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RD Sharma Class 12 Exercise 8.2 Continuity Solutions Maths - Download PDF Free Online
Edited By Kuldeep Maurya | Updated on Jan 27, 2022 02:13 PM IST
RD Sharma is one of the most well-known books in the country. RD Sharma's books are detailed, informative, and also contain step-by-step solutions for their problems. Class 12 RD Sharma chapter 8 exercise 8.2 solution deals with the chapter 'Continuity.' The book has plenty of examples that the students can practice to develop their skills. Still, when it comes to solving a complex chapter like Continuity, students also need to exercise problems. This is where RD Sharma class 12th exercise 8.2 solution comes to light.
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This Story also Contains
- RD Sharma Class 12 Solutions Chapter 8 Continuity- Other Exercise
- Continuity Excercise:8.2
- RD Sharma Chapter wise Solutions
Also Read - RD Sharma Solution for Class 9 to 12 Maths
RD Sharma Class 12 Solutions Chapter 8 Continuity- Other Exercise
- Chapter 8 - Continuity -Ex-8.1
- Chapter 8 - Continuity -Ex-FBQ
- Chapter 8 - Continuity -Ex-MCQ
- Chapter 8 - Continuity -Ex-VSA
Continuity Excercise:8.2
Continuity exercise 8.2 question 1
Answer:f(x) is everywhere continuous.
Hint:
A function is everywhere continuous when it is continuous at every x$\in$IR
Given:
$f(x)=\left\{\begin{array}{cc} \frac{\sin x}{x} & x<0 \\ x+1 & x \geq 0 \end{array}\right.$
Explanation:
Now, at x = 0
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \frac{\sin x}{x} \quad\left[\because \lim _{h \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} x+1=1 \end{aligned}$
And
$f(x) = 0 + 1 = 1$
As,
$\left.\lim _{x \rightarrow 0} f(x)=f(0) \quad \text { [for continuity } \lim _{x \rightarrow 0} f(x)=f(0)\right]$
Hence, f(x) is everywhere continuous.
Note: sine function, identity function, polynomial functions are everywhere continuous.
Continuity exercise 8.2 question 2
Answer:Discontinuous at x = 0.
Hint:
If a function is not continuous at one point then it is discontinuous. As at that point
$\lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x \rightarrow a^{+}} f(x) \neq f(0)$
and this is the definition of discontinuity.
Given:
$f(x)= \begin{cases}\frac{x}{|x|} & x \neq 0 \\ 0 & x=0\end{cases}$
Explanation:
Now consider at x = 0
$\begin{aligned} &\text { L.H.L } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{-h}{-h \mid}=-1 \\ &\text { R.H.L } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{h}{|h|}=1 \end{aligned}$
So
$L.H.S \neq R.H.L$
Function is discontinuous at X = 0
Answer:
x = 1
Hint:
Show LHS $\neq$ RHS or LHL $\neq$ the value of function at given point or RHL $\neq$ the value of function at given point.
Given:
$f(x)=\left\{\begin{array}{rr} x^{3}-x^{2}+2 x-2 & x \neq 1 \\ 4 & x=1 \end{array}\right.$
Explanation:
Now consider the point x = 1
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} x^{3}-x^{2}+2 x-2 \\ &=1-1+2 \times 1-2 \\ &=0 \end{aligned}$
$\begin{aligned} &\text { R.}{\text {H.L }}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(1+h)^{3}-(1-h)^{2}+2(1+h)-2 \\ &=1-1+2-2=0 \\ &\qquad f(1)=4 \end{aligned}$
As
$\begin{aligned} &\text { L.H.L } = \text { R.H.L }\neq f(1) \end{aligned}$
f(x) is discontinuous at x =1
x = 1
Hint:
Show LHS $\neq$ RHS or LHL $\neq$ the value of function at given point or RHL $\neq$ the value of function at given point.
Given:
$f(x)=\left\{\begin{array}{rr} x^{3}-x^{2}+2 x-2 & x \neq 1 \\ 4 & x=1 \end{array}\right.$
Explanation:
Now consider the point x = 1
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} x^{3}-x^{2}+2 x-2 \\ &=1-1+2 \times 1-2 \\ &=0 \end{aligned}$
$\begin{aligned} &\text { R.}{\text {H.L }}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(1+h)^{3}-(1-h)^{2}+2(1+h)-2 \\ &=1-1+2-2=0 \\ &\qquad f(1)=4 \end{aligned}$
As
$\begin{aligned} &\text { L.H.L } = \text { R.H.L }\neq f(1) \end{aligned}$
f(x) is discontinuous at x =1
Continuity exercise 8.2 question 3 (ii)
Answer:
x = 2
Hint:
To check the continuity of such type of function we check at the breaking point.
Given:
$f(x)= \begin{cases}\frac{x^{4}-16}{x-2} & x \neq 2 \\ 16 & x=2\end{cases}$
Explanation:
At point x =2
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} \frac{(2-h)^{4}-16}{(2-h)-2} \\ &=\lim _{h \rightarrow 0} \frac{2^{4}-4.8 h+6.4 h^{2}-4.2 h^{3}+h^{4}-16}{-h} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{32-32 h+24 h^{2}-8 h^{3}+h^{4}-16}{-h} \\ &=\lim _{h \rightarrow 0} 32-24 h+8 h^{2}-h^{3}=32 \end{aligned}$
$\text { R.H.L }=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{(2+h)-2}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{2^{4}+4.8 h+6.4 h^{2}+4.2 h^{3}+h^{4}-16}{h}=32 \\ &\text { Also } f(2)=16 \\ &\text { Thus } L . H .L=R . H . L \neq f(2) \end{aligned}$
Therefore, f(x) is discontinuous at x = 2
x = 2
Hint:
To check the continuity of such type of function we check at the breaking point.
Given:
$f(x)= \begin{cases}\frac{x^{4}-16}{x-2} & x \neq 2 \\ 16 & x=2\end{cases}$
Explanation:
At point x =2
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} \frac{(2-h)^{4}-16}{(2-h)-2} \\ &=\lim _{h \rightarrow 0} \frac{2^{4}-4.8 h+6.4 h^{2}-4.2 h^{3}+h^{4}-16}{-h} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{32-32 h+24 h^{2}-8 h^{3}+h^{4}-16}{-h} \\ &=\lim _{h \rightarrow 0} 32-24 h+8 h^{2}-h^{3}=32 \end{aligned}$
$\text { R.H.L }=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} \frac{(2+h)^{4}-16}{(2+h)-2}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{2^{4}+4.8 h+6.4 h^{2}+4.2 h^{3}+h^{4}-16}{h}=32 \\ &\text { Also } f(2)=16 \\ &\text { Thus } L . H .L=R . H . L \neq f(2) \end{aligned}$
Therefore, f(x) is discontinuous at x = 2
Continuity exercise 8.2 question 3 (iii)
Answer:
x = 0
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin x}{x} & x<0 \\ 2 x+3 & x \geq 0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}=1 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sinh }{h}=1 \\ &\text { Also } f(0)=2 \times 0+3=3 \\ &\text { Thus L.H.L}=\text {R.H.L} \neq \mathrm{f}(0) \end{aligned}$
f is discontinuous at x = 0
x = 0
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin x}{x} & x<0 \\ 2 x+3 & x \geq 0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}=1 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sinh }{h}=1 \\ &\text { Also } f(0)=2 \times 0+3=3 \\ &\text { Thus L.H.L}=\text {R.H.L} \neq \mathrm{f}(0) \end{aligned}$
f is discontinuous at x = 0
Continuity exercise 8.2 question 3 (iv)
Answer:
x = 0
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin 3 x}{x} & x \neq 0 \\ 4 & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin 3(-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sin 3 \mathrm{~h}}{-h}=1 \\ &=\lim _{h \rightarrow 0} \frac{-3 \sin 3 \mathrm{~h}}{3 h}=3 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sin 3 \mathrm{~h}}{h}=3 \\ &\text { Also } f(0)=4 \\ &\text { Thus L.H.L }=\text { R.H.L } \neq \mathrm{f}(0) \end{aligned}$
$\therefore$ f is not continuous at x = 0.
x = 0
Hint:
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Given:
$f(x)= \begin{cases}\frac{\sin 3 x}{x} & x \neq 0 \\ 4 & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin 3(-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sin 3 \mathrm{~h}}{-h}=1 \\ &=\lim _{h \rightarrow 0} \frac{-3 \sin 3 \mathrm{~h}}{3 h}=3 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sin 3 \mathrm{~h}}{h}=3 \\ &\text { Also } f(0)=4 \\ &\text { Thus L.H.L }=\text { R.H.L } \neq \mathrm{f}(0) \end{aligned}$
$\therefore$ f is not continuous at x = 0.
