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RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

Updated on 27 Jan 2022, 01:56 PM IST

The RD Sharma Solutions Class 12 Chapter 8 – Continuity is planned by our specialists to support certainty among students in understanding the ideas canvassed in this chapter and strategies to take care of issues in a more limited period. At Career360 RD Sharma class 12th exercise FBQ helps students who seek to get a decent scholarly score in the exam.

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RD Sharma Class 12 Solutions Chapter 8 FBQ Continuity - Other Exercise
Continuity Excercise: FBQ
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 8 FBQ Continuity - Other Exercise

Continuity Excercise: FBQ

Continuity exercise Fill in the blanks question 1

Answer: $3a^{2}$
Hint:
Use the formula $(x^{3}-a^{3})$ so that $f(x)\neq \frac{0}{0}$ at $x\neq a$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{x^{3}-a^{3}}{x-a} & , x \neq a \\ b & , x=a \end{array}\right.$ is continuous at $x=a$
Solution:
If $f(x)$ is continuous at $x=a$ , then RHL= LHL
LHL,
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a} \\ &=\lim _{x \rightarrow a^{-}} \frac{(x-a)\left(x^{2}+a x+a^{2}\right)}{(x-a)} \\ &=\lim _{x \rightarrow a^{-}} x^{2}+a x+a^{2} \\ &=\lim _{\mathbb{\Xi} \rightarrow 0} a^{2}+a^{2}+a^{2} \\ &=3 a^{2} \end{aligned}$$\left [ \because x=a \right ]$
RHL,
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} b \\ &=b \end{aligned}$
As$f(x)$ is continuous at $x=a$, RHL = LHL
$b=3a^{2}$

Continuity exercise Fill in the blanks question 2

Answer: $\pm 1$
Hint:
Use the formula of $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ to solve $\frac{\sin ^{2} a x}{x^{2}}$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{\sin ^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x=0 \end{array}\right.$is continuous at $x=0$
Solution:
If $f(x)$ is continuous at $x=0$, then $f(x)=\frac{\sin ^{2} a x}{x^{2}}$ and $g(x)=1$ are equal.
LHL,
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{x^{2}} \\ &=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{a^{2} x^{2}}\left(a^{2}\right) \\ \end{aligned}$
$=\lim_{x\rightarrow 0^{-}}\left ( \frac{\sin ax}{ax} \right )^{2}\left ( a^{2} \right )$
$=a^{2}$ $\left[\because \frac{\sin ^{2} a x}{a^{2} x^{2}}=1\right]$
RHL,
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 1=1$
As $f(x)$ is continuous at $x=0$, LHL =RHL
$\begin{aligned} &a^{2}=1 \\ &a=\pm 1 \end{aligned}$

Continuity exercise Fill in the blanks question 3

Answer: 2
Hint: You must know about the concept of continuous function
Given:
$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1<x \leq 2 \end{array} \quad \text { is continuous at } x=1\right.$
Solution:
If $f(x)$ is continuous at $x=1$, then
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ &\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}$
$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$
$\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}$

Continuity exercise Fill in the blanks question 4

Answer: 1
Hint: If $f(x)$is continuous, then all the functions are equal
Given:
$f(x)=\left\{\begin{array}{l} x+k, x<3 \\ 4, x=3 \quad \text { is continuous at } x=3 \\ 3 x-5, x>3 \end{array}\right.$
Solution:
We know that as function is continuous at $x=3$ ,
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} f(x) \\\\ &\lim _{x \rightarrow 3^{-}} x+k=\lim _{x \rightarrow 3^{+}} 3 x-5=\lim _{x \rightarrow 3} 4 \end{aligned}$
$\lim _{h \rightarrow 0}(3-h)+k=\lim _{h \rightarrow 0} 3(3+h)-5=\lim _{x \rightarrow 3} 4$
$\begin{aligned} &3+k=3(3)-5=4 \\ &3+k=4=4 \\ &k=1 \end{aligned}$

