Exercise 10.5 - Chapter 10 Differentiability and Methods of Differentiation 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\operatorname{Ex} 10.5$
Choose the correct or the most suitable answer from the given four alternative Question 1.
$\frac{d}{d x}\left(\frac{2}{\pi} \sin x^{\circ}\right)$ is
(a) $\frac{\pi}{180} \cos x^{\circ}$
(b) $\frac{1}{90} \cos x^{\circ}$
(c) $\frac{\pi}{90} \cos x^{\circ}$
(d) $\frac{2}{\pi} \cos x^{\circ}$
Solution:
(b) $\frac{1}{90} \cos x^{\circ}$
$
\begin{aligned}
y & =\frac{2}{\pi} \sin x \\
\frac{d y}{d x} & =\frac{2}{180^{\circ}}\left(\cos x^{\circ}\right)=\frac{1}{90} \cos x^{\circ}
\end{aligned}
$
Question 2.
If $y=f\left(x^2+2\right)$ and $f^{\prime}(3)=5$, then $\frac{d y}{d x}$ at $x=1$ is
(a) 5
(b) 25
(c) 15
(d) 10
Solution:
(d) 10
$
\begin{aligned}
f^{\prime}(3) & =5 \\
y & =f\left(x^2+2\right) \\
\frac{d y}{d x} & =f^{\prime}\left(x^2+2\right)(2 x)
\end{aligned}
$
So
$
\begin{aligned}
\frac{d y}{d x}(a t x=1) & =f^{\prime}(3)(2) \\
& =(5)(2) \\
& =10
\end{aligned}
$
Question 3.
If $\mathrm{y}=\frac{1}{4} \mathrm{u}^4, \mathrm{u}=\frac{2}{3} \mathrm{x}^3+5$, then $\frac{d y}{d x}$ is

(a) $\frac{1}{27} x^2\left(2 x^3+15\right)^3$
(b) $\frac{2}{27} x\left(2 x^3+5\right)^3$
(c) $\frac{2}{27} x^2\left(2 x^3+15\right)^3$
(d) $-\frac{2}{27} x\left(2 x^3+5\right)^3$
Solution:
(c) $\frac{2}{27} x^2\left(2 x^3+15\right)^3$
$
\begin{aligned}
y & =\frac{1}{4} u^4 ; u=\frac{2}{3} x^3+5 \\
\frac{d y}{d u} & =\frac{1}{4}\left(4 u^3\right) ; \frac{d u}{d x}=\frac{2}{3}\left(3 x^2\right)=2 x^2 \\
\frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x}=u^3\left(2 x^2\right) \\
& =\left(\frac{2}{3} x^3+5\right)^3\left(2 x^2\right) \\
\frac{d y}{d x} & =2 x^2\left[\frac{1}{3}\left(2 x^3+15\right)\right]^3 \\
& =2 x^2\left(\frac{1}{27}\right)\left(2 x^3+15\right)^3=\frac{2 x^2}{27}\left(2 x^3+15\right)^3
\end{aligned}
$

Question 4.
If $f(x)=x^2-3 x$, then the points at which $f(x)=f(x)$ are
(a) both positive integers
(b) both negative integers
(c) both irrational
(d) one rational and another irrational
Solution:
(c) both irrational
$
\begin{aligned}
& f(x)=x^2-3 x \\
& f^{\prime}(x)=2 x-3
\end{aligned}
$
Given $f(x)=f^{\prime}(x)$
$
\begin{aligned}
& \Rightarrow \mathrm{x}^2-3 \mathrm{x}=2 \mathrm{x}-3 \\
& \Rightarrow \mathrm{x}^2-5 \mathrm{x}+3=0 \\
& \mathrm{x}=\frac{5 \pm \sqrt{25-12}}{2}=\frac{5 \pm \sqrt{13}}{2} \\
& \Rightarrow \text { The roots are irrational }
\end{aligned}
$
Question 5.
If $\mathrm{y}=\frac{1}{a-z}$, then $\frac{d z}{d y}$ is
(a) $(a-z)^2$
(b) $-(z-a)^2$
(c) $(z+a)^2$
(d) $-(z+a)^2$