Continuity exercise 8.2 question 3 (v)
Answer:
x = 0
Hint:
$\lim _{x \rightarrow a} f(x)+g(x)=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)$
Given:
$f(x)= \begin{cases}\frac{\sin x}{x}+\cos x & x \neq 0 \\ 5 & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H. } L=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)+\cos (-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}+\cosh =1+1=2 \end{aligned}$
$\begin{aligned} &R . H \cdot L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sinh }{h}+\cosh =1+1=2 \\ &\text { Also } f(0)=5 \\ &\text { Thus } L . H . L=R . H . L \neq \mathrm{f}(0) \end{aligned}$
As
$\lim _{x \rightarrow 0} f \rightarrow(x) \neq x f(0)$
f is not continuous at x = 0
x = 0
Hint:
$\lim _{x \rightarrow a} f(x)+g(x)=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)$
Given:
$f(x)= \begin{cases}\frac{\sin x}{x}+\cos x & x \neq 0 \\ 5 & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\text { L.H. } L=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)+\cos (-h)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}+\cosh =1+1=2 \end{aligned}$
$\begin{aligned} &R . H \cdot L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{\sinh }{h}+\cosh =1+1=2 \\ &\text { Also } f(0)=5 \\ &\text { Thus } L . H . L=R . H . L \neq \mathrm{f}(0) \end{aligned}$
As
$\lim _{x \rightarrow 0} f \rightarrow(x) \neq x f(0)$
f is not continuous at x = 0
Continuity exercise 8.2 question 3 (vi)
Answer:
x = 0
Hint:
Use L-Hospital Rule when we find 0/0 form
Given:
$f(x)= \begin{cases}\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x} & x \neq 0 \\ 10 & x=0\end{cases}$
Explanation:
at x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{(-h)^{4}+(-h)^{3}+2(-h)^{2}}{\tan ^{-1}(-h)} \\ &=\lim _{h \rightarrow 0} \frac{h^{4}-h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{h^{4}+h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \\ &\text { Also } f(0)=10 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}$
Hence, f(x) is discontinuous at x = 0.
x = 0
Hint:
Use L-Hospital Rule when we find 0/0 form
Given:
$f(x)= \begin{cases}\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x} & x \neq 0 \\ 10 & x=0\end{cases}$
Explanation:
at x = 0
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{(-h)^{4}+(-h)^{3}+2(-h)^{2}}{\tan ^{-1}(-h)} \\ &=\lim _{h \rightarrow 0} \frac{h^{4}-h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \end{aligned}$
$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{h^{4}+h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \\ &\text { Also } f(0)=10 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}$
Hence, f(x) is discontinuous at x = 0.
Continuity exercise 8.2 question 3 (vii)
Answer:
x = 0
Hint:
Use L-Hospital Rule when we find 0/0 form
Given:
$f(x)=\left\{\begin{array}{cl} \frac{e^{x}-1}{\log _{e}(1-2 x)} & x \neq 0 \\ 7 & x=0 \end{array}\right.$
Explanation:
At x = 0
$\text { L.H.L } =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{\log _{e}(1-2 h)}$
$=\lim _{h \rightarrow 0} \frac{\frac{e^{-h}-1}{-h}}{\frac{\log _{e}(1-2 h)}{-2 h} \times 2}=\frac{1}{2}$
$\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{e^{h}-1}{\log _{e}(1+2 h)}$
$=\lim _{h \rightarrow 0} \frac{\frac{e^{h}-1}{h}}{\frac{\log _{e}(2 h)}{2 h} \times 2}=\frac{1}{2}$
$\begin{aligned} &\text { Also } f(0)=7 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}$
$\therefore$f(x) is not continuous at x = 0
x = 0
Hint:
Use L-Hospital Rule when we find 0/0 form
Given:
$f(x)=\left\{\begin{array}{cl} \frac{e^{x}-1}{\log _{e}(1-2 x)} & x \neq 0 \\ 7 & x=0 \end{array}\right.$
Explanation:
At x = 0
$\text { L.H.L } =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{\log _{e}(1-2 h)}$
$=\lim _{h \rightarrow 0} \frac{\frac{e^{-h}-1}{-h}}{\frac{\log _{e}(1-2 h)}{-2 h} \times 2}=\frac{1}{2}$
$\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{e^{h}-1}{\log _{e}(1+2 h)}$
$=\lim _{h \rightarrow 0} \frac{\frac{e^{h}-1}{h}}{\frac{\log _{e}(2 h)}{2 h} \times 2}=\frac{1}{2}$
$\begin{aligned} &\text { Also } f(0)=7 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}$
$\therefore$f(x) is not continuous at x = 0
Continuity exercise 8.2 question 3 (viii)
Answer:
f(x) is nowhere discontinuous
Hint:
$f(x)=|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases}$
Given:
$f(x)=|x|= \begin{cases}|x-3| & x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{15}{4} & x<1\end{cases}$
Explanation:
$\begin{aligned} &f(x)= \begin{cases}x-3 & x \geq 3 \\ -(x-3) & 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & x<1\end{cases} \end{aligned}$
Now, we check the limit at x = 1 and x = 3
At x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}-(x-3)=0 \\ &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} x-3=3-3=0 \\ &f(3)=3-3=0 \end{aligned}$
As
$\lim _{x \rightarrow 3} f(x)=f(3)$
Hence, f(x) is continuous at x = 3
Now, at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} \\ &=\frac{1}{4}-\frac{3}{2}+\frac{13}{4} \\ &=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=-(1-3)=2 \\ &f(1)=-(1-3)=2 \end{aligned}$
As
$\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore$f(x) is continuous at x = 1
As polynomial function is everywhere continuous & greater integer function is continuous except its end point &
$\lim _{x \rightarrow 1} f(x) \text { \& } \lim _{x \rightarrow 3} f(x)$
is continuous.
Hence, f(x) is everywhere continuous.
f(x) is nowhere discontinuous
Hint:
$f(x)=|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases}$
Given:
$f(x)=|x|= \begin{cases}|x-3| & x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{15}{4} & x<1\end{cases}$
Explanation:
$\begin{aligned} &f(x)= \begin{cases}x-3 & x \geq 3 \\ -(x-3) & 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & x<1\end{cases} \end{aligned}$
Now, we check the limit at x = 1 and x = 3
At x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}-(x-3)=0 \\ &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} x-3=3-3=0 \\ &f(3)=3-3=0 \end{aligned}$
As
$\lim _{x \rightarrow 3} f(x)=f(3)$
Hence, f(x) is continuous at x = 3
Now, at x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} \\ &=\frac{1}{4}-\frac{3}{2}+\frac{13}{4} \\ &=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=-(1-3)=2 \\ &f(1)=-(1-3)=2 \end{aligned}$
As
$\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore$f(x) is continuous at x = 1
As polynomial function is everywhere continuous & greater integer function is continuous except its end point &
$\lim _{x \rightarrow 1} f(x) \text { \& } \lim _{x \rightarrow 3} f(x)$
is continuous.
Hence, f(x) is everywhere continuous.
Continuity exercise 8.2 question 3 (ix)
Answer:
at x = 3
Hint:
Check the continuity of function at breaking points.
Given:
$f(x)=\left\{\begin{array}{cc} |x|+3 & x \leq-3 \\ -2 x & -3<x<3 \\ 6 x+2 & x>3 \end{array}\right.$
Explanation:
at x = -3
$\begin{aligned} &\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3}-2 x=6 \\ &f(-3)=0+3+3=6 \\ &\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3}-x+3 \\ &=6 \end{aligned}$
Now at x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}-2 x=-6 \\ &\quad \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} 6 x+2=20 \\ \end{aligned}$
As
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x) \end{aligned}$
f(x) is not continuous at x = 3.
at x = 3
Hint:
Check the continuity of function at breaking points.
Given:
$f(x)=\left\{\begin{array}{cc} |x|+3 & x \leq-3 \\ -2 x & -3<x<3 \\ 6 x+2 & x>3 \end{array}\right.$
Explanation:
at x = -3
$\begin{aligned} &\lim _{x \rightarrow-3^{+}} f(x)=\lim _{x \rightarrow-3}-2 x=6 \\ &f(-3)=0+3+3=6 \\ &\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3}-x+3 \\ &=6 \end{aligned}$
Now at x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}-2 x=-6 \\ &\quad \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} 6 x+2=20 \\ \end{aligned}$
As
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x) \end{aligned}$
f(x) is not continuous at x = 3.
Continuity exercise 8.2 question 3 (x)
Answer:
x = 1
Hint:
Check the continuity of function at end points.
Given:
$f(x)= \begin{cases}x^{10}-1 & x \leq 1 \\ x^{2} & x>1\end{cases}$
Explanation:
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} x^{10}-1=\lim _{x \rightarrow 1} x^{10}-1 \\ &=(1)^{10}-1=1-1=0 \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} x^{2} \\ &=(1)^{2}=1 \\ \end{aligned}$
As
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x) \end{aligned}$
f(x) is not continuous at x = 1
x = 1
Hint:
Check the continuity of function at end points.
Given:
$f(x)= \begin{cases}x^{10}-1 & x \leq 1 \\ x^{2} & x>1\end{cases}$
Explanation:
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} x^{10}-1=\lim _{x \rightarrow 1} x^{10}-1 \\ &=(1)^{10}-1=1-1=0 \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} x^{2} \\ &=(1)^{2}=1 \\ \end{aligned}$
As
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x) \end{aligned}$
f(x) is not continuous at x = 1
Continuity exercise 8.2 question 3 (xi)
Answer:
x = 1
Hint:
Check at end points.
Given:
$f(x)=\left\{\begin{array}{cc} 2 x & x<0 \\ 0 & 0 \leq x \leq 1 \\ 4 x & x>1 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} 2 x=2 \times 0=0-(1) \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 0=0-(2) \\ &f(0)=0-(3) \end{aligned}$
As (1)=(2)=(3)
Hence, f(x) is continuous at x = 0
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 4 x=4 \times 1=4 \\ &f(1)=0 \end{aligned}$
As
$\lim _{x \rightarrow 1^{+}} f(x) \neq f(1)$
Hence, f(X) is discontinuous at x = 1
x = 1
Hint:
Check at end points.