Continuity exercise Fill in the blanks question 5

Answer: $b=-1, a=1,a+b=0$

Hint: $\left | x \right |=x$, When $x< 0$ and $\left | x \right |=x$ , when $x> 0$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a, x<4 \\ a+b, x=4 \quad \text { is continuous at } x=4 \\ \frac{x-4}{|x-4|}+b, x>4 \end{array}\right.$

Solution:
As the function is continuous at $x=4$,
$\begin{aligned} &\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4} f(x) \\ &\lim _{x \rightarrow 4^{-}} \frac{x-4}{|x-4|}+a=\lim _{x \rightarrow 4^{+}} \frac{x-4}{|x-4|}+b=\lim _{x \rightarrow 4} a+b \end{aligned}$
$\lim _{h \rightarrow 0} \frac{(4-h)-4}{|(4-h)-4|}+a=\lim _{h \rightarrow 0} \frac{(4+h)-4}{|(4+h)-4|}+b=\lim _{x \rightarrow 4} a+b$
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{-h}{|h|}+a=\lim _{x \rightarrow 4^{+}} \frac{h}{|h|}+b=a+b \\ &-1+a=1+b=a+b \end{aligned}$
$\begin{aligned} &a+b=-1+a \quad, \quad a+b=1+b \\ &b=-1 \quad, \quad a=1 \end{aligned}$
Now, $a+b=1-1=0$

Continuity exercise Fill in the blanks question 6
Answer: -4

Hint: Use the formula of $\cos 3 x=4 \cos ^{3} x-3 \cos x$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{\cos 3 x-\cos x}{x^{2}}, & x \neq 0 \\ \lambda & , x=0 \end{array}\right.$ is continuous at $x=0$
Solution:
We know that
$\lim _{x \rightarrow 0} f(x)=f(0),$ when function is continuous
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{\left(4 \cos ^{3} x-3 \cos x\right)-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos ^{3} x-4 \cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(\cos ^{2} x-1\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(-\sin ^{2} x\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0}-4 \cos x=\lambda \\ &-4(1)=\lambda \end{aligned}$
$\lambda=-4 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \cos 3 x=4 \cos ^{3} x-3 \cos x \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \because \cos 0^{\circ}=1 \end{array}\right]$

Continuity exercise Fill in the blanks question 7

Answer: 0
Hint: Use L-Hospital’s rule
Given:$f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ k & , x=\frac{\pi}{2} \end{array}\right.$ is continuous at $x=\frac{\pi }{2}$

Solution:
If $f(x)$ is continuous at $x=\frac{\pi }{2}$ , then
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\pi-2 x}=k \end{aligned}$
Applying L-Hospital’s rule on LHS that is differentiating both numerator and denominator,
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} \frac{0-(\cos x)}{0-2(1)}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-2}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{2}=k \\ &\frac{\cos \frac{\pi}{2}}{2}=k \end{aligned}$
$k=0 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{2}=0\right]$

Continuity exercise Fill in the blanks question 8

Answer:$f(0)=\frac{-1}{8}$

Hint: Use L-Hospital’s rule

Given:$f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$and $f(x)$ is continuous at $x=0$

Solution:
As $f(x)$ is continuous at $x=0$,
$\begin{aligned} &f(0)=\lim _{x \rightarrow 0} f(x) \\ &f(0)=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x} \end{aligned}$
Using L-Hospital’s rule,
$f(0)=\lim _{x \rightarrow 0} \frac{0-\frac{1}{2 \sqrt{x+4}}}{2 \cos 2 x}$
$\begin{aligned} &f(0)=\lim _{x \rightarrow 0} \frac{-1}{(2 \sqrt{x+4})(2 \cos 2 x)} \\ &f(0)=\lim _{x \rightarrow 0} \frac{-1}{4(\sqrt{x+4})(\cos 2 x)} \end{aligned}$
$\begin{aligned} &f(0)=\frac{-1}{4(1)(\sqrt{4})} \\ &f(0)=\frac{-1}{4(2)} \\ &f(0)=\frac{-1}{8} \end{aligned}$

