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RD Sharma Solutions Class 12 Mathematics Chapter 10 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 10 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 10 FBQ

Updated on 20 Jan 2022, 06:34 PM IST

The RD Sharma Class 12 chapter 10 exercise FBQ arrangement – Our experts plan Differentiation to help ensure students in understanding the thoughts solicited in this chapter and procedures to deal with issues in a more restricted period.

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RD Sharma Class 12 Solutions Chapter 10 FBQ Differentiation - Other Exercise
Differentiation Excercise: FBQ
RD Sharma Chapter-wise Solutions

In this RD Sharma class 12 solutions FBQ Chapter 10, Differentiation is a process in Maths, where we find the quick rate of function change based on one of its variables. Class 12 RD Sharma chapter 10 exercise FBQ solution consists of various topics that play a major role in finding the rate of displacement. At Careers360, RD Sharma class 12 chapter 10 exercise FBQ Solutions help students who seek to get a decent scholarly score in the exam.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 10 FBQ Differentiation - Other Exercise

Differentiation Excercise: FBQ

Differentiation exercise Fill in the blanks question 1

Answer: $\left(\frac{d y}{d u}\right)_{u=-1}=2$
Hint: $y=\left(\begin{array}{cc} u^{2} & u \geq 0 \\ -u^{2} & u<0 \end{array}\right)$
Given: $\mathrm{y}=\mathrm{u}|\mathrm{u}|$
Solution:
$\begin{aligned} &y=\left(\begin{array}{cc} u^{2} & u \geq 0 \\ -u^{2} & u<0 \end{array}\right) \\\\ &\frac{d y}{d u}=\left(\begin{array}{cc} 2 u^{2} & u \geq \ 0 \\\\ -2 u^{2} & u<0 \end{array}\right) \end{aligned}$
$\begin{aligned} &\left(\frac{d y}{d u}\right)_{-1}=(-2(-1)\\ \end{aligned}$
$=2$

Differentiation exercise Fill in the blanks question 2

Answer: $\left(\frac{d y}{d x}\right)_{-1}=1 \text { and }\left(\frac{d y}{d x}\right)_{x=1}=3$
Hint: $y=\left(\begin{array}{cc} 3 x & x \geq 0 \\ x & x<0 \end{array}\right)$
Given: $y=2 x+|x|$
Solution:
$\begin{aligned} &y=\left(\begin{array}{ll} 2 x+x & x \geq 0 \\ 2 x-x & x<0 \end{array}\right) \\\\ &y=\left(\begin{array}{cc} 3 x & x \geq 0 \\ x & x<0 \end{array}\right) \\\\ &\frac{d y}{d x}=\left(\begin{array}{ll} 3 & x>0 \\ 1 & x<0 \end{array}\right) \end{aligned}$
$\left(\frac{d y}{d x}\right)_{-1}=1 \text { and }\left(\frac{d y}{d x}\right)_{1}=3$

Differentiation exercise Fill in the blanks question 3

Answer: $f^{1}(2)=3$
Hint: $f(x)=\left(\begin{array}{ll} x-x^{2} & x^{2}-x \leq 0 \\ x^{2}-x & x^{2}-x \geq 0 \end{array}\right)$
Given: $f(x)=\left|x^{2}-x\right|$
Solution: $f(x)=\left|x^{2}-x\right|$
$f(x)=\left(\begin{array}{cc} -\left(x^{2}-x\right) & x^{2}-x<0 \\ x^{2}-x & x^{2}-x \geq 0 \end{array}\right)$
$\begin{aligned} &f(x)=\left(\begin{array}{cc} \left(x-x^{2}\right) & x^{2}-x<0 \\ x^{2}-x & x^{2}-x \geq 0 \end{array}\right) \\\\ &f^{{}'}(x)=\left(\begin{array}{cc} (1-2 x) & x^{2}-x<0 \\ 2 x-1 & x^{2}-x \geq 0 \end{array}\right) \end{aligned}$
$\begin{aligned} &f^{{}'}(2)=2 x-1 \\\\ &=2(2)-1 \\\\ &=3 \end{aligned}$