Given:
$f(x)=\left\{\begin{array}{cc} 2 x & x<0 \\ 0 & 0 \leq x \leq 1 \\ 4 x & x>1 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} 2 x=2 \times 0=0-(1) \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 0=0-(2) \\ &f(0)=0-(3) \end{aligned}$
As (1)=(2)=(3)
Hence, f(x) is continuous at x = 0
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 4 x=4 \times 1=4 \\ &f(1)=0 \end{aligned}$
As
$\lim _{x \rightarrow 1^{+}} f(x) \neq f(1)$
Hence, f(X) is discontinuous at x = 1
Continuity exercise 8.2 question 3 (xii)
Answer:
f(x) is continuous everywhere.
Hint:
$\lim _{x \rightarrow 0} f(x)-g(x)=\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0} g(x)$
Given:
$f(x)=\left\{\begin{array}{cc} \sin x-\cos x & x \neq 0 \\ -1 & x=0 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \sin x-\cos x \\ &=\lim _{x \rightarrow 0} \sin x-\lim _{x \rightarrow 0} \cos x \\ &=\sin 0-\cos 0 \\ &=-1 \end{aligned}$
And
$\begin{aligned} &f(x)=-1 \end{aligned}$
Constant function,
Sine and cosine function is everywhere continuous and f(x) is continuous at x = 0
$\therefore$f(x) is everywhere continuous.
f(x) is continuous everywhere.
Hint:
$\lim _{x \rightarrow 0} f(x)-g(x)=\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0} g(x)$
Given:
$f(x)=\left\{\begin{array}{cc} \sin x-\cos x & x \neq 0 \\ -1 & x=0 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \sin x-\cos x \\ &=\lim _{x \rightarrow 0} \sin x-\lim _{x \rightarrow 0} \cos x \\ &=\sin 0-\cos 0 \\ &=-1 \end{aligned}$
And
$\begin{aligned} &f(x)=-1 \end{aligned}$
Constant function,
Sine and cosine function is everywhere continuous and f(x) is continuous at x = 0
$\therefore$f(x) is everywhere continuous.
Continuity exercise 8.2 question 3 (xiii)
Answer:
everywhere continuous.
Hint:
Constant and identity function is everywhere continuous.
Given:
$f(x)=\left\{\begin{array}{rr} -2 & x \leq-1 \\ 2 x & -1<x<1 \\ 2 & x \geq 1 \end{array}\right.$
Explanation:
Constant and identity function are everywhere continuous. So, we only have to check at x = -1, x = 1 (end points) -(1)
At x = -1
$\begin{aligned} &\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}-2=-2 \\ &f(-1)=-2 \\ &\lim _{\left(x \rightarrow 1^{+}\right)} f(x)=\lim _{x \rightarrow 1^{-}} 2 \times x \\ &=2 \times-1=-2 \end{aligned}$
As
$\begin{gathered} \lim _{x \rightarrow-1} f(x)=f(-1) \\ f(x)_{\text {is continuous at }} x=-1-(2) \end{gathered}$
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 2 x=2 \times 1=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2=2 \\ &f(1)=2 \end{aligned}$
As
$\lim _{x \rightarrow 1} f(x)=f(1)-(3)$
So from (1), (2) & (3)
f(x) is everywhere continuous.
everywhere continuous.
Hint:
Constant and identity function is everywhere continuous.
Given:
$f(x)=\left\{\begin{array}{rr} -2 & x \leq-1 \\ 2 x & -1<x<1 \\ 2 & x \geq 1 \end{array}\right.$
Explanation:
Constant and identity function are everywhere continuous. So, we only have to check at x = -1, x = 1 (end points) -(1)
At x = -1
$\begin{aligned} &\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}-2=-2 \\ &f(-1)=-2 \\ &\lim _{\left(x \rightarrow 1^{+}\right)} f(x)=\lim _{x \rightarrow 1^{-}} 2 \times x \\ &=2 \times-1=-2 \end{aligned}$
As
$\begin{gathered} \lim _{x \rightarrow-1} f(x)=f(-1) \\ f(x)_{\text {is continuous at }} x=-1-(2) \end{gathered}$
At x = 1
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 2 x=2 \times 1=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2=2 \\ &f(1)=2 \end{aligned}$
As
$\lim _{x \rightarrow 1} f(x)=f(1)-(3)$
So from (1), (2) & (3)
f(x) is everywhere continuous.
Answer:
$\frac{2}{15}$
Hint:
f(x) is continuous so it is also continuous at end points. Put LHL= RHL
Given:
$f(x)= \begin{cases}\frac{\sin 2 x}{5 x} & x \neq 0 \\ 3 k & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x} \\ &=\frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times 2 \\ &=\frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}\left[\operatorname{as} \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=\frac{2}{5} \times 1=\frac{2}{5} \\ &\qquad f(0)=3 k \end{aligned}$
As f(X) is continuous at x = 0 when
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\frac{2}{5}=3 k \\ &k=2 / 15 \end{aligned}$
$\frac{2}{15}$
Hint:
f(x) is continuous so it is also continuous at end points. Put LHL= RHL
Given:
$f(x)= \begin{cases}\frac{\sin 2 x}{5 x} & x \neq 0 \\ 3 k & x=0\end{cases}$
Explanation:
At x = 0
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x} \\ &=\frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times 2 \\ &=\frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}\left[\operatorname{as} \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=\frac{2}{5} \times 1=\frac{2}{5} \\ &\qquad f(0)=3 k \end{aligned}$
As f(X) is continuous at x = 0 when
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\frac{2}{5}=3 k \\ &k=2 / 15 \end{aligned}$
Continuity exercise 8.2 question 4 (ii)
Answer:
$k = -2$
Hint:
Put LHL = RHL = f(2)
Given:
$f(x)= \begin{cases}k x+5 & x \leq 0 \\ x-1 & x>0\end{cases}$
Explanation:
LHL = RHL = f(2) ....(1)
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} k(2-h)+5 \\ &=2 k+5 \end{aligned}$
$R . H . L=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} k(2+h)-1=1$
Using (1) we get
$\begin{aligned} &\therefore 2 k+5=1 \\ &2 k=1-5 \\ &=-4 \\ &k=-2 \end{aligned}$
$k = -2$
Hint:
Put LHL = RHL = f(2)
Given:
$f(x)= \begin{cases}k x+5 & x \leq 0 \\ x-1 & x>0\end{cases}$
Explanation:
LHL = RHL = f(2) ....(1)
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} k(2-h)+5 \\ &=2 k+5 \end{aligned}$
$R . H . L=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} k(2+h)-1=1$
Using (1) we get
$\begin{aligned} &\therefore 2 k+5=1 \\ &2 k=1-5 \\ &=-4 \\ &k=-2 \end{aligned}$
Continuity exercise 8.2 question 4 (iii)
Answer:
No value of k can make +
Hint:
Put L.H.L = R.H.L at x = 0
Given:
$f(x)= \begin{cases}k\left(x^{2}+3 x\right) & x<0 \\ \cos 2 x & x \geq 0\end{cases}$
Explanation:
At x = 0
L.H.L = R.H.L = f(0) ....(1)
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} k\left[(-h)^{2}+3(-h)\right] \\ &=\lim _{h \rightarrow 0} k\left[h^{2}-3 h\right]=0 \\ &f(0)=\cos 2 \times 0=\cos 0=1 \\ &\text { L.H.L } \neq \mathrm{f}(0) \end{aligned}$
Hence no value of k can make f continuous.
No value of k can make +
Hint:
Put L.H.L = R.H.L at x = 0
Given:
$f(x)= \begin{cases}k\left(x^{2}+3 x\right) & x<0 \\ \cos 2 x & x \geq 0\end{cases}$
Explanation:
At x = 0
L.H.L = R.H.L = f(0) ....(1)
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} k\left[(-h)^{2}+3(-h)\right] \\ &=\lim _{h \rightarrow 0} k\left[h^{2}-3 h\right]=0 \\ &f(0)=\cos 2 \times 0=\cos 0=1 \\ &\text { L.H.L } \neq \mathrm{f}(0) \end{aligned}$
Hence no value of k can make f continuous.