Continuity exercise Fill in the blanks question 9

Answer: $k=0$

Hint:Use the formula of $(a^{2}-b^{2})$

Given:
$f(x)=\left\{\begin{array}{l} \frac{x^{2}-9}{x-3}, x \neq 3 \\ 2 x+k, x=3 \end{array}\right.$ is continuous at $x=3$
Solution:
If $f(x)$ is continuous at $x=3$ , then
$\begin{aligned} &\lim _{x \rightarrow 3} f(x)=f(3) \\ &\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=2(3)+k \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}=2(3)+k \\ &\lim _{x \rightarrow 3} x+3=6+k \end{aligned}$
$\begin{aligned} &3+3=6+k \\ &6=6+k \\ &k=0 \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right] \end{aligned}$

Continuity exercise Fill in the blanks question 10

Answer: 2
Hint: Use the formula of $(a^{2}-b^{2})$
Given:$f(x)=\left\{\begin{array}{cl} \frac{x^{2}-1}{x-1}, & x \neq 1 \\ k & , x=1 \end{array} \text { is continuous at } x=1\right.$
Solution:
If $f(x)$ is continuous at $x=1$, then
$\begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\ &\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}=k \end{aligned}$
$\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}=k \; \; \; \; \; \; \; \; \quad\left[\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$
$\begin{aligned} &\lim _{x \rightarrow 1} x+1=k \\ &1+1=k \\ &k=2 \end{aligned}$

Continuity exercise Fill in the blanks question 12

Answer: $f\left(\frac{\pi}{4}\right)=\frac{-1}{2}$
Hint: You must know about L-hospital’s rule
Given:$f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right] \text { and } f(x) \text { is continuous in }\left[0, \frac{\pi}{2}\right]$
Solution:
$f(x)$ is continuous in $\left [ 0,\frac{\pi }{2} \right ]$
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right) \\ &\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=f\left(\frac{\pi}{4}\right) \end{aligned}$
Applying L-Hospital’s rule
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^{2} x}{4}=f\left(\frac{\pi}{4}\right) \\ &\frac{-2}{4}=f\left(\frac{\pi}{4}\right) \end{aligned}$ $\quad\left [\because \sec ^{2} \frac{\pi}{4}=2\right]$

Continuity exercise Fill in the blanks question 13

Answer: $\pi$
Hint: Use the identity $\frac{\sin x}{x}=1$
Given:
$f(x)=x\sin \left ( \frac{\pi }{x} \right )$ is continuous everywhere
Solution:
$f(x)=x\sin \left ( \frac{\pi }{3} \right )$ is continuous
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} x \sin \left(\frac{\pi}{x}\right)=f(0) \\ &\lim _{x \rightarrow 0} \frac{x \sin \left(\frac{\pi}{x}\right)}{\frac{\pi}{x}} \times \frac{\pi}{x}=f(0) \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}$$\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}$$\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}$

Continuity exercise Fill in the blanks question 14

Answer: 0,1,-1
Hint: Function is discontinuous where denominator is zero
Given: $f(x)=\frac{1}{\log |x|}$
Solution:
$\log x$ is not defined at $x=0$.
So,$x=0$, $f(x)$ is discontinuous.
$\log \left | x \right |=0 , when \left | x \right |=1$
i.e. $x=1,-1$
When $x=1,-1$ , value of $\frac{1}{\log \left | x \right |}=\infty$
$x=1,-1$ act as vertical asymptotes
Therefore,$f(x)$ is discontinuous at $x=-1,0,1$

Continuity exercise Fill in the blanks question 15

Answer: 2
Hint: $f(x)$ is continuous when $LHL=RHL$
Given: $f(x)=\left\{\begin{aligned} a x+1, & \text { if } x \geq 1 \\ x+2, & \text { if } x<1 \end{aligned}\right.$
Solution:
$f(x)=\left\{\begin{aligned} a x+1, & \text { if } x \geq 1 \\ x+2, & \text { if } x<1 \end{aligned}\right.$
$f(x)$ is continuous when $LHL=RHL$
$\lim _{x \rightarrow 1^{+}} a x+1=\lim _{x \rightarrow 1^{-}} x+2 \text { at } x=1$
$\lim _{h \rightarrow 0} a(1+h)+1=\lim _{h \rightarrow 0}(1-h)+2$
$\begin{aligned} &a+1=3 \\ &a=2 \end{aligned}$

