 1}}$

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

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

Left-Hand limit is,

$\begin{array}{r}\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\left(x-\left[x\right]\right)\\ =1-\left(0\right)\\ =1\end{array}$

Middle Value is,

$\begin{array}{r}f\left(1\right)=\left(1\right)-\left(1\right)\\ =0\end{array}$

Right-Hand limit is,

$\begin{array}{r}\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}\left(x-\left[x\right]\right)\\ =1-\left(1\right)\\ =0\end{array}$

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, $$\mathbf{\text{f}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\{\begin{array}{r}\text{3}{\text{x}}^{\text{2}}\text{ - kx + 5}\text{0}⩽\text{x < 2}\\ \text{1 - 3x}\text{2}⩽\text{x}⩽\text{3}\end{array}$$ is continuous $\mathrm{\forall }\mathbf{\text{x}}\in \left[\mathbf{\text{0,3}}\right]$.

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,

$\begin{array}{r}\underset{x\to 2^{-}}{lim}3x^2-kx+5=\underset{x\to 2^{+}}{lim}1-3x\\ \Rightarrow 3{\left(2\right)}^2-k\left(2\right)+5=1-3\left(2\right)\\ \Rightarrow 12-2k+5=-5\\ \Rightarrow 2=2k\\ \Rightarrow k=1\end{array}$

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 $\mathrm{\forall }x\in \left[0,3\right]$.

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

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

$\underset{x\to -2^{-}}{lim}\left({\displaystyle \frac{x+2}{\left|x+2\right|}}+a\right)=a+b=\underset{x\to -2^{+}}{lim}\left({\displaystyle \frac{x+2}{\left|x+2\right|}}+2b\right)$

Left-Hand limit is,

$\begin{array}{r}\Rightarrow \underset{x\to -2^{-}}{lim}\left({\displaystyle \frac{x+2}{\left|x+2\right|}}+a\right)=a+b\\ \Rightarrow \underset{x\to -2^{-}}{lim}\left({\displaystyle \frac{x+2}{-\left(x+2\right)}}+a\right)=a+b\\ \Rightarrow -1+a=a+b\\ \Rightarrow b=-1\end{array}$

Right-Hand limit is,

$\begin{array}{r}\Rightarrow a+b=\underset{x\to -2^{+}}{lim}\left({\displaystyle \frac{x+2}{\left|x+2\right|}}+2b\right)\\ \Rightarrow a+b=\underset{x\to -2^{+}}{lim}\left({\displaystyle \frac{x+2}{x+2}}+2b\right)\\ \Rightarrow a+b=1+2b\\ \Rightarrow a-1=1-2\\ \Rightarrow a=0\end{array}$

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

14. Prove that $\mathbf{\text{f}}\left(\mathbf{\text{x}}\right)\mathbf{\text{ = }}\left|\mathbf{\text{x + 1}}\right|$ 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,

$\begin{array}{r}\underset{x\to -1^{-}}{lim}\left|x+1\right|=\left|-1+1\right|=\underset{x\to -1^{+}}{lim}\left|x+1\right|\\ \Rightarrow \underset{x\to -1^{-}}{lim}-\left(x+1\right)=\left|0\right|=\underset{x\to -1^{+}}{lim}\left(x+1\right)\\ \Rightarrow 0=0=0\end{array}$

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.

$\begin{array}{r}\Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{f\left(-1+h\right)-f\left(-1\right)}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{f\left(-1+h\right)-f\left(-1\right)}{h}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{\left|\left(-1+h\right)+1\right|-\left|-1+1\right|}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{\left|\left(-1+h\right)+1\right|-\left|-1+1\right|}{h}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{\left(-1+h+1\right)}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{-\left(-1+h+1\right)}{h}}\\ \Rightarrow 1\ne -1\end{array}$

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

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

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

$\begin{array}{r}\underset{h\to 0^{+}}{lim}{\displaystyle \frac{f\left(0+h\right)-f\left(0\right)}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{f\left(0+h\right)-f\left(0\right)}{h}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{{\left(0+h\right)}^p\sin \left({\displaystyle \frac{1}{0+h}}\right)-0}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{{\left(0+h\right)}^p\sin \left({\displaystyle \frac{1}{0+h}}\right)-0}{h}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{h^p\sin \left({\displaystyle \frac{1}{h}}\right)}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{h^p\sin \left({\displaystyle \frac{1}{h}}\right)}{h}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{h^{p-1}\sin \left({\displaystyle \frac{1}{h}}\right)}{h\cdot {\displaystyle \frac{1}{h}}}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{h^{p-1}\sin \left({\displaystyle \frac{1}{h}}\right)}{h\cdot {\displaystyle \frac{1}{h}}}}\\ \Rightarrow \underset{h\to 0^{+}}{lim}{\displaystyle \frac{h^{p-1}}{h}}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{h^{p-1}}{h}}\end{array}$