Differentiation exercise Fill in the blanks question 4

Answer: $k=1$
Hint: $\frac{d y}{d x}=\cos x^{0}$
Given: $y=\sin x^{0} \quad \text { and } \frac{d y}{d x}=k \cos x^{0}$
Solution: $y=\sin x^{0}$ $\because \hspace{1cm}x^{0}=\frac{\pi}{180}x$
Differentiating y.u.1 to x
$\begin{aligned} &\frac{d y}{d x}= \frac{\pi}{180}\cos x^{0}\ldots \ldots\ldots(i)\\ &\text { And } \frac{d y}{d x}=k \cos x^{0}\ldots\ldots(ii) \end{aligned}$
By (i) and (ii)
$k=\frac{\pi}{180}$

Differentiation exercise Fill in the blanks question 5

Answer: $f^{1}(0)=0$
Hint: $f^{1}(x)=g^{1}(x)>g^{1}(x)$
Solution: $f(x)=e^{x} g(x)$
Differentiating $f(x)$ we get
$\begin{aligned} &f^{1}(x)=e^{x} g^{1}(x)+e^{x} g(x) \\\\ &f^{1}(0)=e^{0} g^{1}(0)+e^{0} g(0) \end{aligned}$
$\begin{aligned} &=1(1)+1(2) \\\\ &=1+2 \\\\ &=3 \end{aligned}$

Differentiation exercise Fill in the blanks question 6

Answer:$f(-3)=3$
Hint:
$\begin{aligned} &|x+2|=\left(\begin{array}{ll} (x+2) & x \geq-2 \\ -x>2 & x < 2 \end{array}\right) \\\\ =& f(x)=3|x+2| \end{aligned}$
Solution: $f(x)=3|x+2|$
$\begin{aligned} &f(x)=\left(\begin{array}{cc} 3(x+2) & x \geq-2 \\ -3(x+2) & x < -2 \end{array}\right) \\\\ &f^{1}(x)=\left(\begin{array}{cc} 3 & x \geq-2 \\ -3 & x < -2 \end{array}\right) \\\\ &f^{1}(-3)=-3 \end{aligned}$

Differentiation exercise Fill in the blanks question 7

Answer: 2
Hint: using chain rule
Given:$f(1)=3$
Solution: $\frac{d}{d x}=\ln {f(e^{x}+2x)}$
$\begin{aligned} &=\frac{1}{f\left(e^{x}+2 x\right)}f^{{}'}\left(e^{x}+2 x\right)\times(e^{x}+2) \\\\ &=\left(e^{x}+2 \right) \frac{f^{{}'}\left(e^{0}+2 x\right)}{f\left(e^{0}+2 x\right)} \end{aligned}$
$\begin{aligned} &=3 \frac{f^{{}'}(1)}{f(1)} \\\\ &=3 \times \frac{2}{3}=2 \end{aligned}$

Differentiation exercise Fill in the blanks question 8

Answer: $f^{{}'}(x)=\left(\begin{array}{cc} 2 x & x \geq 0 \\ -2 x & x^{2} \leq 0 \end{array}\right)$
Hint:$x|x|=\left(\begin{array}{cc} x^{2} & x\geq 0 \\ -x^{2} & x < 0 \end{array}\right)$
Given:$f|x|=x|x|$
Solution:
$f|x|=x|x|$
$\begin{aligned} &f(x)=\left(\begin{array}{cc} x^{2} & x\geq 0 \\ -x^{2} & x < 0 \end{array}\right) \\\\ &f^{1}(x)=\left(\begin{array}{cc} 2 x & x\geq 0 \\ -2 x & x < 0 \end{array}\right) \end{aligned}$

Differentiation exercise Fill in the blanks question 9

Answer: $f^{1}(2)=0$
Hint: the mod function
Given: $f(x)=|x-1|+|x-3|$
Solution:
$\begin{aligned} &|x-1|=\left(\begin{array}{cc} u-1 & u>1 \\ -(u-1) & u<1 \end{array}\right) \\\\ &|x-3|=\left(\begin{array}{cc} x-3 & x>3 \\ -(x-3) & x \leq 3 \end{array}\right) \end{aligned}$
$\begin{aligned} &\text { Case- } 1: x<1 \\ &\begin{aligned} f(x)=&|x-1|+|x-3| \\ &=-x+1-x>3 \\ &=-2 x>4 \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Case- } 2: 1 \leq \mathrm{x}<3 \\ &\begin{aligned} f(x)=&|x-1|-(x-3) \\ &=x-1-x+3=2 \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Case. } 3\\ &f(x)=x-1>x+3=2 x-4\\\\ &f(x)=\left(\begin{array}{cc} -2 x+4 & x<1 \\ 2 & 1 \leq x<3 \\ -(x-3) & x \leq 3 \end{array}\right) \end{aligned}$
$\begin{aligned} &f^{{}'}(x)=\left(\begin{array}{cc} -2 & x<1 \\ 0 & 1 \leq x<3 \\ 2 & x \geq 3 \end{array}\right) \\ &f^{{}'}(x)=0 \end{aligned}$

