RD Sharma Solutions Class 12 Mathematics Chapter 8 MCQ
RD Sharma class 12 Solutions chapter 8 exercise MCQ is considered one of the best books to prepare for the class 12 board exams. It contains in-detail explanations for concepts and questions in every chapter. It is used widely by class 12 CBSE students as well as teachers across the country. This is why it is the best choice for students to prepare for the exams.
This Story also Contains
- RD Sharma Class 12 Solutions Chapter 8 MCQ Continuity - Other Exercise
- Continuity Excercise: MCQ
- RD Sharma Chapter-wise Solutions
RD Sharma Class 12 Solutions Chapter 8 MCQ Continuity - Other Exercise
Continuity Excercise: MCQ
Continuity exercise multiple choice question 1
Answer:The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\frac{4-x^{2}}{4 x-x^{3}}$
Solution:
Since a function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &f(x)=\frac{4-x^{2}}{4 x-x^{3}} \\ &f(x)=\frac{4-x^{2}}{x\left(4-x^{2}\right)} \\ &f(x)=\frac{1}{x} \\ &\text { Where } x \neq 0, \pm 2 \end{aligned}$
f(x) is continuous for all real numbers except (0, $\pm$2).
Therefore, it is continuous exactly at three points (0, $\pm$2)
So, the correct option is (c)
Continuity exercise multiple choice question 2
Answer:
The correct option is (a) and (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\}=f^{\prime}\left(a^{+}\right) \text {for right hand derivative }\\ &\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\}=f^{\prime}\left(a^{-}\right) \text {for left hand derivative } \end{aligned}$
Given:
$f(x)=|x-a| \phi(x)$
Solution:
Using formula (ii)
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{|h+a-a| \phi(a+h)-|a-a| \phi(a)}{h}\right\} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h \phi(a+h)}{h}\\ &=\lim _{h \rightarrow 0} \phi(a+h)=\phi(a)=f^{\prime}\left(a^{+}\right)\; \; \; ....(i) \end{aligned}$
For Left Hand Derivative
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{|a-h-a| \phi(a-h)-|a-a| \phi(a)}{h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{|-h| \phi(a-h)}{-h} \\ &=\lim _{h \rightarrow 0}-\phi(a-h) \end{aligned}$
Therefore
$\begin{aligned} &=\lim _{h \rightarrow 0}-\phi(a-h)\\ &=-\phi(a)\\ &=f^{\prime}\left(a^{-}\right)\; \; \; \; \; \; .....(ii) \end{aligned}$
From (i) and (ii), we get
$f^{\prime}\left(a^{+}\right) \neq f^{\prime}\left(a^{-}\right)$
So, correct option is (a) and (b)
Continuity exercise multiple choice question 3
Answer:The correct option is (a) and (d)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\}=f^{\prime}\left(a^{+}\right) \text {for right hand derivative }\\ &\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\}=f^{\prime}\left(a^{-}\right) \text {for left hand derivative } \end{aligned}$
Given:
$f(x)=\left|\log _{10} x\right|$
Solution:
$f(x)=\left|\log _{10} x\right|$
$=\left|\frac{\log x}{\log _{e} 10}\right|=\left|\log _{\mathrm{e}} x \log _{10} e\right|$
Calculate right hand derivative
Using limit at x = 1
Apply limit at x = 1
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(1+h)-f(1)}{h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{\left|\log _{10} e \log (1+h)\right|-\left|\log _{10} e \log 1\right|}{h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{\log _{10} e|\log (1+h)|}{h} \end{aligned}$
Therefore applying L-Hospital rule
$\begin{aligned} &=\log _{10} e \times 1 \\ &=f^{\prime}\left(1^{+}\right) \end{aligned}$
For left hand limit
$=\lim _{h \rightarrow 0}\left\{\frac{f(1-h)-f(1)}{-h}\right\}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{\left|\log _{10} e \log (1-h)\right|-\left|\log _{10} e \log 1\right|}{-h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{-\log _{10} e|\log (1-h)|}{h} \\ &-\log _{10} e=f^{\prime}\left(1^{-}\right) \end{aligned}$
$\begin{aligned} &\text { The function } f(x)=\left|\log _{10} x\right| \text { is continuous and in the right hand derivative }\\ &f^{\prime}\left(1^{-}\right)=-\log _{10} e \end{aligned}$
So the correct option is (a) and (d)
Continuity exercise multiple choice question 4
Answer:The correct option is (C)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{1}{m} \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m\\ &\text { (iii) } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{l} \frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ k \end{array}\right.$ $\begin{aligned} , \quad &x \neq 0 \\ , \quad &x=0 \end{aligned}$
Solution:
Function f(x) is continuous at x = 0
To calculate the value of k, understand that function is continuous at x = 0
$\begin{aligned} &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{36^{x}-9^{x}-4^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}\right)=k \\ &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\sqrt{2}\left[2 \sin ^{2}\left(\frac{x}{4}\right)\right]}\right)=k \end{aligned}$
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{2 \sqrt{2} x^{2} \cdot \frac{\frac{\sin ^{2} x}{4}}{\frac{x^{2}}{16} \times 16}}\right)=k$
$\Rightarrow \lim _{x \rightarrow 0}\left\{\frac{\left(9^{x}-1\right)\left(4^{x}-1\right)}{\frac{\sqrt{2}}{8} x^{2} \times \frac{\sin ^{2} x y}{\frac{x^{2}}{16}}}\right\}=k$
Using formula (ii) and (iii)
$\begin{aligned} &\left.\Rightarrow \frac{8 \ln 9 \ln 4}{\sqrt{2}}=k \quad \text { [understand that } \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{a^{x}}\right)=\ln a\right] \\ &\Rightarrow \frac{32 \ln 3 \ln 2}{\sqrt{2}}=k \\ &\Rightarrow k=16 \sqrt{2} \ln 2 \ln 3 \end{aligned}$
So correct option is (C)
Answer:
The correct option is (d)
Hint:
Use the given formula:
$\begin{aligned} &\text { (i) If } \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \text { then } f(x) \text { is discontinuous at } x=0 \text { . }\\ &\text { (ii)If } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \text { then } f(x) \text { is continuous at } x=0 \end{aligned}$
$\begin{aligned} &\text { (iii) A function } f(x) \text { is said to be continuous at a point } x=a \text { of its domain, if }\\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$ $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$
Solution:
Simplify the given function
$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$ $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$
$\begin{aligned} &f(x)=\left\{\begin{aligned} &1,\: x>1 \\ &1,\: x \leq 0 \\ &-1,\: 0 \leq x \leq 1 \end{aligned}\right. \\ &\text { Using RHL } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=-1 \\ &\text { Using LHL } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(h)=1 \end{aligned}$
f(x) is discontinuous at x =0
Again, Using RHL,
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\\ \end{aligned}$
Using LHL,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-1 \end{aligned}$
f(x) is discontinuous at x = 1
Therefore,
f(x) is continuous for all except at x = 0 and x = 1
So, the correct option is (d)
The correct option is (d)
Hint:
Use the given formula:
$\begin{aligned} &\text { (i) If } \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \text { then } f(x) \text { is discontinuous at } x=0 \text { . }\\ &\text { (ii)If } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \text { then } f(x) \text { is continuous at } x=0 \end{aligned}$
$\begin{aligned} &\text { (iii) A function } f(x) \text { is said to be continuous at a point } x=a \text { of its domain, if }\\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$ $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$
Solution:
Simplify the given function
$f(x)=\left\{\begin{array}{lr} \frac{\left|x^{2}-x\right|}{x^{2}-x} \\ 1 & \\ -1 & \end{array}\right.$ $\begin{array}{r} , \quad x \neq 0,1 \\ , \quad x=0 \\ , \quad x=1 \end{array}$
$\begin{aligned} &f(x)=\left\{\begin{aligned} &1,\: x>1 \\ &1,\: x \leq 0 \\ &-1,\: 0 \leq x \leq 1 \end{aligned}\right. \\ &\text { Using RHL } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=-1 \\ &\text { Using LHL } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(h)=1 \end{aligned}$
f(x) is discontinuous at x =0
Again, Using RHL,
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=1\\ \end{aligned}$
Using LHL,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-1 \end{aligned}$
f(x) is discontinuous at x = 1
Therefore,
f(x) is continuous for all except at x = 0 and x = 1
So, the correct option is (d)
Continuity exercise multiple choice question 6
Answer:The correct option is (C)
Hint:
Use the given formula:
$\text { (i) } \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
(ii) A function f(x) is said to be continuous at a point x=a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) g(x)\}=1 m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{l} \frac{1-\sin x}{(\pi-2 x)^{2}} \frac{\log \sin x}{\left(\log \left(1+\pi^{2}-4 \pi x+4 x^{2}\right)\right)} \\ \\ k \end{array}\right.$ $\begin{aligned} , \quad &x \neq \frac{\pi}{2} \\ , \quad &x=\frac{\pi}{2} \end{aligned}$
Solution:
Function f(x) is continuous at
$x = \frac{\pi }{2}$
$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$
Using substitution method
$\begin{aligned} &\text { Substitute } \frac{\pi}{2}-x=t \\ &\Rightarrow \lim _{t \rightarrow 0}\left(\frac{\pi}{2}-t\right)=f\left(\frac{\pi}{2}\right) \end{aligned}$
$\Rightarrow \lim _{t \rightarrow 0}\left\{\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{4 t^{2}} \frac{\log \sin \left(\frac{\pi}{2}-t\right)}{\log \left(1+\pi^{2}-4 \pi\left(\frac{\pi}{2}-t\right)+4\left(\frac{\pi}{2}-t\right)^{2}\right.}\right\}=k$
$\Rightarrow \lim _{t \rightarrow 0} \frac{\left(2 \sin ^{2} t / 2\right) \log \cos 1}{4 t^{2} \cdot \log \left(1+\pi^{2}-2 \pi^{2}+4 \pi t+\pi^{2}+4 t^{2}-4 \pi t\right)}=k$
$\begin{aligned} &\Rightarrow \lim _{t \rightarrow 0} \frac{2 \sin ^{2} t / 2 \log \cos t}{4 t^{2} \log \left(1+4 t^{2}\right)}=k \\ &\Rightarrow \lim _{t \rightarrow 0} \frac{2 \sin ^{2} t / 2 \log \sqrt{1-\sin ^{2} t}}{4 \times 4 \frac{t^{2}} {4}} \frac{1}{4 t^{2} \cdot \frac{\log (1+4 t^{2})}{4 t^{2}}}=k \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \end{aligned}$
$\Rightarrow \lim _{t \rightarrow 0} \frac{2}{16}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{\frac{1}{2} \log (1-\sin 2 t)}{4 t^{2} \cdot \frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}}=k$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{-\sin ^{2} f}{2} \cdot \frac{\log \left(1-\sin ^{2} t\right)}{4 t^{2}\left(\sin ^{2} t\right)} \times \frac{1}{\log \left(1+4 t^{2}\right)}=k$
$\Rightarrow \frac{-1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{\sin ^{2} f}{8 t^{2}} \cdot \frac{\log (1-\sin 2 t)}{\left(-\sin ^{2} t-1\right)} \times \frac{1}{\frac{\log (1+4+2)}{4 t^{2}}}=k$
$\Rightarrow-\frac{1}{8 \times 8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \cdot \lim _{t \rightarrow 0} \frac{\log \left(1-\sin ^{2} t\right)}{-\sin 2 t} \cdot \lim _{d \rightarrow 0} \frac{1}{\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}}=k$
Using formula (iii)
$\Rightarrow \frac{1}{64} \lim _{t \rightarrow 0}\left\{\frac{\sin ^{2} \frac{t}{2} \frac{\log \left(1-\sin ^{2} t\right)}{8 t^{2}}}{\left(\frac{t}{2}\right)^{2}} \frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right\}=k$
$\Rightarrow-\frac{1}{64} \lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \lim _{t \rightarrow 0} \log \frac{1-\sin ^{2} t}{-\sin ^{2} t}=k$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=4, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
Using standard limit formula (i)
$\Rightarrow k=-\frac{1}{64}$
So, option (c) is correct.
