RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

Kuldeep MauryaUpdated 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
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ \\&\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}$
$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$
$\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}$

This chapter of RD Sharma class 12th exercise FBQ principally centers around the idea of coherence. To find out about this theme students can download the RD Sharma class 12 solutions FBQ Chapter 8 Continuity. This chapter clarifies Continuity of a point, Continuity of an interval (open or closed) and its applications with tackled examples. RD Sharma class 12 solutions FBQ, the specialists give answer keys as well as some remarkable tips in the book that the students probably won't discover elsewhere. RD Sharma class 12th exercise FBQ has around 20 questions.

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