Differentiation exercise Fill in the blanks question 10

Answer: $\frac{1+\sqrt{3}}{2}$
Hint: $\cos x<\sin x \text { where } 0 \leq x \leq \frac{\pi}{4}$
Given: $f(x)=|\cos x-\sin x|$
Solution: $f(x)=|\cos x-\sin x|$
$\begin{array}{ll} f(x)=\cos x-\sin x & 0 \leq x \leq \pi / 4 \\\\ f(x)=-(\cos x-\sin x) & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \end{array}$
$\begin{aligned} &f(x)=-\cos x-\sin x \quad 0 \leq x \leq \pi / 4 \\\\ &f(x)=\cos x+\sin x \quad \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \\\\ &=\frac{1+\sqrt{3}}{2} \end{aligned}$

Differentiation exercise Fill in the blanks question 11

Answer: $f^{{}'}\left(\frac{\pi}{x}\right)=\left(\frac{1}{\sqrt{2}}\right)$
Hint: $f(x)=\left(\begin{array}{cc} \cos x & 0 \leq x \leq \frac{\pi}{2} \\\\ -\cos x & \frac{\pi}{2}<x<\pi \end{array}\right)$
Given:$f(x)=|\cos x|$
Solution:
$f(x)=|\cos x|$
$f(x)=\left(\begin{array}{cc} \cos x & 0 \leq u \leq \frac{\pi}{2} \\\\ -\cos x & \frac{\pi}{2}<u<\pi \end{array}\right)$
$\begin{aligned} f^{{}'}(x) &=\left(\begin{array}{cc} -\sin x & 0 \leq x \leq \frac{\pi}{2} \\\\ \sin x & \frac{\pi}{2}<x<\pi \end{array}\right) \\\\ f^{1}\left(\frac{\pi}{2}\right) &=-\sin \frac{\pi}{2} \\ &=-\frac{1}{\sqrt{2}} \end{aligned}$

Differentiation exercise Fill in the blanks question 12

Answer: $\frac{2}{3 x}$
Hint: $\text { take } u=x^{2} \text { and } v=x^{3}$
Given:The derivative of $x^{2}$ and $x^{3}$
Solution:
$\begin{aligned} &\text { Let } u=x^{2} \\\\ &\frac{d u}{d x}=2 x \\\\ &\text { And let } v=x^{3} \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=3 x^{2} \\\\ &\frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{2 x}{3 x^{2}}=\frac{2}{3 x} \end{aligned}$

Differentiation exercise Fill in the blanks question 14

Answer: $f^{1}\left(\frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}\right)$
Hint:$f^{1}(x)=\left(\begin{array}{cc} \sin x & 0 \leq x \leq \frac{\pi}{2} \\\\ -\sin x & \frac{\pi}{2}<x<\pi \end{array}\right)$
Given: $f(x)=|\sin x|$
Solution:
$\begin{aligned} &f(x)=\left(\begin{array}{cc} \sin x & 0 \leq x \leq \frac{\pi}{2} \\\\ -\sin x & \frac{\pi}{2}<x<\pi \end{array}\right) \\\\ &f^{{}'}(x)=\left(\begin{array}{cc} \cos x & 0 \leq x \leq \frac{\pi}{2} \\\\ -\cos x & \frac{\pi}{2}<x<\pi \end{array}\right) \end{aligned}$
$\begin{aligned} f^{{}'} \frac{\pi}{2}=& \cos \frac{\pi}{4} \\\\ &=\frac{1}{\sqrt{2}} \end{aligned}$