Answer:
The correct option is (C)
Hint:
Use the given formula:
$\text { (i) } \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1$
(ii) A function f(x) is said to be continuous at a point x = a of its domain
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text {iii}) \lim _{x \rightarrow a}\{f(x) g(x)\}=1 m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)=(x+1)^{\cot x}$
Solution:
$f(x)=(x+1)^{\cot x}$
taking log on both sides
$\log f(x)=(\cot x)(\log (x+1))$
$\lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0}\left(\frac{\frac{\log (x+1)}{x}}{\frac{\tan x}{x}}\right)$
Using formula (iii)
$\lim _{x \rightarrow 0} \log f(x)=\frac{\lim _{x \rightarrow 0} \frac{\log (x+1)}{x}}{\lim _{x \rightarrow 0} \frac{\tan x}{x}}$
Using standard limit formula (i)
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=e \\ &f(0)=e \end{aligned}$
So, the correct option is (C)
The correct option is (C)
Hint:
Use the given formula:
$\text { (i) } \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1$
(ii) A function f(x) is said to be continuous at a point x = a of its domain
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text {iii}) \lim _{x \rightarrow a}\{f(x) g(x)\}=1 m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)=(x+1)^{\cot x}$
Solution:
$f(x)=(x+1)^{\cot x}$
taking log on both sides
$\log f(x)=(\cot x)(\log (x+1))$
$\lim _{x \rightarrow 0} \log f(x)=\lim _{x \rightarrow 0}\left(\frac{\frac{\log (x+1)}{x}}{\frac{\tan x}{x}}\right)$
Using formula (iii)
$\lim _{x \rightarrow 0} \log f(x)=\frac{\lim _{x \rightarrow 0} \frac{\log (x+1)}{x}}{\lim _{x \rightarrow 0} \frac{\tan x}{x}}$
Using standard limit formula (i)
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=e \\ &f(0)=e \end{aligned}$
So, the correct option is (C)
Answer:
The correct option is (b)
Hint:
Use the given formula:
$\text { (i) Standard limit } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$
(ii) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) \pm g(x)\}=1 \pm m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{l} \frac{\log (1+a x)-\log (1-b x)}{x} \\ k \end{array}\right.$ $\begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}$
And f(x) is continuous at x = 0
Solution:
$\because$ f(x) is continuous at x = 0
$\Rightarrow \lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}=k$
Using formula (ii)
$\Rightarrow a \lim _{x \rightarrow 0} \frac{\log (1+a x)}{a x}+b \lim _{x \rightarrow 0} \frac{\log (1-b x)}{-b x}=k$
Using formula (i)
$\Rightarrow a+b=k$
So, the correct option is (b)
The correct option is (b)
Hint:
Use the given formula:
$\text { (i) Standard limit } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$
(ii) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) \pm g(x)\}=1 \pm m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{l} \frac{\log (1+a x)-\log (1-b x)}{x} \\ k \end{array}\right.$ $\begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}$
And f(x) is continuous at x = 0
Solution:
$\because$ f(x) is continuous at x = 0
$\Rightarrow \lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}=k$
Using formula (ii)
$\Rightarrow a \lim _{x \rightarrow 0} \frac{\log (1+a x)}{a x}+b \lim _{x \rightarrow 0} \frac{\log (1-b x)}{-b x}=k$
Using formula (i)
$\Rightarrow a+b=k$
So, the correct option is (b)
Continuity exercise multiple choice question 9
Answer:The correct option is (b)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} \frac{e^{1 / x}-1}{e^{1 / x}+1} \\ 0 \end{array}\right.$ $\begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}$
Solution:
Using substitution method,
$\text { Let } e^{1 / x}=t \text { so, } x \rightarrow 0, t \rightarrow \infty$
$\begin{aligned} &\lim _{t \rightarrow \infty} f(x)=\lim _{t \rightarrow \infty}\left(\frac{t-1}{t+1}\right) \\ &=\frac{1-0}{1+0}=1 \quad \quad[L-\text { Hospital rule }] \end{aligned}$
And,
$\begin{aligned} &f(0)=0 \\ &\text { Therefore, } \lim _{x \rightarrow 0} f(x) \neq f(0) \end{aligned}$
Hence, f(x) is discontinuous at x = 0
So, option (b) is correct.
Answer:
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a \\ a+b \\ \frac{x-4}{|x-4|}+b \end{array}\right.$ $\begin{aligned} , \quad &x<4 \\ , \quad &x=4 \\ , \quad &x>4 \end{aligned}$
and f(x) is continuous at x = 4
Solution:
Using RHL at x = 4
$\begin{aligned} &=\lim _{x \rightarrow 4^{+}} f(x) \\ &=\lim _{h \rightarrow 0} f(4+h) \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right) \\ &=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right) \\ &=1+b \end{aligned}$
Using LHL
$\begin{aligned} &\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h) \\ &=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right) \\ &=\lim _{h \rightarrow \infty}\left(\frac{-h}{|-h|}+a\right) \\ &=a-1 \end{aligned}$
f(x) will be continuous at x = 4, if
$\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &b=-1 \text { and } a=1 \end{aligned}$
So, the correct option is (d)
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a \\ a+b \\ \frac{x-4}{|x-4|}+b \end{array}\right.$ $\begin{aligned} , \quad &x<4 \\ , \quad &x=4 \\ , \quad &x>4 \end{aligned}$
and f(x) is continuous at x = 4
Solution:
Using RHL at x = 4
$\begin{aligned} &=\lim _{x \rightarrow 4^{+}} f(x) \\ &=\lim _{h \rightarrow 0} f(4+h) \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right) \\ &=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right) \\ &=1+b \end{aligned}$
Using LHL
$\begin{aligned} &\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h) \\ &=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right) \\ &=\lim _{h \rightarrow \infty}\left(\frac{-h}{|-h|}+a\right) \\ &=a-1 \end{aligned}$
f(x) will be continuous at x = 4, if
$\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &b=-1 \text { and } a=1 \end{aligned}$
So, the correct option is (d)
Answer:
The correct option is (b)
Hint:
Use the given formula:
$\begin{aligned} &\text { (i) } \lim _{x \rightarrow 0} f(x) g(x)=e \lim _{x \rightarrow 0}(f(x)-1) \cdot g(x)\\ &\text { Where } \lim _{x \rightarrow 0} f(x)=1 \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0} g(x)=0\\ &\text { (ii) } \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=-1 \end{aligned}$
(iii) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} (\cos x)^{\frac{1}{x}} \\ k \end{array}\right.$ $\begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}$
And f(x) is continuous at x = 0
Solution:
$\lim _{x \rightarrow 0} f(x)=f(0)$
Calculate the value of k
$\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=k$
Using formula (i)
$\lim _{x \rightarrow 0} f(x)^{g(x)}=e \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=k$
Apply formula (ii)
$\begin{aligned} &e^{0}=k \\ & k=1 \end{aligned}$
So, option (b) is correct.
The correct option is (b)
Hint:
Use the given formula:
$\begin{aligned} &\text { (i) } \lim _{x \rightarrow 0} f(x) g(x)=e \lim _{x \rightarrow 0}(f(x)-1) \cdot g(x)\\ &\text { Where } \lim _{x \rightarrow 0} f(x)=1 \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0} g(x)=0\\ &\text { (ii) } \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=-1 \end{aligned}$
(iii) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} (\cos x)^{\frac{1}{x}} \\ k \end{array}\right.$ $\begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}$
And f(x) is continuous at x = 0
Solution:
$\lim _{x \rightarrow 0} f(x)=f(0)$
Calculate the value of k
$\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=k$
Using formula (i)
$\lim _{x \rightarrow 0} f(x)^{g(x)}=e \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=k$
Apply formula (ii)
$\begin{aligned} &e^{0}=k \\ & k=1 \end{aligned}$
So, option (b) is correct.
Continuity exercise multiple choice question 12
Answer:The correct option is (a)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=|x|+|x-1|$
Solution:
Both are polynomial function and we know that the polynomial functions are continuous everywhere,
So, according to option,
f(x) is continuous at x = 0, as well as at x = 1
So, option is (a) is correct.
Answer:
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x^{4}-5 x^{2}+4}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$ $\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
Solution:
we write
$\begin{aligned} x^{4}-5 x^{2}+4 &=x^{4}-x^{2}-4 x^{2}+4=x^{2}\left(x^{2}-1\right)-4\left(x^{2}-1\right) \\ &=\left(x^{2}-4\right)\left(x^{2}-1\right) \\ &=(x+2)(x-2)(x+1)(x-1) \end{aligned}$
$\therefore$ we have
$f(x)=\left\{\begin{array}{c} \frac{(x+2)(x-2)(x+1)(x-1)}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$$\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
$f(x)=\left\{\begin{array}{c} \frac{x^{4}-5 x^{2}+4}{(x-1)(x-2)} \\ 6 \\ 12 \end{array}\right.$$\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
$=\left\{\begin{array}{c} \int(x+1)(x+2), x<1 \\ -(x+1)(x+2), 1<x<2 \\ (x+1)(x+2), x>2 \\ 6, x=1 \\ 12, x=2 \end{array}\right.$
Using RHL at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=-(2) \cdot(3)=-6 \end{aligned}$
Using LHL at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=(2)(3)=6 \end{aligned}$
And using RHL at x = 2
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=12$
Using LHL at x = 2
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-12$
f(x) is discontinuous at x=1 and x=2 for f(x) continuous at R[1, 2]
So, the correct option is (d)
The correct option is (d)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x^{4}-5 x^{2}+4}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$ $\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
Solution:
we write
$\begin{aligned} x^{4}-5 x^{2}+4 &=x^{4}-x^{2}-4 x^{2}+4=x^{2}\left(x^{2}-1\right)-4\left(x^{2}-1\right) \\ &=\left(x^{2}-4\right)\left(x^{2}-1\right) \\ &=(x+2)(x-2)(x+1)(x-1) \end{aligned}$
$\therefore$ we have
$f(x)=\left\{\begin{array}{c} \frac{(x+2)(x-2)(x+1)(x-1)}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$$\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
$f(x)=\left\{\begin{array}{c} \frac{x^{4}-5 x^{2}+4}{(x-1)(x-2)} \\ 6 \\ 12 \end{array}\right.$$\begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}$
$=\left\{\begin{array}{c} \int(x+1)(x+2), x<1 \\ -(x+1)(x+2), 1<x<2 \\ (x+1)(x+2), x>2 \\ 6, x=1 \\ 12, x=2 \end{array}\right.$
Using RHL at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=-(2) \cdot(3)=-6 \end{aligned}$
Using LHL at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=(2)(3)=6 \end{aligned}$
And using RHL at x = 2
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=12$
Using LHL at x = 2
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-12$
f(x) is discontinuous at x=1 and x=2 for f(x) continuous at R[1, 2]
So, the correct option is (d)
Continuity exercise multiple choice question 14
Answer:The correct option is (c)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{aligned}$
Given:
f(x) is continuous at x = 0 and
$f(x)=\left\{\begin{array}{cc} \frac{\sin (a+1) x+\sin x}{x} &, x<0 \\ C & , x=0 \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x \sqrt{x}} & , x>0 \end{array}\right.$
Solution:
Using RHL
$\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ =\lim _{h \rightarrow 0} f(h) \\ =\lim _{h \rightarrow 0} \frac{\sqrt{h+b h^{2}}-\sqrt{h}}{b h \sqrt{h}} \end{gathered}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\sqrt{h}(\sqrt{1+b h})-\sqrt{n}}{b h \sqrt{h}} \\ &=\lim _{h \rightarrow 0} \frac{\sqrt{1+b n}-1}{b h} \\ &=\lim _{h \rightarrow 0} \frac{\sqrt{1+b h}-1}{b h} \times \frac{\sqrt{1+b h}+1}{\sqrt{1+b h}+1} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(1+b h)-1}{b h(\sqrt{1+b h}+1)} \\ &=\lim _{h \rightarrow 0} \frac{b h}{b h(\sqrt{1+b h}+1)} \\ &=\lim _{h \rightarrow 0} \frac{1}{\sqrt{1+b h}+1} \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{1+6 \cdot 0}+1} \\ &=\frac{1}{1+1} \\ &=\frac{1}{2} \end{aligned}$
Using L.H.L
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h) \\ &=\lim _{h \rightarrow 0} f(-h) \\ &=\lim _{n \rightarrow 0} \frac{\sin (a+1)(-h)+\sin (-h)}{(-h)} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\sin (a+1) \cdot h-\sinh }{-h} \\ &=\lim _{h \rightarrow 0}\left\{-\frac{\sin (a+1) \cdot h}{-h}+\frac{\sin h}{h}\right\} \quad[\because \sin (-\theta)=-\sin \theta] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{\sin (a+1)}{(a+1)} \times(a+1)+\frac{\sinh }{h}\right\} \\ &=(a+1) \lim _{h \rightarrow 0} \frac{\sin (a+1)}{a+1}+\lim _{h \rightarrow 0} \frac{\sin h}{n} \end{aligned}$
$\begin{aligned} &=(a+1) \cdot 1+1 \quad\left[: \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=a+1+1 \\ &=a+2 \end{aligned}$
Since function f(x) is continuous at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ &=f(0) \\ &\Rightarrow a+2=\frac{1}{2} \\ &\Rightarrow a+2=\frac{1}{2} \\ &\text { or } c=\frac{1}{2} \end{aligned}$
$\begin{aligned} &\begin{aligned} \Rightarrow a &=\frac{1}{2}-2 \\ &=-\frac{3}{2} \text { or } \\ c=\frac{1}{2} \end{aligned}\\ &\text { and from } f(x) \text { , }\\ &b \in \mathbb{R}-\{0\} \end{aligned}$
So, the correct option is (c).