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

Ans: Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dx}}\left({\tan }^{-1}\left({\displaystyle \frac{2x}{1-x^2}}\right)\right)+{\displaystyle \frac{d}{dx}}\left({\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{2\left(1+{\left({\displaystyle \frac{2x}{1-x^2}}\right)}^2\right)}}\cdot {\displaystyle \frac{2\left(1-x^2\right)-2x\left(-2x\right)}{{\left(1-x^2\right)}^2}}+{\displaystyle \frac{1}{1+{\left({\displaystyle \frac{1}{x}}\right)}^2}}\cdot \left(-{\displaystyle \frac{1}{x^2}}\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2\left(1-x^2\right)+4x^2}{2\left({\left(1-x^2\right)}^2+4x^2\right)}}-{\displaystyle \frac{1}{x^2+1}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\left(1+x^2\right)}{{\left(1+x^2\right)}^2}}-{\displaystyle \frac{1}{x^2+1}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=0\end{array}$

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

Ans: Let $x=\cos t$

We have,

$\begin{array}{r}\Rightarrow y=\sin \left[2{\tan }^{-1}\sqrt{{\displaystyle \frac{1-\cos t}{1+\cos t}}}\right]\\ \Rightarrow y=\sin \left[2{\tan }^{-1}\sqrt{\tan \left({\displaystyle \frac{t}{2}}\right)}\right]\\ \Rightarrow y=\sin \left[2\left({\displaystyle \frac{t}{2}}\right)\right]\\ \Rightarrow y=\sin t\\ \Rightarrow y=\sqrt{1-{\cos }^{2}t}\\ \Rightarrow y=\sqrt{1-x^2}\end{array}$

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{2}}{\left(1-x^2\right)}^{-{\displaystyle \frac{1}{2}}}\left(-2x\right)\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{x}{\sqrt{1-x^2}}}\end{array}$

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

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

$\begin{array}{r}{5}^x\ln 5+5^y\ln 5{\displaystyle \frac{dy}{dx}}=5^{x+y}\ln 5\left(1+{\displaystyle \frac{dy}{dx}}\right)\\ \Rightarrow 5^x+{\displaystyle \frac{dy}{dx}}{5}^y=\left(1+{\displaystyle \frac{dy}{dx}}\right)5^{x+y}\\ \Rightarrow 5^x+{\displaystyle \frac{dy}{dx}}{5}^y=5^{x+y}+{\displaystyle \frac{dy}{dx}}{5}^{x+y}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}\left(5^{x+y}-5^y\right)+\left(5^{x+y}-5^x\right)=0\\ \Rightarrow {\displaystyle \frac{dy}{dx}}+{\displaystyle \frac{5^{x+y}-5^x}{5^{x+y}-5^y}}=0\\ \Rightarrow {\displaystyle \frac{dy}{dx}}+{\displaystyle \frac{5^y}{5^x}}=0\\ \Rightarrow {\displaystyle \frac{dy}{dx}}+5^{y-x}=0\end{array}$

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

Ans: Differentiating with respect to $x$

$\begin{array}{r}\Rightarrow -{\displaystyle \frac{2xy}{\sqrt{1-y^2}}}{\displaystyle \frac{dy}{dx}}+\sqrt{1-y^2}-{\displaystyle \frac{2xy}{\sqrt{1-x^2}}}+{\displaystyle \frac{dy}{dx}}\sqrt{1-x^2}=0\\ \Rightarrow {\displaystyle \frac{dy}{dx}}\left(\sqrt{1-x^2}-{\displaystyle \frac{2xy}{\sqrt{1-y^2}}}\right)={\displaystyle \frac{2xy}{\sqrt{1-x^2}}}-\sqrt{1-y^2}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=-{\displaystyle \frac{{\displaystyle \frac{2xy-\left(\sqrt{1-y^2}\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}}}{{\displaystyle \frac{2xy-\left(\sqrt{1-y^2}\sqrt{1-x^2}\right)}{\sqrt{1-y^2}}}}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=-\sqrt{{\displaystyle \frac{1-y^2}{1-x^2}}}\end{array}$

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

Ans: Let,

$\begin{array}{r}x=\sin m\\ m={\sin }^{-1}x\\ y=\sin n\\ n={\sin }^{-1}y\end{array}$

We have,

$\begin{array}{r}\Rightarrow \sqrt{1-{\sin }^{2}m}+\sqrt{1-{\sin }^{2}n}=a\left(\sin m-\sin n\right)\\ \Rightarrow \cos m+\cos n=a\left(2\cos {\displaystyle \frac{m+n}{2}}\sin {\displaystyle \frac{m-n}{2}}\right)\\ \Rightarrow 2\cos {\displaystyle \frac{m+n}{2}}\cos {\displaystyle \frac{m-n}{2}}=2a\cos {\displaystyle \frac{m+n}{2}}\cos {\displaystyle \frac{m-n}{2}}\\ \Rightarrow {\displaystyle \frac{1}{a}}=\tan {\displaystyle \frac{m-n}{2}}\\ \Rightarrow {\tan }^{-1}{\displaystyle \frac{1}{a}}={\displaystyle \frac{m-n}{2}}\\ \Rightarrow 2{\tan }^{-1}{\displaystyle \frac{1}{a}}={\sin }^{-1}x-{\sin }^{-1}y\end{array}$

Differentiating with respect to $x$,

$\begin{array}{r}\Rightarrow 0={\displaystyle \frac{1}{\sqrt{1-x^2}}}-{\displaystyle \frac{1}{\sqrt{1-y^2}}}{\displaystyle \frac{dy}{dx}}\\ \Rightarrow {\displaystyle \frac{dy}{dx}}=\sqrt{{\displaystyle \frac{1-y^2}{1-x^2}}}\end{array}$

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