Continuity exercise Fill in the blanks question 16

Answer: $f(a)$
Hint: $f(x)$ is continuous when $LHL=RHL$
Given: $f(x)$ is continuous at $x=a$
Solution:
$f(x)$ is continuous at $x=a$
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=k \\ &L H L=R H L=f(a) \\ &k=f(a) \end{aligned}$

Continuity exercise Fill in the blanks question 17

Answer: 6
Hint: $f(x)$ is continuous when $LHL=RHL$
Given: $f(x)=\left\{\begin{array}{c} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\ \frac{k}{2}, \text { if } x=0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\ \frac{k}{2}, \text { if } x=0 \end{array}\right.$is continuous at $x=0$
At $x=0$
$LHL=RHL$ $=f(a)$
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=\frac{k}{2} \\ &\lim _{x \rightarrow 0} \frac{3}{3} \frac{\sin 3 x}{x}=\frac{k}{2} \\ &3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=\frac{k}{2} \end{aligned}$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &3(1)=\frac{k}{2} \\ &k=6 \end{aligned}$

Continuity exercise Fill in the blanks question 18

Answer: $(2 n+1) \frac{\pi}{2}, n \in I$
Hint: $f(x)$is discontinuous when $f\rightarrow \infty$
Given: $f(x)=\tan x$
Solution:
$f(x)=\tan x$
The points of discontinuity of $f(x)$ are there at which $\tan x$ is infinite
$\begin{aligned} &\tan x \rightarrow \infty \\ &\tan x=\tan \frac{\pi}{2} \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\begin{array}{l} \because \tan x=\tan y \\ x=(2 n+1) y \end{array}\right]$
$x=(2 n+1) \frac{\pi}{2}, n \in I$

Continuity exercise Fill in the blanks question 19

Answer: All integral points
Hint: $f(x)$ is discontinuous when $f(x)$ is not defined
Given: $f(x)=[x]$
Solution:
$f(x)=[x]$
Continuity will be measured at
$\bullet$$x$ is an integer
$\bullet$ $x$ is not an integer
Case I: $x$ is not an integer
$\begin{aligned} &\lim _{x \rightarrow d} f(x)=f(d) \\ &{[d]=[d]} \end{aligned}$
$f(x)$ is continuous at all non-integer
Case II: $x$ is an integer
$f(x)=[x]$ is continuous if $LHL=RHL=f(c)$
$\begin{aligned} &\lim _{x \rightarrow c^{-}} f(x)=\lim _{k \rightarrow 0} f(c-h)=c-h=c-1 \\ &\lim _{x \rightarrow c^{+}} f(x)=\lim _{k \rightarrow 0} f(c+h)=c+h=c \\ &c \neq c-1 \\ &L H L \neq R H L \end{aligned}$
$f(x)$ is discontinuous at all integer

Continuity exercise Fill in the blanks question 20

Answer:All integer points

Hint:$f(x)$ is discontinuous when $f(x)$ is not defined
Given: $f(x)=\frac{1}{x-[x]}$
Solution:
$f(x)=\frac{1}{x-[x]}$

$\left [ x \right ]$ is discontinuous at all integer points
$x-\left [ x \right ]$ is discontinuous at all integer points
$\frac{1}{x-\left [ x \right ]}$ is discontinuous at each integer value of $x$
All integral points, $f(x)$ is discontinuous

Continuity exercise Fill in the blanks question 3

Answer: 2
Hint: You must know about the concept of continuous function
Given:
$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1<x \leq 2 \end{array} \quad \text { is continuous at } x=1\right.$
Solution:
If $f(x)$ is continuous at $x=1$, then
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ \\&\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}$
$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$
$\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}$

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