Continuity exercise 5 multiple choice question 15
Answer:The correct option is (c)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
f(x) is continuous at x = 0 and
$f(x)=\left\{\begin{array}{l} m x+1, x \leq \frac{\pi}{2} \\ \sin x+n, x>\frac{\pi}{2} \end{array}\right.$
Solution:
Using RHL$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\ &=\lim _{h \rightarrow 0}\left[\sin \left(\frac{\pi}{2}+h\right)+n\right] \\ &=\lim _{h \rightarrow 0}(\cos h+n)=1+n \end{aligned}$
Using LHL
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} m\left(\frac{\pi}{2}-h\right)+1=\frac{m \pi}{2}+1 \end{aligned}$
$\begin{aligned} &\text { Function } f(x) \text { is continuous at } x=\frac{\pi}{2}\\ &\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)\\ &\frac{m \pi}{2}+1=n+1\\ &\frac{m \pi}{2}=n \end{aligned}$
So, option (c) is correct
Continuity exercise multiple choice question 16
Answer:The correct option is (c)
Hint:
Use the given formula:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\frac{\sqrt{a^{2}-a x+x^{2}}-\sqrt{a^{2}+a x+x^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$
Solution:
Using rationalization method,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(a^{2}-a x+x^{2}\right)-\left(a^{2}+a x+x^{2}\right)}{(\sqrt{a+x}-\sqrt{a-x})\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$
$\lim _{x \rightarrow 0} \frac{-2 a x(\sqrt{a+x}+\sqrt{a-x})}{2 x\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$
$\left.\lim _{x \rightarrow 0} \frac{-a(\sqrt{a+x}+\sqrt{a-x})}{\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right.}\right)$
$=\lim _{x \rightarrow 0} \frac{-2 a(\sqrt{a})}{2 a}=-\sqrt{a}$
Understand that, f(x) is continuous for all x, then it is continuous at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=0 \\ &\lim _{x \rightarrow 0}-\sqrt{a}=f(0) \end{aligned}$
So, the correct option is (c)
Answer:
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$
Solution:
Step 1: Simplify the given function
$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$
Case 2: Calculate the RHL
$\begin{gathered} \lim _{x \rightarrow \frac{1}{n}^{+}} f(x)=\lim _{h \rightarrow 0}\left(\frac{1}{n}+h\right) \\ =\left(\frac{1}{h}\right)^{2} \end{gathered}$
Case 3: Calculate the RHL
$\begin{aligned} &\lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\left(\frac{1}{n-1}\right)^{2} \\ &\text { Therefore, } \lim _{x \rightarrow \frac{1}{n}^{+}} f(x) \neq \lim _{x \rightarrow \frac{1}{n}^{-}} f(x) \end{aligned}$
Hence, the correct answer is option (c)
Hence, f(x) is discontinuous only at
$x=\pm \frac{1}{n^{\prime}} n \in Z-\{0\} \text { and } x=0$
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$
Solution:
Step 1: Simplify the given function
$f(x)=\left\{\begin{array}{l} 1,|x| \geq 1 \\ \frac{1}{n^{2}}, \frac{1}{n}<|x|<\frac{1}{n-1}, n=2,3, \ldots \\ 0, x=0 \end{array}\right.$
Case 2: Calculate the RHL
$\begin{gathered} \lim _{x \rightarrow \frac{1}{n}^{+}} f(x)=\lim _{h \rightarrow 0}\left(\frac{1}{n}+h\right) \\ =\left(\frac{1}{h}\right)^{2} \end{gathered}$
Case 3: Calculate the RHL
$\begin{aligned} &\lim _{x \rightarrow \frac{1}{n}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{n}-h\right)=\left(\frac{1}{n-1}\right)^{2} \\ &\text { Therefore, } \lim _{x \rightarrow \frac{1}{n}^{+}} f(x) \neq \lim _{x \rightarrow \frac{1}{n}^{-}} f(x) \end{aligned}$
Hence, the correct answer is option (c)
Hence, f(x) is discontinuous only at
$x=\pm \frac{1}{n^{\prime}} n \in Z-\{0\} \text { and } x=0$
Answer:
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}},(x \neq 0)$
Solution:
Step 1: Understand that, if f(x) to be continuous at x = 0 then,
$\lim _{x \rightarrow 0} f(x)=f(0)$
Therefore,
$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$
$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-(27)^{\frac{1}{3}}}{3\left[3-(243+5 x)^{\frac{1}{5}}\right]}$
$=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-(27)^{4 / 3}}{3\left\{(243)^{15}-(243+5 x)^{1 / 5}\right\}}$
$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{x\left\{(27-2 x)^{1 / 3}-(27)^{4 / 3}\right\}}{\left.x(243)^{1 / 5}-(243+5 x)^{1 / 5}\right\}}$
$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-\left(2+1^{1 / 3}\right.}{x}}{\frac{(243)^{1 / 5}-(243+5 x)^{1 / 5}}{x}}$
$=\frac{2}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{5}}{5 \frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{5 x}}$
$=\frac{2}{15} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{27-2 x-27}}{\frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{243+5 x-243}}$
$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times 27^{-\frac{2}{3}}}{\frac{1}{5} \times 243^{-\frac{4}{5}}}$
$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{27^{\frac{2}{3}}}}{\frac{1}{5} \times \frac{1}{243^{\frac{4}{5}}}}$
$\begin{aligned} &\begin{aligned} &=\frac{2}{9} \frac{(243)^{\frac{4}{5}}}{(27)^{\frac{2}{3}}} \\ =& \frac{2}{9} \frac{\left(3^{5}\right)^{4 / 5}}{\left(3^{3}\right)^{2 / 3}} \\ \therefore f(0) &=\frac{2}{9} \times \frac{3^{4}}{3^{2}} \\ &=2 . \end{aligned}\\ &\text { Therefore, } f(0)=2 \end{aligned}$
Hence, the correct option is (c)
The correct option is (c)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}},(x \neq 0)$
Solution:
Step 1: Understand that, if f(x) to be continuous at x = 0 then,
$\lim _{x \rightarrow 0} f(x)=f(0)$
Therefore,
$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$
$=f(0)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-(27)^{\frac{1}{3}}}{3\left[3-(243+5 x)^{\frac{1}{5}}\right]}$
$=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-(27)^{4 / 3}}{3\left\{(243)^{15}-(243+5 x)^{1 / 5}\right\}}$
$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{x\left\{(27-2 x)^{1 / 3}-(27)^{4 / 3}\right\}}{\left.x(243)^{1 / 5}-(243+5 x)^{1 / 5}\right\}}$
$=\frac{1}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-\left(2+1^{1 / 3}\right.}{x}}{\frac{(243)^{1 / 5}-(243+5 x)^{1 / 5}}{x}}$
$=\frac{2}{3} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{5}}{5 \frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{5 x}}$
$=\frac{2}{15} \lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{27-2 x-27}}{\frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{243+5 x-243}}$
$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times 27^{-\frac{2}{3}}}{\frac{1}{5} \times 243^{-\frac{4}{5}}}$
$=f(0)=\frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{27^{\frac{2}{3}}}}{\frac{1}{5} \times \frac{1}{243^{\frac{4}{5}}}}$
$\begin{aligned} &\begin{aligned} &=\frac{2}{9} \frac{(243)^{\frac{4}{5}}}{(27)^{\frac{2}{3}}} \\ =& \frac{2}{9} \frac{\left(3^{5}\right)^{4 / 5}}{\left(3^{3}\right)^{2 / 3}} \\ \therefore f(0) &=\frac{2}{9} \times \frac{3^{4}}{3^{2}} \\ &=2 . \end{aligned}\\ &\text { Therefore, } f(0)=2 \end{aligned}$
Hence, the correct option is (c)
Answer:
The correct option is (d)
Hint:
Use the given formula :
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &\text { (ii) } \lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{1}{m}, \end{aligned}$
${\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m$
Given:
$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2} \text { is continuous everywhere }$
Solution:
$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
Now,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
Using factorization method
$\Rightarrow \lim _{x \rightarrow 0} \frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}=x(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{x}\left\{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}\right\}=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5}}=f(0)$
$\Rightarrow+\frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{-7 x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5 x}}=f(0)$
$\Rightarrow \frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{256-7 x-256}}{\frac{(5 x+32)^{1 / 5-(32)^{1 / 5}}}{32+5 x-32}}=f(0)$
$\Rightarrow \frac{7 \frac{1}{8}(256)^{1 / 8-1}}{5} \frac{1}{\frac{1}{5}(32)^{1 / 5-1}}=f(0) \quad\left[\therefore \lim _{x \rightarrow a} \frac{x^{n}-a^{4}}{x-a}=n a^{n-1}\right]$
$\begin{aligned} &\Rightarrow \frac{7}{5} \times \frac{1}{8} \times \frac{5}{1} \frac{(256)^{-7 / 8}}{(32)^{-4 / 5}}=f(0) \\ &\Rightarrow \frac{7}{8} \frac{(32)^{\frac{4}{5}}}{(256)^{-\frac{1}{8}}}=f(0) \end{aligned}$
$\begin{aligned} \Rightarrow f(0) &=\frac{7}{8} \times \frac{\left.72^{5}\right)^{\frac{4}{5}}}{\left(2^{8}\right)^{\frac{7}{8}}} \\ &=\frac{7}{2^{3}} \times \frac{2^{4}}{2^{7}} \\ &=\frac{7}{2^{6}}=\frac{7}{64} \end{aligned}$
So the correct-option is (d)
The correct option is (d)
Hint:
Use the given formula :
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &\text { (ii) } \lim _{x \rightarrow a}\left\{\frac{f(x)}{g(x)}\right\}=\frac{1}{m}, \end{aligned}$
${\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m$
Given:
$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2} \text { is continuous everywhere }$
Solution:
$f(x)=\frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
Now,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{\frac{1}{8}}}{(5 x+32)^{\frac{1}{5}}-2}$
Using factorization method
$\Rightarrow \lim _{x \rightarrow 0} \frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}=x(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{x}\left\{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-(32)^{1 / 5}}\right\}=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{(256)^{1 / 8}-(256-7 x)^{1 / 8}}{x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5}}=f(0)$
$\Rightarrow+\frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{-7 x}}{\frac{(5 x+32)^{1 / 5}-(32)^{1 / 5}}{5 x}}=f(0)$
$\Rightarrow \frac{7}{5} \lim _{x \rightarrow 0} \frac{\frac{(256-7 x)^{1 / 8}-(256)^{1 / 8}}{256-7 x-256}}{\frac{(5 x+32)^{1 / 5-(32)^{1 / 5}}}{32+5 x-32}}=f(0)$
$\Rightarrow \frac{7 \frac{1}{8}(256)^{1 / 8-1}}{5} \frac{1}{\frac{1}{5}(32)^{1 / 5-1}}=f(0) \quad\left[\therefore \lim _{x \rightarrow a} \frac{x^{n}-a^{4}}{x-a}=n a^{n-1}\right]$
$\begin{aligned} &\Rightarrow \frac{7}{5} \times \frac{1}{8} \times \frac{5}{1} \frac{(256)^{-7 / 8}}{(32)^{-4 / 5}}=f(0) \\ &\Rightarrow \frac{7}{8} \frac{(32)^{\frac{4}{5}}}{(256)^{-\frac{1}{8}}}=f(0) \end{aligned}$
$\begin{aligned} \Rightarrow f(0) &=\frac{7}{8} \times \frac{\left.72^{5}\right)^{\frac{4}{5}}}{\left(2^{8}\right)^{\frac{7}{8}}} \\ &=\frac{7}{2^{3}} \times \frac{2^{4}}{2^{7}} \\ &=\frac{7}{2^{6}}=\frac{7}{64} \end{aligned}$
So the correct-option is (d)
Answer:
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{c} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x<0 \\ \frac{2 x+1}{x-2} \quad, 0 \leq x \leq 1 \end{array}\right.$
Step 1 : Rationalize the function
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$
$\lim _{h \rightarrow 0} \frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(\sqrt{1-p h}-\sqrt{1+p h}) \times(\sqrt{1-p h}+\sqrt{1+p h})}{(-h)(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{((-p h)-(1+p h)}{(-h) \sqrt{1-p h}+\sqrt{1+p^{n}}} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-2 p h}{-n(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{+2 p}{\sqrt{1-p h}+\sqrt{1+p h}} \end{aligned}$
$\begin{aligned} &=\frac{2 p}{\sqrt{1}+\sqrt{1}} \\ &=\frac{2 p}{1+1} \\ &=\frac{2 p}{2} \\ &=p \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}$
$\begin{aligned} &=\frac{-1}{2} \end{aligned}$
Also,
$\begin{aligned} &f(x)=\frac{2 \cdot 0+1}{0-2} \\ &=\frac{-1}{2} \end{aligned}$
From (i)
$\begin{aligned} &p=-\frac{1}{2}=-\frac{1}{2} \\ &\Rightarrow p=\frac{-1}{2} \end{aligned}$
So, the correct option is (b)
The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{c} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x},-1 \leq x<0 \\ \frac{2 x+1}{x-2} \quad, 0 \leq x \leq 1 \end{array}\right.$
Step 1 : Rationalize the function
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$