Continuity exercise 8.2 question 4 (iv)
Answer:
$a=7 / 2, \: \: b=-17 / 2$
Hint:
Put LHL = RHL at x = 3 , 5
Given:
$f(x)= \begin{cases}2 & x \leq 3 \\ a x+b & 3<x<5 \\ 9 & x \geq 5\end{cases}$
Explanation:
At x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} a x+b\\ &=3 a+b\\ &f(3)=2 \qquad ....(1) \end{aligned}$$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0}(3+h)=\lim _{h \rightarrow 0} a(3+h)+b \\ &=3 a+b \end{aligned}$
$a=7 / 2, \: \: b=-17 / 2$
Hint:
Put LHL = RHL at x = 3 , 5
Given:
$f(x)= \begin{cases}2 & x \leq 3 \\ a x+b & 3<x<5 \\ 9 & x \geq 5\end{cases}$
Explanation:
At x = 3
$\begin{aligned} &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} a x+b\\ &=3 a+b\\ &f(3)=2 \qquad ....(1) \end{aligned}$$\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0}(3+h)=\lim _{h \rightarrow 0} a(3+h)+b \\ &=3 a+b \end{aligned}$
Now, f(x) is continuous at x = 3
If 3a + b =2 .....(A)
Now, at x = 5
$\begin{aligned} &\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5} a x+b \\ &=5 a+b \\ & f(5)=9 \end{aligned}$
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 5} f(x)=\lim _{h \rightarrow 0}(5-h)=\lim _{h \rightarrow 0} a(5-h)+b \\ &=5 a+b \end{aligned}$
Now, f(x) is continuous at x = 5
If 5a + b = 9 .....(B)
on solving A and B we get,
$a=7 / 2, \: \: b=-17 / 2$
Continuity exercise 8.2 question 4 (v)
Answer:
a = 3, b = 1
Hint:
Put LHL = RHL at x = 3 , 5
Given:
$f(x)= \begin{cases}4 & x \leq-1 \\ a x^{2}+b & -1<x<0 \\ \cos x & x \geq 0\end{cases}$
Explanation:
At x = -1
$\begin{aligned} &f(-1)=4\\ &\text { R.H.L }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} a(-1+h)^{2}+b=a+b\\ &\therefore a+b=4 \qquad ....(A) \end{aligned}$
Now at x = 0
$\begin{aligned} &f(0)=\cos 0=1 \\ &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} a(-h)^{2}+b \\ &=b \end{aligned}$
$\begin{aligned} &\therefore f(0)=L . H . L \\ &\Rightarrow b=1 \\ &\text { From }(A) \\ &a=3 \end{aligned}$
Hence, a = 3, b = 1
a = 3, b = 1
Hint:
Put LHL = RHL at x = 3 , 5
Given:
$f(x)= \begin{cases}4 & x \leq-1 \\ a x^{2}+b & -1<x<0 \\ \cos x & x \geq 0\end{cases}$
Explanation:
At x = -1
$\begin{aligned} &f(-1)=4\\ &\text { R.H.L }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} a(-1+h)^{2}+b=a+b\\ &\therefore a+b=4 \qquad ....(A) \end{aligned}$
Now at x = 0
$\begin{aligned} &f(0)=\cos 0=1 \\ &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} a(-h)^{2}+b \\ &=b \end{aligned}$
$\begin{aligned} &\therefore f(0)=L . H . L \\ &\Rightarrow b=1 \\ &\text { From }(A) \\ &a=3 \end{aligned}$
Hence, a = 3, b = 1
Continuity exercise 8.2 question 4 (vi)
Answer:
$p=-1/2$
Hint:
Put LHL = RHL at x = 0
Given:
$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & 0 \leq x \leq 1\end{cases}$
Explanation:
At x = 0
L.H.L
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \\ &=\lim _{x \rightarrow 0} \frac{p}{\sqrt{1+p x}}+\frac{p}{2 \sqrt{1+p x}} \quad\left[0 / 0_{\text {from }}\right] \\ &=\lim _{x \rightarrow 0} \frac{2 p}{\sqrt{1+p x}} \times 2 \end{aligned}$
$\begin{aligned} &=\frac{2 p}{2 \times \sqrt{1+p \times 0}}=\frac{2 p}{2}=p \\ &R . H \cdot L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{2 h+1}{h-2}=\frac{-1}{2} \end{aligned}$
As f(x) is continuous at x = 0
Hence, p = -1/2
$p=-1/2$
Hint:
Put LHL = RHL at x = 0
Given:
$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & 0 \leq x \leq 1\end{cases}$
Explanation:
At x = 0
L.H.L
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \\ &=\lim _{x \rightarrow 0} \frac{p}{\sqrt{1+p x}}+\frac{p}{2 \sqrt{1+p x}} \quad\left[0 / 0_{\text {from }}\right] \\ &=\lim _{x \rightarrow 0} \frac{2 p}{\sqrt{1+p x}} \times 2 \end{aligned}$
$\begin{aligned} &=\frac{2 p}{2 \times \sqrt{1+p \times 0}}=\frac{2 p}{2}=p \\ &R . H \cdot L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{2 h+1}{h-2}=\frac{-1}{2} \end{aligned}$
As f(x) is continuous at x = 0
Hence, p = -1/2
Continuity exercise 8.2 question 4 (vii)
Answer:
a = 2, b = 1
Hint:
Put LHL = RHL at x = 2 &x = 10
Given:
$f(x)= \begin{cases}5 & x \leq 2 \\ a x+b & 2<x<10 \\ 21 & x \geq 10\end{cases}$
Explanation:
At x = 2
$\begin{aligned} &f(2)=5 \\ &\lim _{x \rightarrow 2^{+}} f(x)=i m_{x \rightarrow 2} a x+b \\ &=2 a+b \end{aligned}$
As, f(x) is continuous at x = 2 when 5 + 2a +b - (1)
At x = 10
$\begin{gathered} \lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10} a x+b \\ =10 a+b \\ f(10)=21 \end{gathered}$
As, f(x) is continuous at x = 10
10a + b =21
From (1) & (2) we have
8a = 16
a = 2
Put in (1)
b = 1
Hence, a = 2 & b = 1
a = 2, b = 1
Hint:
Put LHL = RHL at x = 2 &x = 10
Given:
$f(x)= \begin{cases}5 & x \leq 2 \\ a x+b & 2<x<10 \\ 21 & x \geq 10\end{cases}$
Explanation:
At x = 2
$\begin{aligned} &f(2)=5 \\ &\lim _{x \rightarrow 2^{+}} f(x)=i m_{x \rightarrow 2} a x+b \\ &=2 a+b \end{aligned}$
As, f(x) is continuous at x = 2 when 5 + 2a +b - (1)
At x = 10
$\begin{gathered} \lim _{x \rightarrow 10^{-}} f(x)=\lim _{x \rightarrow 10} a x+b \\ =10 a+b \\ f(10)=21 \end{gathered}$
As, f(x) is continuous at x = 10
10a + b =21
From (1) & (2) we have
8a = 16
a = 2
Put in (1)
b = 1
Hence, a = 2 & b = 1
Continuity exercise 8.2 question 4 (viii)
Answer:
k = 6
Hint:
Put LHL = RHL =
$f(\frac{\pi }{2})$
Given:
$f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x} & x<\pi / 2 \\ 3 & x=\pi / 2 \\ \frac{3 \tan ^{2} x}{2 x-\pi} & x \geq 10\end{cases}$
Explanation:
$At\: \: x = \frac{\pi }{2}$
$\begin{aligned} &\text { L.H.L }=f(x) \text { at } x=\frac{\pi}{2} \text { is } \\ &=\lim _{x \rightarrow \frac{\pi}{2}} f(x) \\ &=\lim _{h \rightarrow 0}\left(h-\frac{\pi}{2}\right) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{k \cos \left(h-\frac{\pi}{2}\right)}{\pi-2\left(h-\frac{\pi}{2}\right)}=\frac{k}{2}$
$\begin{aligned} &\text { Again } \mathrm{f}\left(\frac{\pi}{2}\right)=3 \\ &\text { L.H.L } =f\left(\frac{\pi}{3}\right) \\ &\therefore \frac{k}{2}=3 \\ &k=6 \end{aligned}$
k = 6
Hint:
Put LHL = RHL =
$f(\frac{\pi }{2})$
Given:
$f(x)= \begin{cases}\frac{k \cos x}{\pi-2 x} & x<\pi / 2 \\ 3 & x=\pi / 2 \\ \frac{3 \tan ^{2} x}{2 x-\pi} & x \geq 10\end{cases}$
Explanation:
$At\: \: x = \frac{\pi }{2}$
$\begin{aligned} &\text { L.H.L }=f(x) \text { at } x=\frac{\pi}{2} \text { is } \\ &=\lim _{x \rightarrow \frac{\pi}{2}} f(x) \\ &=\lim _{h \rightarrow 0}\left(h-\frac{\pi}{2}\right) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{k \cos \left(h-\frac{\pi}{2}\right)}{\pi-2\left(h-\frac{\pi}{2}\right)}=\frac{k}{2}$
$\begin{aligned} &\text { Again } \mathrm{f}\left(\frac{\pi}{2}\right)=3 \\ &\text { L.H.L } =f\left(\frac{\pi}{3}\right) \\ &\therefore \frac{k}{2}=3 \\ &k=6 \end{aligned}$
Continuity exercise 8.2 question 5
Answer:
$a=-1, b=1 \text { or } a=1, b=1 \pm \sqrt{2}$
Hint:
p at LHL = RHL at
$x =1 ,\: \sqrt{2}$
Given:
$f(x)= \begin{cases}\frac{x^{2}}{9} & 0 \leq x<1 \\ 9 & 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}} & \sqrt{2} \leq x<\infty\end{cases}$
Explanation:
$\begin{aligned} &\text { At } x=1, L . H . L=R . H . L=f(1) \quad ...(A)\\ &f(1)=A \quad ...(1) \end{aligned}$