Solution:
(a) $(a-z)^2$
$
\begin{aligned}
y & =\frac{1}{a-z} \Rightarrow a-z=\frac{1}{y} \Rightarrow z=a-\frac{1}{y} \\
\frac{d z}{d y} & =-\left(\frac{-1}{y^2}\right)=\frac{1}{y^2}=(a-z)^2
\end{aligned}
$
Question 6.
If $y=\cos \left(\sin x^2\right)$, then $\frac{d y}{d x}$ at $x=\sqrt{\frac{\pi}{2}}$ is
(a) -2
(b) 2
(c) $-2 \sqrt{\frac{\pi}{2}}$
(d) 0
Solution:
(d) 0
$
\begin{aligned}
& \mathrm{y}=\cos \left(\sin \mathrm{x}^2\right) \\
& \frac{d y}{d x}=-\sin \left(\sin \mathrm{x}^2\right)\left[\cos \left(\mathrm{x}^2\right)\right](2 \mathrm{x}) \\
& \therefore \frac{d y}{d x} \text { at } \mathrm{x}=\sqrt{\frac{\pi}{2}}=-\sin (1)[0]=0
\end{aligned}
$
Question 7.
If $y=m x+c$ and $f(0)=f^{\prime}(0)=1$, then $f(2)$ is
(a) 1
(b) 2
(c) 3
(d) -3
Solution:
(c) 3
$
\begin{aligned}
& \mathrm{y}=\mathrm{dx}+\mathrm{c} \\
& \frac{d y}{d x}=\mathrm{m} \\
& \mathrm{y}=\mathrm{x}+\mathrm{c} \text { (i.e.) } f(\mathrm{x})=\mathrm{x}+\mathrm{c} \\
& \mathrm{y}(\mathrm{atx}=0)=\mathrm{f}(0) 0+\mathrm{c}=1 \Rightarrow \mathrm{c}=1 \\
& \mathrm{y}=\mathrm{x}+1 \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}+1 \\
& \mathrm{f}(2)=2+1=3
\end{aligned}
$
Question 8.
If $f(x)=x \tan ^{-1} x$, then $f^{\prime}(1)$ is
(a) $1+\sqrt{\frac{\pi}{4}}$
(b) $\frac{1}{2}+\frac{\pi}{4}$
(c) $\frac{1}{2}-\frac{\pi}{4}$
(d) 2
Solution:
(b) $\frac{1}{2}+\frac{\pi}{4}$

$
\begin{aligned}
& f(x)=x \tan ^{-1} x \\
& f(x)=x\left(\frac{1}{1+x^2}\right)+\tan ^{-1} x(1) \\
& f^{\prime}(1)=\frac{1}{1+1^2}+\tan ^{-1}(1)=\frac{1}{2}+\frac{\pi}{4}
\end{aligned}
$
Question 9.