$\lim _{h \rightarrow 0} \frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(\sqrt{1-p h}-\sqrt{1+p h}) \times(\sqrt{1-p h}+\sqrt{1+p h})}{(-h)(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{((-p h)-(1+p h)}{(-h) \sqrt{1-p h}+\sqrt{1+p^{n}}} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-2 p h}{-n(\sqrt{1-p h}+\sqrt{1+p h})} \\ &=\lim _{h \rightarrow 0} \frac{+2 p}{\sqrt{1-p h}+\sqrt{1+p h}} \end{aligned}$
$\begin{aligned} &=\frac{2 p}{\sqrt{1}+\sqrt{1}} \\ &=\frac{2 p}{1+1} \\ &=\frac{2 p}{2} \\ &=p \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}$
$\begin{aligned} &\text { And } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &=\lim _{n \rightarrow 0} \frac{2 h+1}{h-2} \\ &=\frac{2 \times 0+1}{0-2} \\ &=\frac{1}{-2} \end{aligned}$
$\begin{aligned} &=\frac{-1}{2} \end{aligned}$
Also,
$\begin{aligned} &f(x)=\frac{2 \cdot 0+1}{0-2} \\ &=\frac{-1}{2} \end{aligned}$
From (i)
$\begin{aligned} &p=-\frac{1}{2}=-\frac{1}{2} \\ &\Rightarrow p=\frac{-1}{2} \end{aligned}$
So, the correct option is (b)
Continuity exercise multiple choice question 21
Answer:The correct option is (c)
Hint:
If a function is continuous at x=2, then
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
Given:
The function f(x)
$=\left\{\begin{array}{l} x^{2} a, 0 \leq x<1 \\ a, 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}}, \sqrt{2 \leq x<\infty} \end{array}\right. \text { is continuous for } 0 \leq x<\infty$
Solution:
Step 1: Calculate f(x) is continuous for
$x = 1, \: \sqrt{2}$
Calculate for f(x) is continuous at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1+} f(x)=f(1) \cdots(i) \\ &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{a} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left(1+h^{2}-2 h\right)}{a} \cdot \quad\left[\because(\boldsymbol{a}-\boldsymbol{b})^{2}{=} \boldsymbol{a}^{2}+\boldsymbol{b}^{2}-\mathbf{2 a b}\right] \\ &=\lim _{h \rightarrow 0} \frac{1}{a}+\frac{1}{a} h^{2}-\frac{2 h}{a} \end{aligned}$
$\begin{aligned} &=\frac{1}{a}+0-2 \frac{1}{a} \times 0 \\ &=\frac{3}{a} \end{aligned}$
f(a) = a
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+n)=\lim _{n \rightarrow 0} a=a \\ &\text { from (i) } \Rightarrow \frac{1}{a}=a=a \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{a}=a \\ &\Rightarrow a^{2}=1 \\ &\Rightarrow a=\pm 1 \end{aligned}$
Step 2: Calculate for f(x) is continuous at
$\begin{aligned} x=\sqrt{2} \end{aligned}$
Therefore,
$\begin{aligned} &\lim _{x \rightarrow \sqrt{2}}-f(x)=\lim _{x \rightarrow \sqrt{2}} f(x)=f(\sqrt{2})-\text { (ii) } \\ &\Rightarrow \lim _{x \rightarrow \sqrt{2}}-f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}-h)=\lim _{h \rightarrow 0} a=a \end{aligned}$
$\begin{aligned} &\text { and } \lim _{x \rightarrow \sqrt{2}+} f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}+h) \\ &=\lim _{h \rightarrow 0} \frac{2 b^{2}-4 b}{(\sqrt{2}+h)^{2}} \\ &=\left(2 b^{2}-4 b\right) \lim _{h \rightarrow 0} \frac{1}{(\sqrt{2}+4)^{2}} \end{aligned}$
$\begin{aligned} &=\left(2 b^{2}-4 b\right) \cdot \frac{1}{(\sqrt{2}+0)^{2}} \\ &=\left(2 b^{2}-4 b\right) \cdot \frac{1}{(\sqrt{2})^{2}} \\ &=\frac{2 b^{2}-4 b}{2} \\ &\text { also, } f(\sqrt{2})=\frac{2 b^{2}-4 b}{(\sqrt{2})^{2}} \end{aligned}$
$\begin{aligned} &=\frac{2 b^{2}-4 b}{2} \end{aligned}$
from(ii), we have
$\begin{aligned} &\Rightarrow a=\frac{2 b^{2}-4 b}{2} \\ &=\frac{2 b^{2}-4 b}{2} \\ &\Rightarrow a=\frac{2 b^{2}-4 b}{2} \\ &\Rightarrow 2 a-2 b^{2}+4 b=0 \\ &\Rightarrow b^{2}-a-2 b=0 \end{aligned}$
$\begin{aligned} &\text { When }-a=1 \text { then } \\ &\Rightarrow b^{2}-2 b-1=0 \\ &\Rightarrow b=\frac{2 \pm \sqrt{4-(-4)}}{2} \\ &=\frac{2 \pm \sqrt{4+4}}{2} \\ &=\frac{2 \pm \sqrt{8}}{2} \\ &=1 \pm \sqrt{2} \end{aligned}$
$\begin{aligned} &\text { Also, when } a=-1 \text { then }\\ &\Rightarrow b^{2}-2 b+1=0\\ &\Rightarrow b=\frac{2 \pm \sqrt{4-4}}{2}\\ &=\frac{2}{2}\\ &=1\\ &\text { Thus } \mathrm{a}=-1 \text { and } \mathrm{b}=1 \end{aligned}$
The correct option is (c)
Answer:
The correct option is (a)
Hint:
$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$
Given:
$\begin{aligned} &\text { The function } f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}, \text { when } x \neq \frac{\pi}{2} \text { and } f\left(\frac{\pi}{2}\right)=\lambda, \text { then } f(x) \text { will be continuous function }\\ &\text { for } x=\frac{\pi}{2} \end{aligned}$
Solution:
Step 1: Calculate, f(x) is continuous for
$x=\frac{\pi }{2}$
Therefore,
$\begin{aligned} &\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &=\lim _{x \rightarrow \frac{\pi}{2}(\pi-2 x)^{2}}=f\left(\frac{\pi}{2}\right) \end{aligned}$
$\begin{aligned} &\text { Assume, }\left(\frac{\pi}{2}-x\right)=t \text { and substitute in equation }(i)\\ &\therefore \lim _{t \rightarrow 0}\left[\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{(2 t)^{2}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}$
$\begin{aligned} &\lim _{t \rightarrow 0}\left(\frac{1-\cos t}{4 t^{2}}\right)=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{t \rightarrow 0}\left[\frac{2 \sin ^{2}\left(\frac{t}{2}\right)}{t^{2}}\right]=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{i \rightarrow 0}\left[\frac{\frac{2}{4} \sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}$
Simplify further,
$\begin{aligned} &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \\ &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin \left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}=f\left(\frac{\pi}{2}\right) \end{aligned}$
Apply formula (i) i.e
$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$
$\begin{aligned} &=\frac{1}{8} \times(1)^{2}=f\left(\frac{\pi}{2}\right) \\ &=f\left(\frac{\pi}{2}\right)=\frac{1}{8} \\ &\therefore f\left(\frac{\pi}{2}\right)=\lambda \\ &\therefore \Rightarrow \lambda=\frac{1}{8} \end{aligned}$
Hence, the value of
$\begin{aligned} &\lambda \text { is }\frac{1}{8} \end{aligned}$
Hence, the correct answer is option (a)
The correct option is (a)
Hint:
$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$
Given:
$\begin{aligned} &\text { The function } f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}, \text { when } x \neq \frac{\pi}{2} \text { and } f\left(\frac{\pi}{2}\right)=\lambda, \text { then } f(x) \text { will be continuous function }\\ &\text { for } x=\frac{\pi}{2} \end{aligned}$
Solution:
Step 1: Calculate, f(x) is continuous for
$x=\frac{\pi }{2}$
Therefore,
$\begin{aligned} &\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &=\lim _{x \rightarrow \frac{\pi}{2}(\pi-2 x)^{2}}=f\left(\frac{\pi}{2}\right) \end{aligned}$
$\begin{aligned} &\text { Assume, }\left(\frac{\pi}{2}-x\right)=t \text { and substitute in equation }(i)\\ &\therefore \lim _{t \rightarrow 0}\left[\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{(2 t)^{2}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}$
$\begin{aligned} &\lim _{t \rightarrow 0}\left(\frac{1-\cos t}{4 t^{2}}\right)=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{t \rightarrow 0}\left[\frac{2 \sin ^{2}\left(\frac{t}{2}\right)}{t^{2}}\right]=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{i \rightarrow 0}\left[\frac{\frac{2}{4} \sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}$
Simplify further,
$\begin{aligned} &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \\ &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin \left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}=f\left(\frac{\pi}{2}\right) \end{aligned}$
Apply formula (i) i.e
$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$
$\begin{aligned} &=\frac{1}{8} \times(1)^{2}=f\left(\frac{\pi}{2}\right) \\ &=f\left(\frac{\pi}{2}\right)=\frac{1}{8} \\ &\therefore f\left(\frac{\pi}{2}\right)=\lambda \\ &\therefore \Rightarrow \lambda=\frac{1}{8} \end{aligned}$
Hence, the value of
$\begin{aligned} &\lambda \text { is }\frac{1}{8} \end{aligned}$
Hence, the correct answer is option (a)
Continuity exercise multiple choice question 23
Answer:The correct option is (d)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)=1$
$\text { (ii) } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1$
$\text { (iii) } \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)=\log a$
Given:
The function
$f(x)= \begin{cases}\frac{\left(4^{x}-1\right)^{3}}{\sin (x a) \log \left(1+x^{2} 3\right)} &, x \neq 0 \\ 12(\log 4)^{3} & , x=0\end{cases}$
Solution:
Step 1: Calculate, f(x) is continuous for x = 0
$\begin{aligned} &\therefore \lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0}\left[\frac{\left(4^{x}-1\right)^{3}}{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}\right]=12(\log 4)^{3} \end{aligned}$
$\lim _{x \rightarrow 0}\left[\frac{\frac{\left(4^{x}-1\right)^{3}}{x^{3}}}{\frac{\sin \frac{x}{a} \log \left(1+\frac{x^{2}}{3}\right)}{x^{3}}}\right]=12(\log 4)^{3}$
$\lim _{x \rightarrow 0}\left[\frac{a\left(\frac{4^{x}-1}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \log \left(1+\frac{x^{2}}{3}\right)} x^{2}\right]=12(\log 4)^{3}$
$3 a \lim _{x \rightarrow 0}\left[\frac{\left(\frac{4^{x}-1}{x}\right)^{3}}{\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}$
$3 a\left[\frac{\lim_{x\rightarrow 0}\left(\frac{4^{x}-1}{x}\right)^{3}}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right) \lim_{x \rightarrow 0} \frac{\log \left(1+\frac{x^{2}}{3}\right)}{\left(\frac{x^{2}}{3}\right)}}\right]=12(\log 4)^{3}$
Apply formula (i), formula (ii) and formula (iii)
$\begin{aligned} &\therefore 3 a(\log 4)^{3}=12(\log 4)^{3} \\ &a=4 \end{aligned}$
Hence, the correct answer is option (d)
Continuity exercise multiple choice question 24
Answer:The correct option is (C)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a-} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
Given:
the function
$f(x)= tan\: x$
Solution:
Step 1: Check continuity of the function
$f(x)=\tan x \text { on the set }\{n \pi: n \in Z\}$
$\therefore f(x)=\tan n \pi$
is defined at the integral point
Check continuity of the function
$f(x)=\tan x \text { on the set }\{2 n \pi: n \in Z\}$
$f(x)=\tan 2 n \pi$
is defined at the integral point
Check continuity of the function
$f(x)=\tan x \text { on the set }\left\{(2 n+1) \frac{\pi}{2}: n \in Z\right\}$
$f(x)=\tan (2 n+1) \frac{\pi}{2}$
is defined at the integral point
$\begin{aligned} &f(x)=\tan \left(n \pi+\frac{\pi}{2}\right) \\ &f(x)=-\cot (n \pi) \end{aligned}$
is not defined at the integral point
Check continuity of the function
$f(x)=\tan x \text { on the set }\left\{\frac{n \pi}{2}: n \in Z\right\}$
$f(x)=\tan \frac{n \pi}{2}$
is defined at the integral point
Hence, the correct answer is option (c)
Answer:
The correct option is (b)
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin 3 x}{x} &, x \neq 0 \\ \frac{k}{2} &, x \neq 0\end{cases}$
Solution:
is continuous at x = 0, then
LHL = RHL
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ &\lim _{x \rightarrow 0^{-}} \frac{\sin 3 x}{x}=f(0) \end{aligned}$
$\begin{aligned} &\Rightarrow 3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=\frac{k}{2} \\ &\Rightarrow 3.1=\frac{k}{2} \quad\left[\frac{\sin x}{x}=1 \therefore \frac{\sin 3 x}{3 x}=1\right] \end{aligned}$
Hence, k = 6
So, the correct option is (b)
The correct option is (b)
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin 3 x}{x} &, x \neq 0 \\ \frac{k}{2} &, x \neq 0\end{cases}$
Solution:
is continuous at x = 0, then
LHL = RHL
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ &\lim _{x \rightarrow 0^{-}} \frac{\sin 3 x}{x}=f(0) \end{aligned}$
$\begin{aligned} &\Rightarrow 3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=\frac{k}{2} \\ &\Rightarrow 3.1=\frac{k}{2} \quad\left[\frac{\sin x}{x}=1 \therefore \frac{\sin 3 x}{3 x}=1\right] \end{aligned}$
Hence, k = 6
So, the correct option is (b)
Answer:
The correct option is (b)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)=1$
$\text { (ii) } \lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)=1$
Given:
the function
$f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$
is continuous at each point of its domain
Solution:
Step 1: Understand that according to question, f(x) is continuous at x = 0