$L . H . L=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{a}=\frac{1}{a}$
Using (A)
$\begin{aligned} &a=\frac{1}{a} \Rightarrow a^{2}=1 \Rightarrow a=\pm 1\\ &\text { At } x=\sqrt{2}, \quad \text { L.H.L }=\text { R.H.L }=f(\sqrt{2}) \quad ....(B) \end{aligned}$
$f(\sqrt{2})=\frac{2 b^{2}-4 b}{(\sqrt{2})^{2}}=\frac{2 b^{2}-4 b}{2}=b^{2}-2 b \quad ....(2)$
$\text { L.H.L }=\lim _{x \rightarrow \sqrt{2}^{2}} f(x)=\lim _{h \rightarrow 0}(\sqrt{2}-h)=\lim _{h \rightarrow 0} a=a$
So using (B) we get
$\begin{aligned} &b^{2}-2 b=\pm 1 \\ &\qquad b^{2}-2 b=1 \text { or } \quad b^{2}-2 b=-1 \\ &\qquad b^{2}-2 b-1=0 \quad \text { or } \quad b^{2}-2 b+1=0 \end{aligned}$
$\begin{aligned} &b=\frac{1 \pm \sqrt{2}}{1} \quad \text { Or } \quad(b-1)^{2}=0\\ &b=1 \pm \sqrt{2} \quad \text { Or } \quad b=1 \end{aligned}$
Hence,
$a=-1, b=1 \quad \text { or } \quad a=1, b=1 \pm \sqrt{2}$
$a=-1, b=1 \text { or } a=1, b=1 \pm \sqrt{2}$
Hint:
p at LHL = RHL at
$x =1 ,\: \sqrt{2}$
Given:
$f(x)= \begin{cases}\frac{x^{2}}{9} & 0 \leq x<1 \\ 9 & 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}} & \sqrt{2} \leq x<\infty\end{cases}$
Explanation:
$\begin{aligned} &\text { At } x=1, L . H . L=R . H . L=f(1) \quad ...(A)\\ &f(1)=A \quad ...(1) \end{aligned}$
$L . H . L=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{a}=\frac{1}{a}$
Using (A)
$\begin{aligned} &a=\frac{1}{a} \Rightarrow a^{2}=1 \Rightarrow a=\pm 1\\ &\text { At } x=\sqrt{2}, \quad \text { L.H.L }=\text { R.H.L }=f(\sqrt{2}) \quad ....(B) \end{aligned}$
$f(\sqrt{2})=\frac{2 b^{2}-4 b}{(\sqrt{2})^{2}}=\frac{2 b^{2}-4 b}{2}=b^{2}-2 b \quad ....(2)$
$\text { L.H.L }=\lim _{x \rightarrow \sqrt{2}^{2}} f(x)=\lim _{h \rightarrow 0}(\sqrt{2}-h)=\lim _{h \rightarrow 0} a=a$
So using (B) we get
$\begin{aligned} &b^{2}-2 b=\pm 1 \\ &\qquad b^{2}-2 b=1 \text { or } \quad b^{2}-2 b=-1 \\ &\qquad b^{2}-2 b-1=0 \quad \text { or } \quad b^{2}-2 b+1=0 \end{aligned}$
$\begin{aligned} &b=\frac{1 \pm \sqrt{2}}{1} \quad \text { Or } \quad(b-1)^{2}=0\\ &b=1 \pm \sqrt{2} \quad \text { Or } \quad b=1 \end{aligned}$
Hence,
$a=-1, b=1 \quad \text { or } \quad a=1, b=1 \pm \sqrt{2}$
Continuity exercise 8.2 question 6
Answer:
$a = \pi /6,\; b=-\pi /12$
Hint:
Put at LHL = RHL at
$x = \pi /4,\; x=\pi /2$
Given:
$f(x)= \begin{cases}x+a \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+b & \pi / 4 \leq x<\pi / 2 \\ a \cos 2 x-b \sin x & \pi / 2 \leq x<\pi\end{cases}$
Explanation:
At x = $\pi$/4
$L.H.L = R.H.L =f(\frac{\pi }{4}) \qquad ....(A)$
Now,
$\begin{aligned} &f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cdot \cot \left(\frac{\pi}{4}\right)+b\\ &=\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b \qquad ....(1) \end{aligned}$
L.H.L
$\begin{aligned} &\lim _{x \rightarrow \pi / 4^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right) \\ &=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=\frac{\pi}{4}+a \end{aligned}$
Using (A)
$\begin{aligned} &\frac{1}{2}+b=\frac{\pi}{4}+a\\ &a-b=\frac{\pi}{4} \qquad ....(B) \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{\pi}{2} \\ &\text { L.H.L }=\text { R.H.L }=f\left(\frac{\pi}{2}\right) \quad \ldots(C) \end{aligned}$
Now
$f\left(\frac{\pi}{2}\right)=a \cos 2 \cdot \frac{\pi}{2}-b \sin \frac{\pi}{2}=-a-b \qquad ....(2)$
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b=b \end{aligned}$
Using (C)
$-a-b=b \Rightarrow 2 b=-a \Rightarrow b=\frac{-a}{2}$
From (B)
$\begin{aligned} &a+\frac{a}{2}=\frac{\pi}{4} \\ &\frac{3}{2} a=\frac{\pi}{4} \\ &\Rightarrow a=\frac{\pi}{6} \end{aligned}$
$\begin{aligned} &b=\frac{-a}{2}=-\frac{\pi}{12} \\ \end{aligned}$
Hence
$\begin{aligned} &a=\pi / 6, b=-\pi / 12 \end{aligned}$
$a = \pi /6,\; b=-\pi /12$
Hint:
Put at LHL = RHL at
$x = \pi /4,\; x=\pi /2$
Given:
$f(x)= \begin{cases}x+a \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+b & \pi / 4 \leq x<\pi / 2 \\ a \cos 2 x-b \sin x & \pi / 2 \leq x<\pi\end{cases}$
Explanation:
At x = $\pi$/4
$L.H.L = R.H.L =f(\frac{\pi }{4}) \qquad ....(A)$
Now,
$\begin{aligned} &f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cdot \cot \left(\frac{\pi}{4}\right)+b\\ &=\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b \qquad ....(1) \end{aligned}$
L.H.L
$\begin{aligned} &\lim _{x \rightarrow \pi / 4^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right) \\ &=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=\frac{\pi}{4}+a \end{aligned}$
Using (A)
$\begin{aligned} &\frac{1}{2}+b=\frac{\pi}{4}+a\\ &a-b=\frac{\pi}{4} \qquad ....(B) \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{\pi}{2} \\ &\text { L.H.L }=\text { R.H.L }=f\left(\frac{\pi}{2}\right) \quad \ldots(C) \end{aligned}$
Now
$f\left(\frac{\pi}{2}\right)=a \cos 2 \cdot \frac{\pi}{2}-b \sin \frac{\pi}{2}=-a-b \qquad ....(2)$
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b=b \end{aligned}$
Using (C)
$-a-b=b \Rightarrow 2 b=-a \Rightarrow b=\frac{-a}{2}$
From (B)
$\begin{aligned} &a+\frac{a}{2}=\frac{\pi}{4} \\ &\frac{3}{2} a=\frac{\pi}{4} \\ &\Rightarrow a=\frac{\pi}{6} \end{aligned}$
$\begin{aligned} &b=\frac{-a}{2}=-\frac{\pi}{12} \\ \end{aligned}$
Hence
$\begin{aligned} &a=\pi / 6, b=-\pi / 12 \end{aligned}$
Continuity exercise 8.2 question 6
Answer:
$a = \pi /6,\; b=-\pi /12$
Hint:
Put at LHL = RHL at
$x = \pi /4,\; x=\pi /2$
Given:
$f(x)= \begin{cases}x+a \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+b & \pi / 4 \leq x<\pi / 2 \\ a \cos 2 x-b \sin x & \pi / 2 \leq x<\pi\end{cases}$
Explanation:
At x = $\pi$/4
$L.H.L = R.H.L =f(\frac{\pi }{4}) \qquad ....(A)$
Now,
$\begin{aligned} &f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cdot \cot \left(\frac{\pi}{4}\right)+b\\ &=\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b \qquad ....(1) \end{aligned}$
L.H.L
$\begin{aligned} &\lim _{x \rightarrow \pi / 4^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right) \\ &=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=\frac{\pi}{4}+a \end{aligned}$
Using (A)
$\begin{aligned} &\frac{1}{2}+b=\frac{\pi}{4}+a\\ &a-b=\frac{\pi}{4} \qquad ....(B) \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{\pi}{2} \\ &\text { L.H.L }=\text { R.H.L }=f\left(\frac{\pi}{2}\right) \quad \ldots(C) \end{aligned}$
Now
$f\left(\frac{\pi}{2}\right)=a \cos 2 \cdot \frac{\pi}{2}-b \sin \frac{\pi}{2}=-a-b \qquad ....(2)$
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b=b \end{aligned}$
Using (C)
$-a-b=b \Rightarrow 2 b=-a \Rightarrow b=\frac{-a}{2}$
From (B)
$\begin{aligned} &a+\frac{a}{2}=\frac{\pi}{4} \\ &\frac{3}{2} a=\frac{\pi}{4} \\ &\Rightarrow a=\frac{\pi}{6} \end{aligned}$
$\begin{aligned} &b=\frac{-a}{2}=-\frac{\pi}{12} \\ \end{aligned}$
Hence,
$\begin{aligned} &a=\pi / 6, b=-\pi / 12 \end{aligned}$
$a = \pi /6,\; b=-\pi /12$
Hint:
Put at LHL = RHL at
$x = \pi /4,\; x=\pi /2$
Given:
$f(x)= \begin{cases}x+a \sqrt{2} \sin x & 0 \leq x<\pi / 4 \\ 2 x \cot x+b & \pi / 4 \leq x<\pi / 2 \\ a \cos 2 x-b \sin x & \pi / 2 \leq x<\pi\end{cases}$
Explanation:
At x = $\pi$/4
$L.H.L = R.H.L =f(\frac{\pi }{4}) \qquad ....(A)$
Now,
$\begin{aligned} &f\left(\frac{\pi}{4}\right)=2 \cdot \frac{\pi}{4} \cdot \cot \left(\frac{\pi}{4}\right)+b\\ &=\frac{\pi}{2} \cdot 1+b=\frac{\pi}{2}+b \qquad ....(1) \end{aligned}$
L.H.L
$\begin{aligned} &\lim _{x \rightarrow \pi / 4^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{4}-h\right)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right)+a \sqrt{2} \sin \left(\frac{\pi}{4}-h\right) \\ &=\frac{\pi}{4}+a \sqrt{2} \cdot \frac{1}{\sqrt{2}}=\frac{\pi}{4}+a \end{aligned}$
Using (A)
$\begin{aligned} &\frac{1}{2}+b=\frac{\pi}{4}+a\\ &a-b=\frac{\pi}{4} \qquad ....(B) \end{aligned}$
$\begin{aligned} &\text { At } x=\frac{\pi}{2} \\ &\text { L.H.L }=\text { R.H.L }=f\left(\frac{\pi}{2}\right) \quad \ldots(C) \end{aligned}$
Now
$f\left(\frac{\pi}{2}\right)=a \cos 2 \cdot \frac{\pi}{2}-b \sin \frac{\pi}{2}=-a-b \qquad ....(2)$
$\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{\pi}{2}-h\right) \cot \left(\frac{\pi}{2}-h\right)+b=b \end{aligned}$
Using (C)
$-a-b=b \Rightarrow 2 b=-a \Rightarrow b=\frac{-a}{2}$
From (B)
$\begin{aligned} &a+\frac{a}{2}=\frac{\pi}{4} \\ &\frac{3}{2} a=\frac{\pi}{4} \\ &\Rightarrow a=\frac{\pi}{6} \end{aligned}$
$\begin{aligned} &b=\frac{-a}{2}=-\frac{\pi}{12} \\ \end{aligned}$
Hence,
$\begin{aligned} &a=\pi / 6, b=-\pi / 12 \end{aligned}$
Continuity exercise 8.2 question 7
Answer:
a = 3, b = -2
Hint:
Put at LHL = RHL at x = 2, x = 4
Given:
$f(x)= \begin{cases}x^{2}+a x+b & 0 \leq x<2 \\ 3 x+2 & 2 \leq x \leq 4 \\ 2 a x+5 b & 4<x \leq 8\end{cases}$
Explanation:
At x = 2
$\begin{aligned} &L . H . L=R . H . L=f(2) \qquad \qquad \qquad...(A)\\ &f(2)=3 \times 2+2=8 \qquad \qquad \qquad...(1)\\ &\text { L.H. } L=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)\\ &=\lim _{h \rightarrow 0}(2-h)^{2}+a(2-h)+b\\ &=4+2 a+b \end{aligned}$
from (B)
$\begin{aligned} &4+2 a+b=8\\ &\Rightarrow 2 a+b=4 \qquad \ldots(B)\\ &\text { Now at } x=4\\ &L . H . L=R . H . L=f(4) \quad \ldots (C)\\ &f(4)=3 \times 4+2=8 \quad \ldots (2) \end{aligned}$
$\begin{aligned} &R . H . L=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0}(4+h) \\ &=\lim _{h \rightarrow 0} 2 a(4+h)+5 b=8 b+5 h \end{aligned}$
From (C)
$8 b+5 h=14 \quad \dots(D)$
Solving (B) & (D) we get
Hence, a = 3, b = -2
a = 3, b = -2
Hint:
Put at LHL = RHL at x = 2, x = 4
Given:
$f(x)= \begin{cases}x^{2}+a x+b & 0 \leq x<2 \\ 3 x+2 & 2 \leq x \leq 4 \\ 2 a x+5 b & 4<x \leq 8\end{cases}$
Explanation:
At x = 2
$\begin{aligned} &L . H . L=R . H . L=f(2) \qquad \qquad \qquad...(A)\\ &f(2)=3 \times 2+2=8 \qquad \qquad \qquad...(1)\\ &\text { L.H. } L=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)\\ &=\lim _{h \rightarrow 0}(2-h)^{2}+a(2-h)+b\\ &=4+2 a+b \end{aligned}$
from (B)
$\begin{aligned} &4+2 a+b=8\\ &\Rightarrow 2 a+b=4 \qquad \ldots(B)\\ &\text { Now at } x=4\\ &L . H . L=R . H . L=f(4) \quad \ldots (C)\\ &f(4)=3 \times 4+2=8 \quad \ldots (2) \end{aligned}$
$\begin{aligned} &R . H . L=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0}(4+h) \\ &=\lim _{h \rightarrow 0} 2 a(4+h)+5 b=8 b+5 h \end{aligned}$
From (C)
$8 b+5 h=14 \quad \dots(D)$
Solving (B) & (D) we get
Hence, a = 3, b = -2
Continuity exercise 8.2 question 8
Answer:
$\frac{1}{2}$
Given:
$f(x)= \begin{cases}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x} & x \neq \pi / 4 \\ k & x=\pi / 4\end{cases}$
Hint:
Apply L-Hospital Rule when you get 0/0 from.
Explanation:
At x = $\pi$/4
$\begin{aligned} &\text { L.H.L } =\text { R.H.L }=f\left(\frac{\pi}{4}\right) \quad \ldots(A) \\ &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)}{\cot \left(\frac{\pi}{4}-h\right)}=\lim _{h \rightarrow 0} \frac{\tanh }{\tan 2 h}$
$=\lim _{h \rightarrow 0} \frac{\frac{\tanh }{h}}{\frac{\tan 2 h}{2 h} \times 2}=\frac{1}{2}$
Hence F (x) will be continious on
$[0, \frac{\pi }{2}]$
If
$f(\frac{\pi }{4})=\frac{1}{2}$
$\frac{1}{2}$
Given:
$f(x)= \begin{cases}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x} & x \neq \pi / 4 \\ k & x=\pi / 4\end{cases}$
Hint:
Apply L-Hospital Rule when you get 0/0 from.
Explanation:
At x = $\pi$/4
$\begin{aligned} &\text { L.H.L } =\text { R.H.L }=f\left(\frac{\pi}{4}\right) \quad \ldots(A) \\ &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right) \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)}{\cot \left(\frac{\pi}{4}-h\right)}=\lim _{h \rightarrow 0} \frac{\tanh }{\tan 2 h}$
$=\lim _{h \rightarrow 0} \frac{\frac{\tanh }{h}}{\frac{\tan 2 h}{2 h} \times 2}=\frac{1}{2}$
Hence F (x) will be continious on
$[0, \frac{\pi }{2}]$
If
$f(\frac{\pi }{4})=\frac{1}{2}$
Continuity exercise 8.2 question 9
Answer:
Everywhere continuous
Given:
$f(x)=\left\{\begin{array}{cc} 2 x-1 & x<2 \\ \frac{3 x}{2} & x \geq 2 \end{array}\right.$
Hint:
Polynomial & identity functions are everywhere continuous.
Explanation:
As polynomial & identity function are everywhere continuous.
So, we only have to check at end point i.e. x = 2
$f(2)=\frac{3 \times 2}{2}=3$
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} 2(2-h)-1=3 \\ &\text { R.H.L }=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} \frac{3(2+h)}{2}=3 \\ &\text { L.H.L }=\text { R.H.L }=f(2)=3 \end{aligned}$
f(x) is continuous at x = 2 and everywhere
Everywhere continuous
Given:
$f(x)=\left\{\begin{array}{cc} 2 x-1 & x<2 \\ \frac{3 x}{2} & x \geq 2 \end{array}\right.$
Hint:
Polynomial & identity functions are everywhere continuous.
Explanation:
As polynomial & identity function are everywhere continuous.
So, we only have to check at end point i.e. x = 2
$f(2)=\frac{3 \times 2}{2}=3$
$\begin{aligned} &\text { L.H.L } =\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0}(2-h)=\lim _{h \rightarrow 0} 2(2-h)-1=3 \\ &\text { R.H.L }=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0}(2+h)=\lim _{h \rightarrow 0} \frac{3(2+h)}{2}=3 \\ &\text { L.H.L }=\text { R.H.L }=f(2)=3 \end{aligned}$
f(x) is continuous at x = 2 and everywhere
Continuity exercise 8.2 question 10
Answer:
everywhere continuous.
Hint:
sine function is everywhere continuous.
Given:
$f(x)=sin(x)$
Explanation:
$f(x)= \begin{cases}\sin x & x \geq 0 \\ \sin (-x) & x<0\end{cases}$
$= \begin{cases}\sin x & x \geq 0 \\ -\sin x & x<0\end{cases}$
As sine function is everywhere continuous &
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}-\sin 0=0 \\ &\text { and } \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \sin 0=0 \\ &f(0)=\sin 0=0 \end{aligned}$
Hence, f(x) is everywhere continuous.
everywhere continuous.
Hint:
sine function is everywhere continuous.
Given:
$f(x)=sin(x)$
Explanation:
$f(x)= \begin{cases}\sin x & x \geq 0 \\ \sin (-x) & x<0\end{cases}$
$= \begin{cases}\sin x & x \geq 0 \\ -\sin x & x<0\end{cases}$
As sine function is everywhere continuous &
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}-\sin 0=0 \\ &\text { and } \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \sin 0=0 \\ &f(0)=\sin 0=0 \end{aligned}$
Hence, f(x) is everywhere continuous.