Differentiation exercise Fill in the blanks question 15

Answer: $f^{1}\left(\frac{\pi}{6}\right)=\frac{1+\sqrt{3}}{2}$
Hint:
Given: $f(x)=|\sin x-\cos x|$
Solution: $f(x)=|\sin x-\cos x|$
$\begin{aligned} &f(x)=\left(\begin{array}{cc} \sin x-\cos x & 0 \leq x \leq \frac{\pi}{4} \\\\ -(\sin x-\cos x) & \frac{\pi}{4}<x \leq \frac{\pi}{4} \end{array}\right) \\\\ &f^{{}'}(x)=\left(\begin{array}{cc} \sin x+\cos x & 0 \leq x \leq \frac{\pi}{4} \\\\ -(\sin x+\cos x) & \frac{\pi}{4}<x \leq \frac{\pi}{4} \end{array}\right) \end{aligned}$
$\begin{aligned} f^{{}'}\left(\frac{\pi}{6}\right)=& \sin \frac{\pi}{6}+\cos \frac{\pi}{6} \\\\ &=\frac{1}{2}+\frac{\sqrt{3}}{2} \\\\ \end{aligned}$
$=\frac{\sqrt{3}+1}{2}$

Differentiation exercise Fill in the blanks question 16

Answer: 2
Hint: $\frac{d(\tan x)}{d x}=\sec ^{2} x$
Given: $y=\tan x^{0}$
$\\ \\ \because x^{0}=\frac{\pi}{180}x\\ \\ \therefore \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\pi}{180} \cdot\sec^{2}\frac{\pi}{180}x=\frac{\pi}{180}\sec^{2}x^{0}$

Differentiation exercise Fill in the blanks question 17

Answer: $\frac{d y}{d x}=0$
Hint: $\frac{d}{d x}=\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}$
Given: $y=\sin ^{-1}\left(e^{x}\right)+\cos ^{-1}\left(e^{x}\right)$
Solution: $y=\sin ^{-1}\left(e^{x}\right)+\cos ^{-1}\left(e^{x}\right)$
Differentiating (i) we get;
$\begin{aligned} \frac{d y}{d x} &=\frac{1}{\sqrt{1+e^{2x}}} e^{x}-\frac{1}{\sqrt{1+e^{2x}}} e^{x} \\\\ &=0 \end{aligned}$

Differentiation exercise Fill in the blanks question 18

Answer: $\frac{3}{\sqrt{1-x^{2}}}$
Hint: $\text { put } x=\sin \theta$
Given: $y=\sin ^{-1}\left(3 x-4 x^{3}\right)$
Solution: By putting $x=\sin \theta \text { in } \mathrm{y}$
$\begin{aligned} &y=\sin ^{-1}(3 \sin \theta-4 \sin \theta) \\\\ &y=\sin ^{-1}(\sin 3 \theta) \end{aligned}$
$\begin{aligned} &=3 \theta \\\\ &=3 \sin ^{-1} x \end{aligned}$
$\frac{d y}{d x}=\frac{3}{\sqrt{1-x^{2}}}$

Differentiation exercise Fill in the blanks question 19

Answer: 0
Hint: $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$
Given: $y=\sec ^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
Solution:
$y=\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
$\begin{aligned} &=\frac{\pi}{2} \\\\ &=0 \end{aligned}$

Differentiation exercise Fill in the blanks question 20

Answer: $\text { tan } u$
Hint: $\text { put } u=\cos x \text { and } v=\sin x$
Given: The derivative of $\cos \; x$ w.r.t $\sin x$
Solution:
$\begin{aligned} &\text { Let } u=\cos x \text { and } v=\sin x \\\\ &\frac{d u}{d v}=-\sin x &\frac{d v}{d u}=\cos x \end{aligned}$
Therefore:
$\begin{aligned} &\frac{d u}{d v}=\frac{d u \mid d v}{d v \mid d u}=\frac{-\sin x}{\cos x} \\\\ &=-\tan x \end{aligned}$

Differentiation exercise Fill in the blanks question 21

Answer: $\frac{\log _{10} e}{x}$
Hint: $\text { use: }\left[\frac{d}{d x} \log _{e} x=\frac{1}{x}\right]$
Given: $\text { If } \quad y=\log _{10} x$
Solution: $\log _{10} x$
$\begin{aligned} &y=\log _{10} e \cdot \log _{e} x \\\\ &\frac{d y}{d x}=\log _{10} e\times \frac{1}{x} \end{aligned}$

So the answer is $\frac{\log _{10} e}{x}$

Differentiation exercise Fill in the blanks question 22

Answer: the correct answer is $\frac{3 x^{2}}{1+x^{6}}$
Hint: integrate the function.
Given: $\frac{d}{\mathrm{dx}} f(x)=\frac{1}{1+x^{2}}$
Solution: integrating we get
$\begin{aligned} &\int \frac{d}{d x}(f(x)) d x=\int\left[\frac{1}{1+x^{2}}\right] d x \\\\ &f(x)=\tan ^{-1} x+e \end{aligned}$
$\begin{aligned} &f\left(x^{3}\right)=\tan ^{-1}\left(x^{3}\right)+c \\\\ &f^{1}\left(x^{3}\right)=\frac{3 x^{2}}{1+x^{6}} \end{aligned}$
So the answer is $\frac{3 x^{2}}{1+x^{6}}$