$\begin{aligned} &\therefore \lim f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0) \end{aligned}$
$\lim _{x \rightarrow \infty} \frac{x\left(2-\frac{\sin ^{-1} x}{x}\right)}{x\left(2+\frac{\tan ^{-1} x}{x}\right)}=f(0)$
Simplify further,
$\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=f(0)$
$\frac{2-\lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)}{2+\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)}=f(0)$
Apply formula (i) and formula (ii)
$\begin{aligned} &\frac{2-1}{2+1}=f(0) \\ &f(0)=\frac{1}{3} \end{aligned}$
Hence, the correct answer is option (b)
The correct option is (b)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)=1$
$\text { (ii) } \lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)=1$
Given:
the function
$f(x)=\frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}$
is continuous at each point of its domain
Solution:
Step 1: Understand that according to question, f(x) is continuous at x = 0
$\begin{aligned} &\therefore \lim f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{2 x-\sin ^{-1} x}{2 x+\tan ^{-1} x}=f(0) \end{aligned}$
$\lim _{x \rightarrow \infty} \frac{x\left(2-\frac{\sin ^{-1} x}{x}\right)}{x\left(2+\frac{\tan ^{-1} x}{x}\right)}=f(0)$
Simplify further,
$\lim _{x \rightarrow 0} \frac{2-\frac{\sin ^{-1} x}{x}}{2+\frac{\tan ^{-1} x}{x}}=f(0)$
$\frac{2-\lim _{x \rightarrow 0}\left(\frac{\sin ^{-1} x}{x}\right)}{2+\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x}{x}\right)}=f(0)$
Apply formula (i) and formula (ii)
$\begin{aligned} &\frac{2-1}{2+1}=f(0) \\ &f(0)=\frac{1}{3} \end{aligned}$
Hence, the correct answer is option (b)
Continuity exercise multiple choice question 27
Answer:The correct option is (a)
Hint:
If a function is continuous at x = a, then
$\lim _{x \rightarrow t} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(a)$
Solution: the function
is continuous at every point of its domain
Step 1: Understand that according to question, f(x) is continuous at every point of its domain.
Therefore, it is continuous at x = 1
$\begin{aligned} &\therefore \lim _{x \rightarrow 1} f(x)=f(1) \\ &\lim _{h \rightarrow 0} f(1+h)=f(1) \\ &\lim _{h \rightarrow 0}\left[4(1+h)^{2}+3 b(1+h)\right]=5(1)-4 \\ &4+3 b=1 \end{aligned}$
Simplify further,
$\begin{aligned} &3 b=-3 \\ &b=-1 \end{aligned}$
Hence, the correct answer is option (a)
Continuity exercise multiple choice question 28
Answer:The correct option is (b)
Hint:
If a function f is continuous at x = a then
$\lim _{x \rightarrow 0^{-}} f(a)=\lim _{x \rightarrow 0^{+}} f(x)=f(a)$
Given:
The function
$f(x)=\frac{1}{1-x}$
Step 1: Understand that
$\{f: R-\{1\} \rightarrow R\}$
$\begin{aligned} &\text { Therefore } f(f(x))=f\left(\frac{x-1}{x}\right) \\ &\begin{array}{c} f(f(x))=f\left(\frac{1}{1-\left(\frac{1}{x-1}\right)}\right) \\ \\=\frac{x-1}{x} \end{array} \end{aligned}$
$\text { Step 2: Understand that } \therefore f \circ f: R-\{0,1\} \rightarrow R$
$\therefore f(f(f(x)))=f\left(\frac{x-1}{x}\right)$
$\begin{aligned} &f(f(f(x)))=f\left(\frac{x-1}{1-\left(\frac{x-1}{x}\right)}\right) \\ &=x \end{aligned}$
$\begin{aligned} &\text { Step 3: Understand that } \therefore f \circ f \circ f: R-\{0,1\} \rightarrow R \\ &\text { Therefore, } f(f(f(x))) \text { is not defined at } x=0,1 \\ &\text { Hence, } f(f(f(x))) \text { is discontinuous at } \{0,1\} \\ &\text { The set of point discontinuous of the function } f(f(f(x))) \text { is }\{0,1\} \end{aligned}$
Hence, the correct answer is option (b)
Answer:
The correct option is (b)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow 4-} f(x)=\lim _{x \rightarrow 0+} f(x)=f(a)$
$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)=1$
Given:
The function
$f(x)=\frac{\tan \left(\frac{\pi}{4} x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}$
Step 1: Understand that, if f(x) is continuous a
$x=\frac{\pi }{4}$
Understand that, if
$\frac{\pi }{4}-x=y$
Therefore,
$x\rightarrow \frac{\pi }{4} \text { and }y\rightarrow 0$
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}=f\left(\frac{\pi}{4}\right) \\ &\therefore \lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot 2\left(\frac{\pi}{4}-y\right)}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}$
$\lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot \left(\frac{\pi}{2}-2 y\right)}\right)=f\left(\frac{\pi}{4}\right)$
$\begin{aligned} &\lim _{y \rightarrow 0}\left(\frac{\tan y}{\tan 2 y}\right)=f\left(\frac{\pi}{4}\right) \\ &\frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}$
Apply formula (i)
$\begin{aligned} &\therefore \frac{1}{2}\left(\frac{1}{1}\right)=f\left(\frac{\pi}{4}\right) \\ &f\left(\frac{\pi}{4}\right)=\frac{1}{2} \end{aligned}$
Hence, the correct answer is option (b)
The correct option is (b)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow 4-} f(x)=\lim _{x \rightarrow 0+} f(x)=f(a)$
$\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)=1$
Given:
The function
$f(x)=\frac{\tan \left(\frac{\pi}{4} x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}$
Step 1: Understand that, if f(x) is continuous a
$x=\frac{\pi }{4}$
Understand that, if
$\frac{\pi }{4}-x=y$
Therefore,
$x\rightarrow \frac{\pi }{4} \text { and }y\rightarrow 0$
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}=f\left(\frac{\pi}{4}\right) \\ &\therefore \lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot 2\left(\frac{\pi}{4}-y\right)}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}$
$\lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot \left(\frac{\pi}{2}-2 y\right)}\right)=f\left(\frac{\pi}{4}\right)$
$\begin{aligned} &\lim _{y \rightarrow 0}\left(\frac{\tan y}{\tan 2 y}\right)=f\left(\frac{\pi}{4}\right) \\ &\frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}$
Apply formula (i)
$\begin{aligned} &\therefore \frac{1}{2}\left(\frac{1}{1}\right)=f\left(\frac{\pi}{4}\right) \\ &f\left(\frac{\pi}{4}\right)=\frac{1}{2} \end{aligned}$
Hence, the correct answer is option (b)
Answer:
The correct option is (a)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a+} f(x)=\lim _{x \rightarrow a-} f(x)=f(a)$
Given:
the function
$f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$
is not defined for x = 2. In order to make f(x) continuous at x = 2
Step 1: Take
$f(x)={x^{3}+x^{2}-16 x+20}$
and calculate further
$\therefore f(x)={x^{3}+x^{2}-16 x+20}$
$\begin{aligned} &=x^{2}(x-2)+3 x(x-2)-10(x-2)\\ \end{aligned}$
Simplify further,
$\begin{aligned} &=(x-2)(x-2)(x+5)\\ &=(x-2)^{2}(x+5) \end{aligned}$
Therefore, the function can be rewritten as
$\begin{aligned} &\therefore f(x)=\frac{(x-2)^{2}(x+5)}{(x-2)} \\ &f(x)=(x-2)(x+5) \end{aligned}$
Step 2: Understand that if f(x) is continuous at x = 2
Therefore,
$\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(x) \\ &=\lim _{h \rightarrow 0}(x-2)(x+5)=f(x) \\ &f(2)=0 \end{aligned}$
Hence, the correct answer is option (a)
The correct option is (a)
Hint:
If a function f is continuous at x = a, then
$\lim _{x \rightarrow a+} f(x)=\lim _{x \rightarrow a-} f(x)=f(a)$
Given:
the function
$f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$
is not defined for x = 2. In order to make f(x) continuous at x = 2
Step 1: Take
$f(x)={x^{3}+x^{2}-16 x+20}$
and calculate further
$\therefore f(x)={x^{3}+x^{2}-16 x+20}$
$\begin{aligned} &=x^{2}(x-2)+3 x(x-2)-10(x-2)\\ \end{aligned}$
Simplify further,
$\begin{aligned} &=(x-2)(x-2)(x+5)\\ &=(x-2)^{2}(x+5) \end{aligned}$
Therefore, the function can be rewritten as
$\begin{aligned} &\therefore f(x)=\frac{(x-2)^{2}(x+5)}{(x-2)} \\ &f(x)=(x-2)(x+5) \end{aligned}$
Step 2: Understand that if f(x) is continuous at x = 2
Therefore,
$\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(x) \\ &=\lim _{h \rightarrow 0}(x-2)(x+5)=f(x) \\ &f(2)=0 \end{aligned}$
Hence, the correct answer is option (a)
Answer:
The correct option is (a)
Hint:
f(x) is continuous at x = 0
$(\text { LHL of } f(x) \text { at } x=0)=(\text { RHL of } f(x) \text { at } x=0)=f(0)$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\: \: \: \: \: \: \: \: \: \: .....(i)$
$\begin{aligned} \text { Now, }&\lim _{x \rightarrow 0^{-}} f(x) \\ &=\lim _{x \rightarrow 0} a \sin \frac{\pi}{2}(x+1) \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} a \sin \left(\frac{\pi}{2}+\frac{\pi}{2} x\right) \\ &=\lim _{x \rightarrow 0} a \cos \frac{\pi x}{2}=a \end{aligned}$ $\begin{aligned} &{\left[\therefore f(x)=a \sin \frac{\pi}{2}(x+1), \text { if } x \leq 0\right]} \\ &{[\cos 0={1}]} \end{aligned}$
$\begin{aligned} \text { Again, }&\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
$=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}} \quad\left[\because f(x)=\frac{\tan x-\sin x}{x^{3}} \text { if } x>0\right]$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\frac{\sin x}{\cos x}-\sin x}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x-\sin x \cos x}{\cos x \times x^{3}}=\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{\cos x \times x^{3}} \end{aligned}$
$=\lim _{x \rightarrow 0} \frac{1}{\cos x} \lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{2 \sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4} \times 4} \quad\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$
$=\frac{1}{1} \times 1 \times \frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$
$=\frac{1}{2}\left[\lim _{\frac{x}{2} \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]=\frac{1}{2} \times 1=\frac{1}{2}$
$\begin{gathered} \text { Also, } f(0)=a \sin \frac{\pi}{2}(0+1) \\ \qquad=a \sin \frac{\pi}{2}=a \end{gathered}$
Putting above values in (i) we get,
$\begin{gathered} a=\frac{1}{2} \end{gathered}$
So, the correct option is (a)
The correct option is (a)
Hint:
f(x) is continuous at x = 0
$(\text { LHL of } f(x) \text { at } x=0)=(\text { RHL of } f(x) \text { at } x=0)=f(0)$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\: \: \: \: \: \: \: \: \: \: .....(i)$
$\begin{aligned} \text { Now, }&\lim _{x \rightarrow 0^{-}} f(x) \\ &=\lim _{x \rightarrow 0} a \sin \frac{\pi}{2}(x+1) \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} a \sin \left(\frac{\pi}{2}+\frac{\pi}{2} x\right) \\ &=\lim _{x \rightarrow 0} a \cos \frac{\pi x}{2}=a \end{aligned}$ $\begin{aligned} &{\left[\therefore f(x)=a \sin \frac{\pi}{2}(x+1), \text { if } x \leq 0\right]} \\ &{[\cos 0={1}]} \end{aligned}$
$\begin{aligned} \text { Again, }&\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
$=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}} \quad\left[\because f(x)=\frac{\tan x-\sin x}{x^{3}} \text { if } x>0\right]$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\frac{\sin x}{\cos x}-\sin x}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x-\sin x \cos x}{\cos x \times x^{3}}=\lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{\cos x \times x^{3}} \end{aligned}$
$=\lim _{x \rightarrow 0} \frac{1}{\cos x} \lim _{x \rightarrow 0} \frac{\sin x}{x} \times \frac{2 \sin ^{2} \frac{x}{2}}{\frac{x^{2}}{4} \times 4} \quad\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$
$=\frac{1}{1} \times 1 \times \frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$
$=\frac{1}{2}\left[\lim _{\frac{x}{2} \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]=\frac{1}{2} \times 1=\frac{1}{2}$
$\begin{gathered} \text { Also, } f(0)=a \sin \frac{\pi}{2}(0+1) \\ \qquad=a \sin \frac{\pi}{2}=a \end{gathered}$
Putting above values in (i) we get,
$\begin{gathered} a=\frac{1}{2} \end{gathered}$
So, the correct option is (a)
Answer:
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)= \begin{cases}a x^{2}+b & , 0 \leq x<1 \\ 4 & , \quad x=1 \\ x+3 &, 1<x \leq 2\end{cases}$
Step 1: Understand that, f(x) is continuous at x = 1
Therefore,
$\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(1) \\ &\lim _{h \rightarrow 0} f(1-h)=4 \\ &\lim _{h \rightarrow 0} a(1-h)^{2}+b=4 \end{aligned}$
Therefore, the possible values for (a, b) can be (2, 2), (3, 1), (4, 0) but (a,b) ≠ (5, 2)
Hence, the correct answer is option (d).