Continuity exercise 8.2 question 11
Answer:
Everywhere continuous
Hint:
sine function, polynomial function and identity are everywhere continuous.
Given:
$f(x)=\left\{\begin{array}{cc} \frac{\sin x}{x} & x<0 \\ x+1 & x \geq 0 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &f(0)=0+1=1 \\ &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h) \\ &=\lim _{h \rightarrow 0} \frac{\sin (-h)}{-h}=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}=1 \\ &\text { R.H.L } =\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} h+1=1 \\ &\text { L.H.L } =\text { R.H.L }=f(0)=1 \end{aligned}$
So, f(x) is everywhere continuous.
Everywhere continuous
Hint:
sine function, polynomial function and identity are everywhere continuous.
Given:
$f(x)=\left\{\begin{array}{cc} \frac{\sin x}{x} & x<0 \\ x+1 & x \geq 0 \end{array}\right.$
Explanation:
At x = 0
$\begin{aligned} &f(0)=0+1=1 \\ &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h) \\ &=\lim _{h \rightarrow 0} \frac{\sin (-h)}{-h}=\lim _{h \rightarrow 0} \frac{-\sinh }{-h}=1 \\ &\text { R.H.L } =\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} h+1=1 \\ &\text { L.H.L } =\text { R.H.L }=f(0)=1 \end{aligned}$
So, f(x) is everywhere continuous.
Continuity exercise 8.2 question 12
Answer:
Discontinuous at all integral points.
Hint:
Greatest integer function is discontinuous at integral points.
Given:
$g(x)=x-[x]$
Explanation:
$g(x)=x-[x]$
It is defined at all integral points
let n be an integer.
Then,
$\begin{aligned} &g(n)=n-[n]=0\\ &\text { The left hand limit of } \mathrm{f} \text { at } \mathrm{x}=\mathrm{n} \text { is }\\ &\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{n}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]\\ &=n-(n-1)=1 \end{aligned}$
$\begin{aligned} &\text { The right hand limit of } f \text { at } x=n \text { is }\\ &\lim _{x \rightarrow n^{+}} g(x)=\lim _{x \rightarrow n^{+}}(x-[x])=\lim _{x \rightarrow n^{+}}(x)-\lim _{x \rightarrow n^{+}}[x]\\ &=n-n=0 \end{aligned}$
$\begin{aligned} &\text { L.H. } L \neq R . H . L\\ &\therefore g \text { is not continious at } x=n \end{aligned}$
Hence g is discontinious at all integral points.
Discontinuous at all integral points.
Hint:
Greatest integer function is discontinuous at integral points.
Given:
$g(x)=x-[x]$
Explanation:
$g(x)=x-[x]$
It is defined at all integral points
let n be an integer.
Then,
$\begin{aligned} &g(n)=n-[n]=0\\ &\text { The left hand limit of } \mathrm{f} \text { at } \mathrm{x}=\mathrm{n} \text { is }\\ &\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{n}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]\\ &=n-(n-1)=1 \end{aligned}$
$\begin{aligned} &\text { The right hand limit of } f \text { at } x=n \text { is }\\ &\lim _{x \rightarrow n^{+}} g(x)=\lim _{x \rightarrow n^{+}}(x-[x])=\lim _{x \rightarrow n^{+}}(x)-\lim _{x \rightarrow n^{+}}[x]\\ &=n-n=0 \end{aligned}$
$\begin{aligned} &\text { L.H. } L \neq R . H . L\\ &\therefore g \text { is not continious at } x=n \end{aligned}$
Hence g is discontinious at all integral points.
Continuity exercise 8.2 question 13 (i)
Answer:
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin\: x+cox\: x$
Explanation:
As sine and cosine function are everywhere continuous and addition of two continuous functions is again continuous.
Hence, f(x) is everywhere continuous.
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin\: x+cox\: x$
Explanation:
As sine and cosine function are everywhere continuous and addition of two continuous functions is again continuous.
Hence, f(x) is everywhere continuous.
Continuity exercise 8.2 question 13 (ii)
Answer:
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin \: x-cos\: x$
Explanation:
As sine and cosine function are everywhere continuous and subtraction of two continuous functions is continuous.
Hence, f(x) is everywhere continuous.
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin \: x-cos\: x$
Explanation:
As sine and cosine function are everywhere continuous and subtraction of two continuous functions is continuous.
Hence, f(x) is everywhere continuous.
Continuity exercise 8.2 question 13 (iii)
Answer:
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin\: x \times cos\: x$
Explanation:
As sine and cosine function are everywhere continuous and product of two continuous functions is continuous.
Hence, f(x) is everywhere continuous.
everywhere continuous.
Hint:
sine and cosine function are everywhere continuous.
Given:
$f(x)=sin\: x \times cos\: x$
Explanation:
As sine and cosine function are everywhere continuous and product of two continuous functions is continuous.
Hence, f(x) is everywhere continuous.
Continuity exercise 8.2 question 14
Answer:
cos2 x is continuous
Given:
f(x) = cos x2
Explanation:
f(x) = cos x2
Let a be any real number then,
L.H.L
$\begin{aligned} &\lim _{x \rightarrow a-} f(x)=\lim _{h \rightarrow 0} f(a-h) \\ &=\lim _{h \rightarrow 0} \cos (a-h)^{2} \\ &=\cos a^{2} \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a+h) \\ &=\lim _{h \rightarrow 0} \cos (a+h)^{2} \\ &=\cos a^{2} \end{aligned}$
Also, L.H.L=R.H.L=f(a)
f(x) is continuous everywhere
cos2 x is continuous
Given:
f(x) = cos x2
Explanation:
f(x) = cos x2
Let a be any real number then,
L.H.L
$\begin{aligned} &\lim _{x \rightarrow a-} f(x)=\lim _{h \rightarrow 0} f(a-h) \\ &=\lim _{h \rightarrow 0} \cos (a-h)^{2} \\ &=\cos a^{2} \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a+h) \\ &=\lim _{h \rightarrow 0} \cos (a+h)^{2} \\ &=\cos a^{2} \end{aligned}$
Also, L.H.L=R.H.L=f(a)
f(x) is continuous everywhere
Continuity exercise 8.2 question 15
Answer:
|cos x| is continuous.
Hint:
Continuous of two continuous functions is continuous.
Given:
f(x) = |cos x|
Explanation:
f(x) = |cos x|
Let a be any real number
L.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h) \\ &=\lim _{h \rightarrow 0}|\cos (a-h)| \\ &=|\cos a| \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h) \\ &=\lim _{h \rightarrow 0}|\cos (a+h)| \\ &=\cos a \end{aligned}$
L.H.L = R.H.L = f(a)
f(x) is continuous everywhere
|cos x| is continuous.
Hint:
Continuous of two continuous functions is continuous.
Given:
f(x) = |cos x|
Explanation:
f(x) = |cos x|
Let a be any real number
L.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h) \\ &=\lim _{h \rightarrow 0}|\cos (a-h)| \\ &=|\cos a| \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h) \\ &=\lim _{h \rightarrow 0}|\cos (a+h)| \\ &=\cos a \end{aligned}$
L.H.L = R.H.L = f(a)
f(x) is continuous everywhere
Continuity exercise 8.2 question 16
Answer:
No point of discontinuity.
Given:
$f(x)=\left | x \right |-\left | x+1 \right |$
Explanation:
$f(x)=\left | x \right |-\left | x+1 \right |$
$\begin{gathered} =\left\{\begin{array}{c} -x+x+1, x<-1 \\ -x-(x+1),-1 \leq x<0 \\ x-(x+1), x \geq 0 \end{array}\right\} \\ =\left\{\begin{array}{c} 1, x<-1 \\ -2 x-1,-1 \leq x<0 \\ -1, x \geq 0 \end{array}\right\} \end{gathered}$
Continuity at x = -1
L.H.L
$\begin{aligned} &\lim _{x \rightarrow-1^{-}} f(x) \\ &=\lim _{x \rightarrow-1^{-}} 1=1 \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow-1^{+}} f(x) \\ &\lim _{x \rightarrow-1^{+}}(-2 x-1) \\ &=-2 \times-1-1=1 \end{aligned}$
And
$\begin{aligned} &f(-1)=-2 \times-1-1=1 \\ &\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \end{aligned}$
F(x) is continuous at x=-1
Continuous at x=0
L.H.L
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-2 x-1) \\ &=-2 \times 0-1=-1 \end{aligned}$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}-1=-1$
And
$f(0)=-2\times 0-1=-1$
L.H.L=R.H.L=f(0)
F(x) is continuous at x=0, hence continuous everywhere
No point of discontinuity.
Given:
$f(x)=\left | x \right |-\left | x+1 \right |$
Explanation:
$f(x)=\left | x \right |-\left | x+1 \right |$
$\begin{gathered} =\left\{\begin{array}{c} -x+x+1, x<-1 \\ -x-(x+1),-1 \leq x<0 \\ x-(x+1), x \geq 0 \end{array}\right\} \\ =\left\{\begin{array}{c} 1, x<-1 \\ -2 x-1,-1 \leq x<0 \\ -1, x \geq 0 \end{array}\right\} \end{gathered}$
Continuity at x = -1
L.H.L
$\begin{aligned} &\lim _{x \rightarrow-1^{-}} f(x) \\ &=\lim _{x \rightarrow-1^{-}} 1=1 \end{aligned}$
R.H.L
$\begin{aligned} &\lim _{x \rightarrow-1^{+}} f(x) \\ &\lim _{x \rightarrow-1^{+}}(-2 x-1) \\ &=-2 \times-1-1=1 \end{aligned}$
And
$\begin{aligned} &f(-1)=-2 \times-1-1=1 \\ &\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \end{aligned}$
F(x) is continuous at x=-1
Continuous at x=0
L.H.L
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-2 x-1) \\ &=-2 \times 0-1=-1 \end{aligned}$
R.H.L
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}-1=-1$
And
$f(0)=-2\times 0-1=-1$
L.H.L=R.H.L=f(0)
F(x) is continuous at x=0, hence continuous everywhere
Continuity exercise 8.2 question 17
Answer:
f(x) is continuous.
Hint:
Check at x = 0
Given:
$f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.$
Explanation:
$f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.$
f is defined at all points of the real line
Let c be a real number
Case 1:
$\begin{aligned} &\text { If } c \neq 0 \text { then } \mathrm{f}(c)=c^{2} \sin \frac{1}{c}\\ &\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow c}\left(x^{2}\right) \cdot \lim _{x \rightarrow c}\left(\sin \frac{1}{x}\right)\\ \end{aligned}$
$\begin{aligned} &=c^{2} \sin \frac{1}{c}\\ &\therefore \lim _{x \rightarrow c} f(x)=f(c)\\ &\therefore f \text { is continious at all points } \mathrm{x} \neq 0 \end{aligned}$
Case 2:
$\begin{aligned} &\text { If } \mathrm{c}=0 \text { then } \mathrm{f}(0)=0\\ &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)\\ &\text { It is known that }-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0\\ &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \end{aligned}$
$\begin{aligned} &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \\ &\Rightarrow \lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2} \end{aligned}$
$\begin{aligned} &\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0 \\ &\Rightarrow \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=0 \end{aligned}$
Similarly,
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right) \\ &=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
Therefore f is continuous at x=0
From the above observations.
It can be conclude that f is continuous at every point of the real line.
Thus f is a continuous function.
f(x) is continuous.
Hint:
Check at x = 0
Given:
$f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.$
Explanation:
$f(x)=\left\{\begin{array}{cc} x^{2} \sin 1 / x & x \neq 0 \\ 0 & x=0 \end{array}\right.$
f is defined at all points of the real line
Let c be a real number
Case 1:
$\begin{aligned} &\text { If } c \neq 0 \text { then } \mathrm{f}(c)=c^{2} \sin \frac{1}{c}\\ &\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\lim _{x \rightarrow c}\left(x^{2}\right) \cdot \lim _{x \rightarrow c}\left(\sin \frac{1}{x}\right)\\ \end{aligned}$
$\begin{aligned} &=c^{2} \sin \frac{1}{c}\\ &\therefore \lim _{x \rightarrow c} f(x)=f(c)\\ &\therefore f \text { is continious at all points } \mathrm{x} \neq 0 \end{aligned}$
Case 2:
$\begin{aligned} &\text { If } \mathrm{c}=0 \text { then } \mathrm{f}(0)=0\\ &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(x^{2} \sin \frac{1}{x}\right)\\ &\text { It is known that }-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0\\ &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \end{aligned}$
$\begin{aligned} &\Rightarrow-x^{2} \leq x^{2} \sin \frac{1}{x} \leq x^{2} \\ &\Rightarrow \lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2} \end{aligned}$
$\begin{aligned} &\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0 \\ &\Rightarrow \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=0 \end{aligned}$
Similarly,
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(x^{2} \sin \frac{1}{x}\right) \\ &=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}$
Therefore f is continuous at x=0
From the above observations.
It can be conclude that f is continuous at every point of the real line.
Thus f is a continuous function.
Continuity exercise 8.2 question 18
Answer:
$x=-2, -5/2$
Hint:
$f \circ f(x)=f(f(x))$
Given:
$f(x)=1 / x+2$
Explanation:
Clearly,
$f(x)=1 / x+2$
is discontinuous at
X = -2
Also, it is not defined at x = -2
For x ≠ -2
$\begin{aligned} &f(x)=f\left(\frac{1}{x+2}\right) \\ &=\frac{1}{\frac{1}{x+2}+2} \\ &=\frac{\frac{1}{2 x+5}}{x+2} \\ &=\frac{x+2}{2 x+5} \end{aligned}$
We observe that,
f(f(x)) is discontinuous and not defined a
$\begin{aligned} x=-\frac{5}{2} \end{aligned}$
Hence
f(f(x)) is not continuous at x = -2 and
$\begin{aligned} x=-\frac{5}{2} \end{aligned}$
$x=-2, -5/2$
Hint:
$f \circ f(x)=f(f(x))$
Given:
$f(x)=1 / x+2$
Explanation:
Clearly,
$f(x)=1 / x+2$
is discontinuous at
X = -2
Also, it is not defined at x = -2
For x ≠ -2
$\begin{aligned} &f(x)=f\left(\frac{1}{x+2}\right) \\ &=\frac{1}{\frac{1}{x+2}+2} \\ &=\frac{\frac{1}{2 x+5}}{x+2} \\ &=\frac{x+2}{2 x+5} \end{aligned}$
We observe that,
f(f(x)) is discontinuous and not defined a
$\begin{aligned} x=-\frac{5}{2} \end{aligned}$
Hence
f(f(x)) is not continuous at x = -2 and
$\begin{aligned} x=-\frac{5}{2} \end{aligned}$
Continuity exercise 8.2 question 19
Answer:
Discontinuous at x = 1/2 , 1 , 2
Hint:
f(x)/g(x) is continuous at every point hen f(x) & g(x) are continuous except
$g(x) \neq 0$
Given:
$f(x)=\frac{1}{t^{2}+t-2}, t=\frac{1}{x-1}$
Explanation:
$\begin{aligned} &f(t)=\frac{1}{t^{2}+t-2} \text { where } t=\frac{1}{x-1}\\ &\text { Clearly } t=\frac{1}{x-1} \text { is discontinuous at } \mathrm{x}=1 \end{aligned}$
$\begin{aligned} &\text { For } \mathrm{x} \neq 1 \text { we have, }\\ &f(t)=\frac{1}{t^{2}+t-2}=\frac{1}{(t+2)(t-1)} \end{aligned}$
$\begin{aligned} &\text { This is discontinuous at } x=-2 \text { and } \mathrm{t}=1\\ &\text { For } t=-2, t=\frac{1}{x-1} \Rightarrow x=\frac{1}{2}\\ &\text { For } t=1, t=\frac{1}{x-1} \Rightarrow x=2 \end{aligned}$
Hence F is discontinuous at
$\begin{aligned} x=\frac{1}{2}, x = 1 \text { and } x = 2 \end{aligned}$
Discontinuous at x = 1/2 , 1 , 2
Hint:
f(x)/g(x) is continuous at every point hen f(x) & g(x) are continuous except
$g(x) \neq 0$
Given:
$f(x)=\frac{1}{t^{2}+t-2}, t=\frac{1}{x-1}$
Explanation:
$\begin{aligned} &f(t)=\frac{1}{t^{2}+t-2} \text { where } t=\frac{1}{x-1}\\ &\text { Clearly } t=\frac{1}{x-1} \text { is discontinuous at } \mathrm{x}=1 \end{aligned}$
$\begin{aligned} &\text { For } \mathrm{x} \neq 1 \text { we have, }\\ &f(t)=\frac{1}{t^{2}+t-2}=\frac{1}{(t+2)(t-1)} \end{aligned}$
$\begin{aligned} &\text { This is discontinuous at } x=-2 \text { and } \mathrm{t}=1\\ &\text { For } t=-2, t=\frac{1}{x-1} \Rightarrow x=\frac{1}{2}\\ &\text { For } t=1, t=\frac{1}{x-1} \Rightarrow x=2 \end{aligned}$
Hence F is discontinuous at
$\begin{aligned} x=\frac{1}{2}, x = 1 \text { and } x = 2 \end{aligned}$
The RD Sharma class 12 solution of Continuity exercise 8.2 consists of 40 questions that cover up almost the majority of all the topics of the chapter continuity. The concept covered in this chapter are-
Continuous function.
Absolute continuous function.
Absolute Continuity of a measure concerning another measure.
Continuous probability distribution.
The questions in this exercise are divided into two parts: level 1 and level 2. With such a vast number of topics to cover, it gets easier for students to study in two parts, understanding the level of difficulty of the question and practicing accordingly.
The solutions in RD Sharma class 12th exercise 8.2 are created by experts all around the country who are excellent in the field of academics, thus providing helpful tips to students which help them solve questions more efficiently. The format of the RD Sharma class 12 solutions chapter 8 exercise 8.2 solution corresponds with the syllabus of NCERT, which makes it more useful for students to prepare for public exams as well.
A few reasons are listed below why the RD Sharma class 12th exercise 8.2 are helpful in the preparation of exams:-
The questions are designed to cover up almost all the topics that have the possible chances to be asked in the exams.
The RD Sharma class 12 chapter 8 exercise 8.2 consists of questions that are frequently asked in board exams and also provides tricks and tips to solve the questions in an alternate and easy way.
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- 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
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