Differentiation exercise Fill in the blanks question 23

Answer: the correct answer is ='0'
Hint: $y=\cos \left(\sin x^{2}\right) \frac{d}{d x} \sin x=-\cos x$
$\frac{dy}{d x}=-\sin \left(\sin x^{2}\right)-\cos x^{2} \cdot 2 x$
Given: $y=\cos \left(\sin x^{2}\right)$
Solution: using $\frac{dy}{dx}$ to get the value at $\frac{\lambda }{2}$
Now at $x=\sqrt{\pi / 2}$
$\frac{d}{d x}=-\sin \left(\sin \sqrt{\frac{\pi}{2}}\right) \cdot \cos \sqrt{\frac{\pi}{2}} \cdot 2 \sqrt{\frac{\pi}{2}}=0$
So the correct answer is ‘0'

Differentiation exercise Fill in the blanks question 24

Answer: The correct answer is $\frac{1}{x}$
Hint: Differentiate the function but u should not be equal to zero
Given: $\mathrm{y}=\log _{\mathrm{e}}|\mathrm{x}|$
Solution:
$\mathrm{y}=\log |\mathrm{x}|=\left(\begin{array}{cc} \log (-x), & x<0 \\ \log (x), & x>0 \end{array}\right)$
$\frac{d y}{d x}=\left(\begin{array}{cc} \frac{-1}{-x}=\frac{1}{x},, & x<0 \\ \frac{1}{x}, & x>0 \end{array}\right)$
Which means it is equal to $\frac{1}{x}$ when $x\neq 0$
$\frac{d}{d x} \log |x|=\frac{1}{x}, x \neq 0$
So the answer is $\frac{1}{x}$

Differentiation exercise Fill in the blanks question 25

Answer: the correct answer is =0
Hint: put 1,4,5 in the equation to get the respective value the use $f^{1}(1)+f^{1}(u)-f^{1}(5)$
Given: $y=a x^{2}+b x+c$
Solution:
$\begin{aligned} &y=a x^{2}+b x+c \\\\ &f^{1}(x)=\frac{d y}{d x}=2 a x+b \\\\ &f^{1}(1)=2 a+b \text { and } \end{aligned}$
$\begin{aligned} &f^{1}(x)=3 a+4 b \\\\ &f^{1}(5)=10 a+5 b \\\\ &f^{1}(1)+f^{1}(u)-f^{1}(5) \end{aligned}$
$=2 a+b+8 a+4 b-10 a-5 b$
So the correct answer is 0

Differentiation exercise Fill in the blanks question 26

Answer: the correct answer is $\frac{1}{\sqrt{2}}$
Hint:
Given: $f^{1}(1)=2 \text { and } g^{1}(\sqrt{2})=4$
Solution:
$\begin{aligned} &y=f(\tan x) \text { and } z=g(\sec x) \\\\ &\frac{d y}{d x}=f^{\prime}(\tan x) \sec ^{2} x \text { and } \frac{d z}{d x}=g^{\prime}(\sec x) \cdot \sec x+\tan x \end{aligned}$
$\begin{aligned} &\frac{d y}{d z}=\frac{f^{1}(\tan x) \cdot \sec ^{2} x}{g^{1}(\sec x) \cdot \sec x \tan x} \\\\ &\frac{d y}{d z}=\frac{f^{\prime}(1) \cdot(\sqrt{2})^{2}}{g^{\prime}(\sqrt{2}) \cdot(\sqrt{2})}=\frac{1}{\sqrt{2}} \end{aligned}$
So the answer is $\frac{1}{\sqrt{2}}$

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Some important concepts used in chapter 10 FBQ:-

Questions related to Differentiation of a function.
Questions related to the Differentiation of inverse trigonometric functions by the chain rule.
Questions related to Differentiation by using trigonometric substitutions.
Solutions on Differentiation of implicit functions.
Questions related to Logarithmic differentiation.
Differentiation of infinite series.
Differentiation of parametric functions.
Differentiation of determinants.