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)= \begin{cases}a x^{2}+b & , 0 \leq x<1 \\ 4 & , \quad x=1 \\ x+3 &, 1<x \leq 2\end{cases}$
Step 1: Understand that, f(x) is continuous at x = 1
Therefore,
$\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(1) \\ &\lim _{h \rightarrow 0} f(1-h)=4 \\ &\lim _{h \rightarrow 0} a(1-h)^{2}+b=4 \end{aligned}$
Therefore, the possible values for (a, b) can be (2, 2), (3, 1), (4, 0) but (a,b) ≠ (5, 2)
Hence, the correct answer is option (d).
Continuity exercise multiple choice question 33
Answer:The correct option is (b)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
(i) Standard limits
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1, \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1$
Given:
$f(x)=\left\{\begin{array}{cc} \frac{\log (1+3 x)-\log (1-2 x)}{x} \quad, x \neq 0 \\ k \qquad \quad, x=0 \end{array} \right.$
Step 1: Understand that, if f(x) is continuous at x = 0
$\begin{aligned} &\therefore \lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+3 x)-\log (1-2 x)}{x}=k \end{aligned}$
$\lim _{x \rightarrow 0} \frac{3 \log (1+3 x)}{3 x}-\frac{2 \log (1-2 x)}{2 x}=k$
Simplifying further,
$3 \lim _{x \rightarrow 0} \frac{\log (1+3 x)}{3 x}+2 \lim _{x \rightarrow 0} \frac{\log (1-2 x)}{-2 x}=k$
Applying formula (i)
$\begin{aligned} &\therefore 3 \times 1+2 \times 1=k \\ &k=3+2 \\ &k=5 \end{aligned}$
So, correct option is (b)
Continuity exercise multiple choice question 34
Answer:The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow a} \frac{sin\: x}{x}=1 \\ \end{aligned}$
Given:
$f(x)= \begin{cases}\frac{1-\cos 10 x}{x^{2}} &, x<0 \\ a & , x=0 \\ \frac{\sqrt{x}}{\sqrt{625+x}-25} &, x>0\end{cases}$
then the value of a so that f(x) may be continuous at x = 0
Using LHL
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(-h) \\ &\quad=\lim _{h \rightarrow 0} \frac{1-\cos (-10 h)}{(-h)^{2}} \\ &\quad=50 \lim _{h \rightarrow 0} \frac{(\sin 5 h)^{2}}{(5 h)^{2}} \\ &\quad=50(\text { Using standard limits }) \end{aligned}$
Function f(x) is continuous at x=0
50 = a
So, the correct option is (b).
Continuity exercise multiple choice question 35
Answer:The correct option is (a)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ &\text { (iii) } \lim _{x \rightarrow 0}\{f(x) \pm g(x)\}=1 \pm m \text {, } \end{aligned}$
$\begin{aligned} &{\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \\ \end{aligned}$
Given:
$\begin{aligned} & f(x)=x \sin \frac{1}{x} \end{aligned}$
Solution:
$\begin{aligned} & f(x)=x \sin \frac{1}{x} \end{aligned}$
$=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$
Function f(x) is continuous at x = 0
$\begin{aligned} &=\lim _{x \rightarrow 0} f(x)=f(0) \\ &=0=f(0) \end{aligned}$
So, option (a) is correct.
Continuity exercise multiple choice question 36
Answer:
The correct option is (d)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \end{aligned}$
Given:
$\begin{aligned} f(x)=\sin \frac{1}{x} \end{aligned}$
Solution:
$\begin{aligned} f(x)=\sin \frac{1}{x} \end{aligned}$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \sin \frac{1}{x}$
Function f(x) is continuous at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \sin \frac{1}{x}=k \end{aligned}$
Value does not exist for fraction to be continuous.
So, the option (d) is correct.
The correct option is (d)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \end{aligned}$
Given:
$\begin{aligned} f(x)=\sin \frac{1}{x} \end{aligned}$
Solution:
$\begin{aligned} f(x)=\sin \frac{1}{x} \end{aligned}$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \sin \frac{1}{x}$
Function f(x) is continuous at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \sin \frac{1}{x}=k \end{aligned}$
Value does not exist for fraction to be continuous.
So, the option (d) is correct.
Continuity exercise multiple choice question 37
Answer:The correct option is (e)
Hint:
Use this definition of continuity to solve this problem. A function f(x) is said to be continuous at x=a
if
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=f(a)$
Given:
$f(x)=\left\{\begin{array}{cl} (1+a x)^{1 / x} & , x<0 \\ b & , x=0 \\ \frac{(x+c)^{1 / 3}-1}{(x+1)^{1 / 2-1}} & , x>0 \end{array}\right.$
may be continuous at x = 0
Solution:
Now at x = 1
$f(x)=\left\{\begin{array}{cl} (1+a x)^{1 / x} & , x<0 \\ b & , x=0 \\ \frac{(x+c)^{1 / 3}-1}{(x+1)^{1 / 2-1}} & , x>0 \end{array}\right.$
To find the value of a, b and c we will use definition of continuous function
So we know if f(x) is continuous at x = 0 then we have,
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)-(1)$
Using equation (1) We have,
$\lim _{x \rightarrow 0^{-}} f(x)=f(0) \Rightarrow \lim _{h \rightarrow 0} f(0-h)=f(0)$
$\begin{aligned} &\left.\Rightarrow \lim _{h \rightarrow 0}\{1+a(-h)\}^{\frac{1}{h}}=f(0) \quad [\because f(x)=(1+a x)^{1 / x}, x<0\right] \\ &\Rightarrow \lim _{h \rightarrow 0}(1-a h)^{-\frac{1}{h}}=f(0)=b \quad[\because \boldsymbol{f}(\boldsymbol{x})=\boldsymbol{b}, \boldsymbol{x}=\mathbf{0}] \end{aligned}$
Taking log both sides, then
$\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \log (1-a b)^{-1 / n}=\log b \\ &\Rightarrow \lim _{h \rightarrow 0}-\frac{1}{n} \log (1-a h)=\log b \quad\left[\because \log a^{m}=\boldsymbol{m l o g} a\right] \end{aligned}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{a \log (1-a h)}{-a h}=\log b \quad \text { [multiply and divide a on L.H.S] }$
$\begin{aligned} &\Rightarrow a \lim _{h \rightarrow 0} \frac{\log \{1+(-a b)\}}{(-a b)}=\log b\\ &\Rightarrow a \cdot 1=\log b &\left[\therefore \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]\\ &\Rightarrow a=\log b \qquad - (2) \end{aligned}$
Again, using equation (1) we get,
$\begin{aligned} &\lim _{x \rightarrow 0+} f(x)=\{(0) \\ &\Rightarrow \lim _{h \rightarrow 0} f(0+h)=f(0) \end{aligned}$
$\begin{aligned} &\left.\Rightarrow \lim _{h \rightarrow 0} f(n)=f^{\prime} (0) \right) \\ &\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1}{(h+1)^{1 / 2}-1}\right)=f(0) \quad\left[\because \boldsymbol{f}(x)=\frac{(x+\boldsymbol{c})^{\mathbf{1} / 3}-4}{(\boldsymbol{x}+\mathbf{1})^{\mathbf{1} / 2}-\mathbf{1}}, x>\mathbf{0}\right] \end{aligned}$
$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1}{(n+1)^{1 / 2}-1} \times \frac{(n+1)^{1 / 2}+1}{(n+1)^{1 / 2}+1}\right)=\mathrm{b} \quad[\because f(x)=b, x=0]$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+r)^{4} 3-1\right) \times\left((n+1)^{42}+1\right)}{\left((h+1)^{1 / 2}-1\right)\left((h+1)^{1 / 2}+1\right)}=b$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)\left((n+1)^{4 / 2}+1\right)}{\left((h+1)^{1 / 2}\right)^{2}-(1)^{2}}=\mathrm{b} \quad\left[\because(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=\boldsymbol{a}^{2}-\boldsymbol{b}^{2}\right]$
$\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)\left((h+1)^{42}+1\right)}{h+1-1}=b \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{\left.\left((h+c)^{1 / 3}-1\right)(n+1)^{4 / 2}+1\right)}{h}=b \end{aligned}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\left((h+c)^{1 / 3}-1\right)}{h} \times \lim _{h \rightarrow 0}\left((n+1)^{1 / 2}+1\right)=b \quad\left[\begin{array}{ll} \because \lim _{x \rightarrow a} f(x) \cdot g(x)= & \lim _{x \rightarrow a} f(x) \\ x \lim _{x \rightarrow a} g(x) \end{array}\right]$
$\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(h+c)^{1 / 3}-1^{4 / 3}}{h}\right) \times(0+1)^{1 / 2}+1=b \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{(h+c)^{1 / 3}-1^{1 / 3}}{(h+c)-c} \times 1^{1 / 2}+1=b \\ &\Rightarrow g \times 2=b \end{aligned}$
$\begin{aligned} &\text { where } g \therefore=\lim _{h \rightarrow 0} \frac{(h+c)^{1 / 3}-1^{1 / 3}}{(h+c)-c}-(a) \\ &\Rightarrow g \cdots=\frac{6}{2} \quad-(3) \end{aligned}$
To find the value of g, we will use
$\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-\mathrm{L}}$
But this formula is applicable on g only if c=L
So we take c = L then we have,
$g=\lim _{h \rightarrow 0} \frac{(h+1)^{1 / 3}-1^{1 / 3}}{(h+1)-1}$ $\left[\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]$
$\begin{aligned} &=\frac{1}{3} \cdot 1^{1 / 3-1} \\ &=\frac{1}{3} 1^{-2 / 3}=\frac{1}{3} \cdot 1=\frac{1}{3} \\ &\therefore g=\frac{1}{3} \end{aligned}$
Equation (3)
$\begin{aligned} &\Rightarrow \frac{1}{3}=\frac{b}{2} \\ &\Rightarrow b=\frac{2}{3} \end{aligned}$
Then equation 2 becomes,
$\begin{aligned} &\Rightarrow a=\log \frac{2}{3} \\ &\therefore a=\frac{\log 2}{3}, b=\frac{2}{3}, c=1 \end{aligned}$
Continuity exercise multiple choice question 38
Answer:The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
Given:
$f(x)= \begin{cases}2 \sqrt{x} &, 0 \leq x \leq 1 \\ 4-2 x &, 1<x<\frac{5}{2} \\ 2 x-7 &, \frac{5}{2} \leq x \leq 4\end{cases}$
Solution:
Now at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0} f(4-2(1+h)) \\ &=2 \end{aligned}$
Again,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0}(2 \sqrt{1-h}) \\ &=2 \\ &=\lim _{x^{+} \rightarrow 0} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x) \end{aligned}$
Therefore, function is continuous at x = 0
Now we will check the continuity of f(x) at
$\begin{aligned} x = \frac{5}{2} \end{aligned}$
$\begin{aligned} &\text { So } \lim _{x \rightarrow \frac{5}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{5}{2}-h\right) \\ &=\lim _{h \rightarrow 0}\left(4-2\left(\frac{5}{2}-h\right)\right) \quad\left[\because f(x)=4-2 x, x<\frac{5}{2}\right] \\ &=4-2\left(\frac{5}{2}-0\right) \end{aligned}$
$\begin{aligned} &=4-2 \times \frac{5}{2} \\ &=4-5 \\ &=-1 \end{aligned}$
$\begin{aligned} \text { And } \lim _{x \rightarrow \frac{5}{2}^{+}} f(x) &=\lim _{h \rightarrow 0} f\left(\frac{5}{2}+h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{5}{2}+h\right)-7 \quad\left[\because f(x)=2 x-7, x>\frac{5}{2}\right] \\ &=2\left(\frac{5}{2}+0\right)-7 \end{aligned}$
$\begin{aligned} &=2 \times \frac{5}{2}-7 \\ &=5-7 \\ &=-2 \\ \lim _{x \rightarrow \frac{5}{2}} f(x) & \neq \lim _{x \rightarrow \frac{5}{2}} f(x) \\ \text { le }-1 \neq-2 & \end{aligned}$
Therefore the function f(x) is discontinuous at