$\frac{d}{d x}\left(\mathrm{e}^{\mathrm{x}+5 \log x}\right)$ is
(a) $e^x \cdot x^4(x+5)$
(b) $e^x \cdot x(x+5)$
(c) $\mathrm{e}^{\mathrm{x}}+\frac{5}{x}$
(d) $\mathrm{e}^{\mathrm{x}}-\frac{5}{x}$
Solution:
(a) $e^x \cdot x^4(x+5)$
$
\begin{aligned}
& \mathrm{y}=\mathrm{e}^{\mathrm{x}+5 \operatorname{logx}}=\mathrm{e}^{\mathrm{x}} \cdot \mathrm{e}^{5 \log x}=\mathrm{e}^{\mathrm{x}} \cdot \mathrm{e}^{\log \mathrm{x}^5} \\
& =\mathrm{x}^5 \mathrm{e}^{\mathrm{x}} \\
& \therefore \frac{d y}{d x}=\mathrm{x}^5\left(\mathrm{e}^{\mathrm{x}}\right)+\mathrm{e}^{\mathrm{x}}\left(5 \mathrm{x}^4\right) \\
& =\mathrm{e}^{\mathrm{x}} \cdot \mathrm{x}^4(\mathrm{x}+5)
\end{aligned}
$
Question 10.
If the derivative of $(a x-5) e^{3 x}$ at $x=0$ is -13 , then the value of a is
(a) 8
(b) -2
(c) 5
(d) 2
Solution:
(d) 2
$
\begin{aligned}
& \mathrm{y}=(\mathrm{ax}-5) \mathrm{e}^{3 \mathrm{x}} \\
& \frac{d y}{d x}=\mathrm{y}^{\prime}=(\mathrm{ax}-5)\left(3 \mathrm{e}^{3 \mathrm{x}}\right)+\mathrm{e}^{3 \mathrm{x}}(\mathrm{a}) \\
& =\mathrm{e}^{3 \mathrm{x}}[3 \mathrm{ax}-15+\mathrm{a}] \\
& \text { Given } \frac{d y}{d x}=-13 \mathrm{at} \mathrm{x}=0 \\
& \Rightarrow[-15+\mathrm{a}]=-13 \\
& \Rightarrow \mathrm{a}=-13+15 \\
& \mathrm{a}=2
\end{aligned}
$

Question 11.
$\mathrm{x}=\frac{1-t^2}{1+t^2}, \mathrm{y}=\frac{2 t}{1+t^2}$ then $\frac{d y}{d x}$ is
(a) $-\frac{y}{x}$
(b) $\frac{y}{x}$
(c) $-\frac{x}{y}$
(d) $\frac{x}{y}$
Solution:
(c)
Given $\mathrm{x}=\frac{1-t^2}{1+t^2}$ and $\mathrm{y}=\frac{2 t}{1+t^2}$
when we put $t=\tan \theta$

Then $x=\cos 2 \theta$ and $y=\sin 2 \theta$
$
\begin{array}{ll}
\text { Now } & \frac{d x}{d \theta}=(-\sin 2 \theta)(2), \frac{d y}{d \theta}=(\cos 2 \theta)(2) \\
\therefore & \frac{d y}{d x}=\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{2 \cos 2 \theta}{-2 \sin 2 \theta}=-\frac{x}{y}
\end{array}
$
Question 12 .
If $\mathrm{x}=\mathrm{a} \sin \theta$ and $\mathrm{y}=\mathrm{b} \cos \theta$, then $\frac{d^2 y}{d x^2}$ is
(a) $\frac{a}{b^2} \sec ^2 \theta$
(b) $-\frac{b}{a} \sec ^2 \theta$
(c) $-\frac{b}{a^2} \sec ^3 \theta$
(d) $-\frac{b^2}{a^2} \sec ^3 \theta$
Solution:
(c) $-\frac{b}{a^2} \sec ^3 \theta$
$
\begin{aligned}
x & =a \sin \theta, y=b \cos \theta \\
\frac{d x}{d \theta} & =a \cos \theta, \frac{d y}{d \theta}=b(-\sin \theta)=-b \sin \theta \\
\frac{d y}{d x} & =\frac{d y}{d \theta} / \frac{d x}{d \theta}=\frac{-b \sin \theta}{a \cos \theta}=\frac{-b}{a} \tan \theta \\
\frac{d^2 y}{d x^2} & =\frac{d}{d \theta}\left(\frac{d y}{d x}\right) / \frac{d x}{d \theta} \\
& =\frac{d}{d \theta}\left(\frac{-b}{a} \tan \theta\right) / a \cos \theta \\
& =\frac{-b}{a} \sec ^2 \theta / a \cos \theta \\
& =\frac{-b}{a^2} \sec ^3 \theta
\end{aligned}
$
Question 13.
The differential coefficient of $\log _{10} \mathrm{x}$ with respect to $\log _{\mathrm{x}} 10$ is
(a) 1
(b) $-\left(\log _{10} \mathrm{x}\right)^2$
(c) $\left(\log _{\mathrm{x}} 10\right)^2$
(d) $\frac{x^2}{100}$
Solution:
(b) $-\left(\log _{10} \mathrm{x}\right)^2$

$
\begin{aligned}
u & =\log _{10} x=\log _e x \log _{10} e \\
\frac{d u}{d x} & =\left(\log _{10} e\right)\left(\frac{1}{x}\right)=\frac{1}{x} \log _{10} e \\
v & =\left(\log _x 10\right)=\frac{\log _e 10}{\log _e x} \\
\frac{d v}{d x} & =\log _e 10\left[\frac{-1}{\left(\log _e x\right)^2} \times \frac{1}{x}\right] \\
& =\frac{-\log _e 10}{x\left(\log _e x\right)^2} \\
& =-\left(\log _{10} e \times \log _{10} e\right) \times\left(\log _e x\right)^2 \\
& =-\left(\log _{10} e \times \log _e x\right)^2=-\left(\log _{10} x\right)^2 \\
y \quad & =\log _{10 x}=\frac{1}{\log _x 10} \\
\frac{d y}{d x} & =\frac{-1}{\left(\log _x 10\right)^2}=-\left(\log _{10} x\right)^2
\end{aligned}
$
Question 14 .
If $f(x)=x+2$, then $f^{\prime}(f(x))$ at $x=4$ is
(a) 8
(b) 1
(c) 4
(d) 5
Solution:
(b) 1
$f(x)=x+2$
$f^{\prime}(x)=1$
$f^{\prime}(x)($ at $x=4)=1$
Question 15.
If $\mathrm{y}=\frac{(1-x)^2}{x^2}$, then $\frac{d y}{d x}$ is
(a) $\frac{2}{x^2}+\frac{2}{x^3}$
(b) $-\frac{2}{x^2}+\frac{2}{x^3}$
(c) $-\frac{2}{x^2}-\frac{2}{x^3}$
(d) $-\frac{2}{x^3}+\frac{2}{x^2}$
Solution:

(d) $-\frac{2}{x^3}+\frac{2}{x^2}$
$
\begin{aligned}
y & =\frac{(1-x)^2}{x^2}=\frac{1-2 x+x^2}{x^2} \\
y & =\frac{1}{x^2}-\frac{2}{x}+1 \\
y^{\prime} & =\frac{d y}{d x}=\frac{-2}{x^3}-2\left(\frac{-1}{x^2}\right) \\
& =-\frac{2}{x^3}+\frac{2}{x^2}
\end{aligned}
$
Question 16.
If $\mathrm{pv}=81$, then $\frac{d p}{d v}$ at $\mathrm{v}=9$ is
(a) 1
(b) -1
(c) 2
(d) -2
Solution:
(b) -1
$
\begin{aligned}
p v & =81 \\
p(1)+v\left(\frac{d p}{d v}\right) & =0 \\
\frac{d p}{d v} & =\frac{-p}{v} \quad\left(\text { Given } p v=81 \text { so at } v=9, p=\frac{81}{9}=9\right) \\
\therefore \quad \frac{d p}{d v}(a t v=9) & =\frac{-9}{9}=-1 \quad
\end{aligned}
$
Question 17.
If $f(x)=\left\{\begin{array}{clc}x-5 & \text { if } & x \leq 1 \\ 4 x^2-9 & \text { if } & 1<x<2 \\ 3 x+4 & \text { if } & x \geq 2\end{array}\right.$ then the right hand derivative of $f(x)$ at $\mathrm{x}=2$ is
(a) 0
(b) 2
(c) 3
(d) 4
Solution:

(c) 3
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{f(x+2)-f(2)}{x} & \\
= & \lim _{h \rightarrow 0} \frac{3(2+h)+4-(6+4)}{h} \\
= & \lim _{h \rightarrow 0} \frac{6+3 h+4-6-4}{h}=\lim _{h \rightarrow 0} \frac{3 h}{h}=3
\end{aligned}
$
Question 18 .
It is given that $\mathrm{f}^{\prime}(\mathrm{a})$ exists, then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ is
(a) $f(a)-a f^{\prime}(a)$
(b) $f^{\prime}$ (a)
(c) $-f^{\prime}(a)$
(d) $f(a)+a f^{\prime}(a)$
Solution:
(a) $f(a)-a f^{\prime}(a)$
Question 19.
If $f(\mathrm{x})=\left\{\begin{array}{lll}x+1, & \text { when } & x<2 \\ 2 x-1 & \text { when } & x \geq 2\end{array}\right.$ then $f^{\prime}(2)$ is
(a) 0
(b) 1
(c) 2
(d) does not exist
Solution:
(d) does not exist

Question 20.
If $g(x)=\left(x^2+2 x+3\right) f(x)$ and $f(0)=5$ and $\lim _{x \rightarrow 0} \frac{f(x)-5}{\boldsymbol{x}}=4$ then $g^{\prime}(\theta)$ is
(a) 20
(b) 22
(c) 18
(d) 12

Solution:
(b) 22
$
\begin{aligned}
& g^{\prime}(0)=\lim _{x \rightarrow 0} \frac{g(o+h)-g(o)}{h}=\lim _{h \rightarrow 0}\left(h^2+2 h\right) f(h)+\left[\frac{3 f(h)-2 f(0)}{h}\right] \\
& =\lim _{h \rightarrow 1} \frac{h(h+2) f(h)}{h}+\lim _{h \rightarrow 0} \frac{3(f(h)-5)}{h} \\
& =2 f(0)+3(4) \\
& =2(5)+12=22
\end{aligned}
$
Question 21.
If $f(x)=\left\{\begin{array}{cc}x+2, & -1<x<3 \\ 5 & \boldsymbol{x}=\mathbf{3} \\ \mathbf{8}-\boldsymbol{x} & \boldsymbol{x}>\mathbf{3}\end{array}\right.$, then at $\mathrm{x}=3, \mathrm{f}^{\prime}(\mathrm{x})$ is
(a) 1
(b) -1
(c) 0
(d) does not exist
Solution:
(d) does not exist

Question 22.
The derivative of $f(x)=x|x|$ at $x=-3$ is
(a) 6
(b) -6
(c) does not exist
(d) 0
Solution:
(a) 6
$f(x)=x|x|$
$f(x)=x(-x) \Rightarrow f(x)=-x^2$
$f^{\prime}(\mathrm{x})=-(2 \mathrm{x})$
$f^{\prime}(-3)=-(2)(-3)=6$

Question 23.
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ccc}2 a-x, & \text { for } & -a<x<a \\ 3 x-2 a & \text { for } & x \geq a\end{array}\right.$, then which one of the following is true?
(a) $f(x)$ is not differentiable at $x=a$
(b) $f(x)$ is discontinuous at $x=a$
(c) $f(x)$ is continuous for all $x$ in $R$
(d) $f(x)$ is differentiable for all $x \geq$ a
Solution:
(a) $f(x)$ is not differentiable at $x=a$

Question 24.
If $f(x)=\left\{\begin{array}{cc}a x^2-b, & -1<x<1 \\ \frac{1}{|x|}, & \text { elsewhere }\end{array}\right.$ is differentiable at $\mathrm{x}=1$, then
(a) $a=\frac{1}{2}, b=\frac{-3}{2}$
(b) $a=\frac{-1}{2}, b=\frac{3}{2}$
(c) $a=-\frac{1}{2}, b=-\frac{3}{2}$
(d) $a=\frac{1}{2}, b=\frac{3}{2}$
Solution:
(c) $a=-\frac{1}{2}, b=-\frac{3}{2}$

Question 25.
Then number of points in $\mathrm{R}$ in which the function $f(x)=|\mathrm{x}-1|+|\mathrm{x}-3|+\sin \mathrm{x}$ is not differentiable, is
(a) 3
(b) 2
(c) 1
(d) 4
Solution:
(b) 2
$f(x)=|x-1|+|x-3|+\sin \mathrm{x}$ is not differentiable at $\mathrm{x}=1$, and $\mathrm{x}=3$