$\begin{aligned} x = \frac{5}{2} \end{aligned}$
So, correct option is (b)
Continuity exercise multiple choice question 39
Answer:The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{1-\sin ^{2} x}{3 \cos ^{2} x} &, x<\frac{\pi}{2} \\ a & , x=\frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\pi-2 x)^{2}} &, x>\frac{\pi}{2}\end{cases}$
Solution:
$f(x)= \begin{cases}\frac{1-\sin ^{2} x}{3 \cos ^{2} x} &, x<\frac{\pi}{2} \\ a & , x=\frac{\pi}{2} \\ \frac{b(1-\sin x)}{(\pi-2 x)^{2}} &, x>\frac{\pi}{2}\end{cases}$
So it is given that the function f(x) is continuous at
$x=\frac{\pi}{2}$
So, we have
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}}+f(x)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)-(1) \\ &\text { L.H.L at } x=\frac{\pi}{2} \\ &{\text { i.e }} \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f(\frac{\pi}{2}-h) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{1-\sin ^{2}(\pi / 2-h)}{3 \cos ^{2}(\pi / 2-h)} \quad\left[\therefore \boldsymbol{f}(x)=\frac{1-\sin ^{2} x}{3 \cos ^{2} x}, x<\pi / 2\right]$
$=\lim _{h \rightarrow 0} \frac{1-\{\sin (\pi / 2-h)\}^{2}}{3\{\cos (\pi / 2-h)\}^{2}}$
$=\lim _{h \rightarrow 0} \frac{1-(\cos h)^{2}}{3(\sin h)^{2}} \quad\left[\begin{array}{c} \because \sin (\pi / 2-\theta)=\cos \theta, \\ \cos (\pi / 2-\theta)=\sin \theta \end{array}\right]$
$=\lim _{h \rightarrow 0} \frac{1-\cos ^{2} h}{3 \sin ^{2} h}$
$\begin{aligned} &=\lim _{n \rightarrow 0} \frac{\sin ^{2} h}{3 \sin ^{2} h} \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=\perp \Rightarrow 1-\cos ^{2} \theta=\sin ^{2} \theta\right] \\ &=\frac{1}{3} \lim _{h \rightarrow 0} 1=\frac{1}{3} . \quad {-(2)} \end{aligned}$
$\begin{aligned} &\text { R.H.I at } x=\pi / 2 \\ &{\text { i.e }} \lim _{x \rightarrow \pi / 2^{+}} f(x)=\lim _{h \rightarrow 0} f(\pi / 2+h) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{b\{1-\sin (\pi / 2+h)\}}{\{\pi-2(\pi / 2+h)\}^{2}} \quad\left[\because {f}(x)=\frac{{b}(\mathbf{1}-\sin x)}{(\pi-2 x)^{2}}, x>\pi / 2\right]$
$=\lim _{h \rightarrow 0} \frac{b\{1-\cosh \}}{\left\{\pi-2\left(\frac{\pi}{2}\right)-2 h\right\}^{2}} \quad[\because \sin (\pi / 2+\theta)=\cos \theta]$
$=\lim _{h \rightarrow 0} \frac{b\left\{2 \sin ^{2} h / 2\right\}}{(\pi-\pi-2 h)^{2}} \quad\left[\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{2 b \sin ^{2} h}{2}}{(-2 h)^{2}} \\ &=\lim _{h \rightarrow 0} \frac{\frac{2 b \sin ^{2} h}{2}}{4 h^{2}} \\ &=\frac{26}{4} \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} n}{2}}{h^{2}} \end{aligned}$
$\begin{aligned} &=\frac{b}{2} \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} h}{2}}{\frac{h^{2}}{4} x^{4}} \\ &=\frac{b}{8} \lim _{h \rightarrow 0}\left(\frac{\frac{\sin h}{2}}{\frac{h}{2}}\right)^{2} \\ &=\frac{6}{8}\left(\lim _{h \rightarrow 0} \frac{\sin n / 2}{h / 2}\right)^{2} \end{aligned}$
$\begin{aligned} &=\frac{b}{8} \times 1^{2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=\frac{b}{8} \times 1 \\ &=\frac{b}{8}-(3) \end{aligned}$
$\begin{aligned} &\text { also, } f(x) a t-x=\pi / 2\\ &{\text { i.e }} f(\pi / 2)=a \quad -(4) &[\because {f}({x})={a}, {x}=\pi / 2] \end{aligned}$
Putting the value of equation 2 ,3 , 4 in (1)
we get ,
$\begin{aligned} &-\frac{1}{3}=\frac{b}{8}=a \\ &\text { Sor } a=\frac{1}{3} \text { or } \frac{b}{8}=\frac{1}{3} \Rightarrow b=\frac{8}{3} \\ &\therefore a=\frac{1}{3}, b=\frac{8}{2} \end{aligned}$
Continuity exercise multiple choice question 40
Answer:The correct option is (b)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
Given:
$f(x)= \begin{cases}\frac{1}{5}\left(2 x^{2}+3\right) &, \quad x \leq 1 \\ 6-5 x & , \quad 1<x<3 \\ x-3 & , \quad x \geq 3\end{cases}$
Solution:
Now at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &\lim _{h \rightarrow 0}[6-5(1+h)]=1 \\ &\lim _{h \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}$
Therefore continuous at x = 1
Now, at x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h) \\ &\lim _{h \rightarrow 0}[(3+h)-3] \\ &=0 \end{aligned}$
For left hand limit,
$\begin{aligned} &=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h) \\ &=\lim _{h \rightarrow 0}[6-5(3-h)]=-9 \end{aligned}$
We have
$=\lim _{x \rightarrow 3^{+}} f(x) \neq \lim _{x \rightarrow 3^{-}} f(x)$
Therefore discontinuous at x = 3
So, the correct option is (b).
Continuity exercise multiple choice question 41
Answer:
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{-}} f(x)=f(a) \end{aligned}$
Given:
$f(x)= \begin{cases}5 x-4 &, 0<x \leq 1 \\ 4 x^{2}+3 a x &, 1<x<2\end{cases}$
is continuous at every point of domain
Solution:
$f(x)= \begin{cases}5 x-4 &, 0<x \leq 1 \\ 4 x^{2}+3 a x &, 1<x<2\end{cases}$
At x = 1 f(x)=1
So,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} s(1-h)-4 \quad[\because {f}(x)=5 x-4 \text { when } x<1] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} 5-5 h-4\\ &=\lim _{h \rightarrow 0} 1-5 h\\ &=1-5 \times 0\\ &=1 \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 1^{+}} f(x) &=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0}\left\{4(1+h)^{2}+3 a(1+n)\right\} &\left[\because {f}({x})=4 {x}^{2}+3 {a} x \text { when } x>\mathbf{1}\right] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{4\left(1+h^{2}+2 b\right)+3 a+3 a h\right\} \quad\left[\because(\boldsymbol{a}+\boldsymbol{b})^{2}=\boldsymbol{a}^{2}+\boldsymbol{b}^{2}+2 \boldsymbol{a} \boldsymbol{b}\right] \\ &=\lim _{h \rightarrow 0}\left(4+4 h^{2}+8 h+3 a+3 a h\right) \end{aligned}$
$\begin{aligned} &=4+4 \times 0^{2}+8 x 0+3 a+3 a \times 0 \\ &=4+0+0+3 a+0=4+3 a \quad-(2) \\ &\text { Also, } f(1)=5 \cdot 1-4 \\ &\quad=5-4 \\ &\qquad=1 \quad-(3) \quad[\because {f}(x)=5 x-4 \text { when } x=1] \end{aligned}$
It is given that the function f(x) is continuous at x=1
$\begin{aligned} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) &=\lim _{x \rightarrow 1^{+}} f(x) \\ &=f(1) \\ \Rightarrow 1=4+3 a & \end{aligned}$
$\begin{aligned} &=1 \quad \text { [using equation (1), (2) and (3) }\\ &\text { So, } \Rightarrow 4+3 a=1 \\ \Rightarrow 3 a &=1-4 \\ &=-3 \\ \Rightarrow a &=\quad \frac{-3}{3}=-1 \\ & \therefore a=-1 \end{aligned}$
Continuity exercise multiple choice question 42
The correct option is (d)
Hint:
A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{-}} f(x)=f(a) \end{aligned}$
Given:
$f(x)= \begin{cases}5 x-4 &, 0<x \leq 1 \\ 4 x^{2}+3 a x &, 1<x<2\end{cases}$
is continuous at every point of domain
Solution:
$f(x)= \begin{cases}5 x-4 &, 0<x \leq 1 \\ 4 x^{2}+3 a x &, 1<x<2\end{cases}$
At x = 1 f(x)=1
So,
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} s(1-h)-4 \quad[\because {f}(x)=5 x-4 \text { when } x<1] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} 5-5 h-4\\ &=\lim _{h \rightarrow 0} 1-5 h\\ &=1-5 \times 0\\ &=1 \end{aligned}$
$\begin{aligned} \lim _{x \rightarrow 1^{+}} f(x) &=\lim _{h \rightarrow 0} f(1+h) \\ &=\lim _{h \rightarrow 0}\left\{4(1+h)^{2}+3 a(1+n)\right\} &\left[\because {f}({x})=4 {x}^{2}+3 {a} x \text { when } x>\mathbf{1}\right] \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0}\left\{4\left(1+h^{2}+2 b\right)+3 a+3 a h\right\} \quad\left[\because(\boldsymbol{a}+\boldsymbol{b})^{2}=\boldsymbol{a}^{2}+\boldsymbol{b}^{2}+2 \boldsymbol{a} \boldsymbol{b}\right] \\ &=\lim _{h \rightarrow 0}\left(4+4 h^{2}+8 h+3 a+3 a h\right) \end{aligned}$
$\begin{aligned} &=4+4 \times 0^{2}+8 x 0+3 a+3 a \times 0 \\ &=4+0+0+3 a+0=4+3 a \quad-(2) \\ &\text { Also, } f(1)=5 \cdot 1-4 \\ &\quad=5-4 \\ &\qquad=1 \quad-(3) \quad[\because {f}(x)=5 x-4 \text { when } x=1] \end{aligned}$
It is given that the function f(x) is continuous at x=1
$\begin{aligned} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) &=\lim _{x \rightarrow 1^{+}} f(x) \\ &=f(1) \\ \Rightarrow 1=4+3 a & \end{aligned}$
$\begin{aligned} &=1 \quad \text { [using equation (1), (2) and (3) }\\ &\text { So, } \Rightarrow 4+3 a=1 \\ \Rightarrow 3 a &=1-4 \\ &=-3 \\ \Rightarrow a &=\quad \frac{-3}{3}=-1 \\ & \therefore a=-1 \end{aligned}$
Continuity exercise multiple choice question 42
Answer:
The correct option is (a)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ &(\text { iii }) \lim _{x \rightarrow 0}\{f(x) \pm g(x)\}=1 \pm m, \\ &{\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2}\end{cases}$
Solution:
$f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2} \end{cases} \qquad -(1)$
$\begin{aligned} &\text { At } x=\frac{\pi}{2}\\ &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)\\ &\Rightarrow \lim _{x \rightarrow \pi / 2} \frac{\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}}=k &\text { [using 1] } \end{aligned}$
Let
$\begin{aligned} &\pi-2 x=2 y\\ &\Rightarrow y=\frac{\pi-2 x}{2}\\ &\Rightarrow y=\frac{\pi}{2}-x\\ &\text { Since, when } x \rightarrow \frac{\pi}{2} \text { then } y \rightarrow 0 \text {. } \end{aligned}$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\cos x)-\cos x}{(2 y)^{2}}=k \quad[\because \pi-2 x=y]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin \{\cos (\pi / 2-y)\}-\cos (\pi / 2-y)}{4 y^{2}}=k \quad\left[\because y=\frac{\pi}{2}-x \Rightarrow x=\frac{\pi}{2}-y\right]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\sin y)-\sin y}{4 y^{2}}=k \quad\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{2 \sin \left(\frac{\sin y-y}{2}\right) \cdot \cos \left(\frac{\sin y+y}{2}\right)}{4 y^{2}}=k \quad\left[\because \sin c-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cdot \cos \left(\frac{c+D}{2}\right)\right]$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\left(\frac{\sin y-y}{2}\right) \cdot \sin \left(\frac{\sin y-y}{2}\right)}{y^{2} \cdot\left(\frac{\sin y-y}{2}\right)} \cdot \cos \left(\frac{\sin y+y}{2}\right)$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\sin y-y}{2}}{y}\right) \cdot\left\{\frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \mid \cdot\left\{\frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}\right\}=k\right.$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\sin y-y}{2 y} \times \lim _{y \rightarrow 0} \frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k$
$\left[\because \lim _{x \rightarrow a} f(x) \cdot g(x) \cdot h(x)=\lim _{x \rightarrow a} f(x) \lim _{x \rightarrow a} g(x) \lim _{x \rightarrow a} h(x)\right]$
$\Rightarrow \frac{1}{4}(1-1) \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &\Rightarrow \frac{1}{4} \times 0 \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y^{\circ}}=R \\ &\Rightarrow 0=k \end{aligned}$
$\begin{aligned} \therefore k=0 \end{aligned}$
So, the correct option is (a)
The correct option is (a)
Hint:
(i) A function f(x) is said to be continuous at a point x = a of its domain, if
$\begin{aligned} &\lim _{x \rightarrow a} f(x)=f(a) \\ &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \end{aligned}$
(ii) Standard limits
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ &(\text { iii }) \lim _{x \rightarrow 0}\{f(x) \pm g(x)\}=1 \pm m, \\ &{\text { Where, }} \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}$
Given:
$f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2}\end{cases}$
Solution:
$f(x)= \begin{cases}\frac{(\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}} &, x \neq \frac{\pi}{2} \\ k & , x<\frac{\pi}{2} \end{cases} \qquad -(1)$
$\begin{aligned} &\text { At } x=\frac{\pi}{2}\\ &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)\\ &\Rightarrow \lim _{x \rightarrow \pi / 2} \frac{\sin (\cos x)-\cos x)}{(\pi-2 x)^{2}}=k &\text { [using 1] } \end{aligned}$
Let
$\begin{aligned} &\pi-2 x=2 y\\ &\Rightarrow y=\frac{\pi-2 x}{2}\\ &\Rightarrow y=\frac{\pi}{2}-x\\ &\text { Since, when } x \rightarrow \frac{\pi}{2} \text { then } y \rightarrow 0 \text {. } \end{aligned}$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\cos x)-\cos x}{(2 y)^{2}}=k \quad[\because \pi-2 x=y]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin \{\cos (\pi / 2-y)\}-\cos (\pi / 2-y)}{4 y^{2}}=k \quad\left[\because y=\frac{\pi}{2}-x \Rightarrow x=\frac{\pi}{2}-y\right]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{\sin (\sin y)-\sin y}{4 y^{2}}=k \quad\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right]$
$\Rightarrow \lim _{y \rightarrow 0} \frac{2 \sin \left(\frac{\sin y-y}{2}\right) \cdot \cos \left(\frac{\sin y+y}{2}\right)}{4 y^{2}}=k \quad\left[\because \sin c-\sin D=2 \sin \left(\frac{C-D}{2}\right) \cdot \cos \left(\frac{c+D}{2}\right)\right]$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\left(\frac{\sin y-y}{2}\right) \cdot \sin \left(\frac{\sin y-y}{2}\right)}{y^{2} \cdot\left(\frac{\sin y-y}{2}\right)} \cdot \cos \left(\frac{\sin y+y}{2}\right)$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\sin y-y}{2}}{y}\right) \cdot\left\{\frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \mid \cdot\left\{\frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}\right\}=k\right.$
$\Rightarrow \frac{1}{2} \lim _{y \rightarrow 0} \frac{\sin y-y}{2 y} \times \lim _{y \rightarrow 0} \frac{\sin \left(\frac{\sin y-y}{2}\right)}{\left(\frac{\sin y-y}{2}\right)} \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k$
$\left[\because \lim _{x \rightarrow a} f(x) \cdot g(x) \cdot h(x)=\lim _{x \rightarrow a} f(x) \lim _{x \rightarrow a} g(x) \lim _{x \rightarrow a} h(x)\right]$
$\Rightarrow \frac{1}{4}(1-1) \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y}=k \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &\Rightarrow \frac{1}{4} \times 0 \times 1 \times \lim _{y \rightarrow 0} \frac{\cos \left(\frac{\sin y+y}{2}\right)}{y^{\circ}}=R \\ &\Rightarrow 0=k \end{aligned}$
$\begin{aligned} \therefore k=0 \end{aligned}$
So, the correct option is (a)
Answer:
The correct option is (d)
Hint:
use this to get
$\frac{f(x)}{g(x)}$
Given:
$\begin{aligned} &f(x)=2 x \\ &g(x)=\frac{x^{2}}{2}+1 \end{aligned}$
For the discontinuous function
$\frac{f(x)}{g(x)} \rightarrow g(x) \neq 0$
If g(x) = 0, then it is a discontinuous function
So, the correct option is (d)
The correct option is (d)
Hint:
use this to get
$\frac{f(x)}{g(x)}$
Given:
$\begin{aligned} &f(x)=2 x \\ &g(x)=\frac{x^{2}}{2}+1 \end{aligned}$
For the discontinuous function
$\frac{f(x)}{g(x)} \rightarrow g(x) \neq 0$
If g(x) = 0, then it is a discontinuous function
So, the correct option is (d)
Answer:
The correct option is (a)
Given:
We have
$f(x)=\cot x=\frac{\cos x}{\sin x}$
Now, f(x) is discontinuous when sinx = 0 we know that
$\sin x=0 \text { at } x=n \pi, n \in \mathbb{Z}$
f(x) = cot x is discontinuous on the set
$\{x: x=n \pi: n \in \mathbb{Z}\}$
So, the correct option is (a)
The correct option is (a)
Given:
We have
$f(x)=\cot x=\frac{\cos x}{\sin x}$
Now, f(x) is discontinuous when sinx = 0 we know that
$\sin x=0 \text { at } x=n \pi, n \in \mathbb{Z}$
f(x) = cot x is discontinuous on the set
$\{x: x=n \pi: n \in \mathbb{Z}\}$
So, the correct option is (a)
Continuity exercise multiple choice question 45
Answer:The correct option is (a)
Given:
The given function f is
$f(x)=x^{2} \sin \frac{1}{x}, \text { where } x \neq 0$
It is evident that f is defined at all points of the real line.
let C be a real number.
$\begin{aligned} &\text { Case } 1 \text { : If } c \neq 0 \text {, then } f(c)=c^{2} \sin \frac{1}{c}\\ &\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\left(\lim _{x \rightarrow c} x^{2}\right)\left(\lim _{x \rightarrow c} \sin \frac{1}{x}\right)=c^{2} \sin \frac{1}{c} \end{aligned}$
$\therefore \lim _{x \rightarrow c} f(x)=f(c)$
Therefore, f is continuous at all points x≠0
Case 2: If c=0, then f(0)=0
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)=c^{2} \sin \frac{1}{c}$
It is known that
$\begin{aligned} &-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 \\ &=-x^{2} \leq \sin \frac{1}{x} \leq x^{2} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2} \\ &=0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0 \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=0 \end{aligned}$
Similarly,
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right)=0$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)$
There fore f is continuous at x = 0
So, the correct option is (a)
Continuity exercise multiple choice question 46
Answer:The correct option is (d)
$f(x)=\frac{1}{x-[x]}$
The greatest integer function becomes discontinuous at every integer value of x. So, the whole function will become discontinuous at each integer value of x.
It has infinite number of points of discontinuity.
So, the correct option is (d)
Continuity exercise multiple choice question 47
Answer:The correct option is (d)
The function f(x) = [x] is continuous for all x except all integral values of x
Hence, it is continuous at x = 1.5, which is not an integer
So, the correct option is (d)
Continuity exercise multiple choice question 48
Answer:The correct option is (b)
Hint:
Standard limit:
$\begin{aligned} (i) \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ (ii) \lim _{x \rightarrow 0} \cos x=1 \end{aligned}$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{\sin x}{x}+\cos x &, \text { if } x \neq 0 \\ k & , \text { if } x=0 \end{array}\right.$
f(x) is continuous at x = 0.
So,$\begin{aligned} &=\lim _{x \rightarrow 0} f(x)=f(0) \\ &=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\cos x\right)=k \\ &=\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \cos x=k \\ &=1+1=k \\ &=k=2 \end{aligned}$
RD Sharma class 12 solution of Continuity ex MCQ contains questions of the chapter Continuity. There are about 48 MCQs in this exercise that are extremely basic and simple to solve if you are aware of the fundamentals of this chapter. However, if you find any question tricky to solve, you can refer to the RD Sharma class 12th exercise MCQ to get a good understanding of all the concepts in this chapter.
JEE Main Highest Scoring Chapters & Topics
Focus on high-weightage topics with this eBook and prepare smarter. Gain accuracy, speed, and a better chance at scoring higher.
Download E-bookThe essential concepts covered in the RD Sharma class 12th exercise MCQ are:-
The intuitive notion of continuity
Testing the continuity of a function
Continuity of a point
Continuity of an interval
Continuity of an open interval
Continuity of an closed interval
Definition and meaning of continuous function
Properties of continuous function
The questions prepared in the RD Sharma class 12th exercise MCQ are created by experts from academics across the country. The experts also provide helpful tips and tricks to solve the questions quickly and alternately, which will be less time-consuming. RD Sharma solutions Therefore, using Class 12 RD Sharma exercise MCQ solution can help prepare for the exams efficiently as the questions mentioned in the book are frequently asked in the board exams. The level of questions in the RD Sharma class 12th exercise MCQ is exactly the same as in the NCERT that are helpful in the preparation of public examinations as well.
The latest version of the RD Sharma class 12th exercise MCQ is available at the Career360 website. As the new version is released, the older materials are replaced. The solutions by Career360 are updated to the latest version. Also, the RD Sharma class 12th exercise MCQ can be downloaded free of cost from the Career360 website, which means it will not cost you a single penny to own the RD Sharma class 12 chapter 8 exercise MCQ.
RD Sharma Chapter-wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
Articles
Upcoming School Exams
Get answers from students and experts
Applications for Admissions are open.
This ebook serves as a valuable study guide for NEET 2025 exam.
NEET Previous 10 Year Questions
Get nowThis e-book offers NEET PYQ and serves as an indispensable NEET study material.
JEE Main Important Physics formulas
Get nowAs per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
JEE Main Important Chemistry formulas
Get nowAs per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
JEE Main high scoring chapters and topics
Get nowAs per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
JEE Main Important Mathematics Formulas
Get nowAs per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters