Limit Continuity And Differentiability Ques 97
- For every pair of continuous functions $f, g:[0,1] \rightarrow R$ such that $\max {f(x): x \in[0,1]}=\max {g(x): x \in[0,1]}$. The correct statement(s) is (are)
(2014 Adv.)
(a) $[f(c)]^{2}+3 f(c)=[g(c)]^{2}+3 g(c)$ for some $c \in[0,1]$
(b) $[f(c)]^{2}+f(c)=[g(c)]^{2}+3 g(c)$ for some $c \in[0,1]$
(c) $[f(c)]^{2}+3 f(c)=[g(c)]^{2}+g(c)$ for some $c \in[0,1]$
(d) $[f(c)]^{2}=[g(c)]^{2}$ for some $c \in[0,1]$
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Answer:
Correct Answer: 97.(a, d)
Solution:
- The function $f(x)=\tan x$ is not defined at $x=\frac{\pi}{2}$, so $f(x)$ is not continuous on $(0, \pi)$.
(b) Since $g(x)=x \sin \frac{1}{x}$ is continuous on $(0, \pi)$ and the integral function of a continuous function is continuous,
$\therefore f(x)=\int _0^{x} t \sin \frac{1}{t} d t$ is continuous on $(0, \pi)$.
(c) Also, $f(x)=2 \sin \frac{2 x}{9}, \frac{3 \pi}{4}<x<\pi$
$$ \begin{aligned} & 1, \quad 0<x < \frac{3 \pi}{4} \\ & \frac{2 x}{9}, \frac{3 \pi}{4}<x<\pi \end{aligned} $$
We have, $\quad \lim _{x \to 3\pi^{-}} f(x)=1$
$$ \lim _{x \rightarrow \frac{3 \pi}{4}^{+}} f(x)=\lim _{x \rightarrow \frac{3 \pi}{4}} 2 \sin \frac{2 x}{9}=1 $$
So, $f(x)$ is continuous at $x=3 \pi / 4$.
$\Rightarrow \quad f(x)$ is continuous at all points.
(d) Finally, $f(x)=\frac{\pi}{2} \sin (x+\pi) \Rightarrow f\left(\frac{\pi}{2}\right)=-\frac{\pi}{2}$
$$ \begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \quad=\lim _{h \rightarrow 0} \frac{\pi}{2} \sin \left(\frac{3 \pi}{2}-h\right)=\frac{\pi}{2} \\ & \text { and } \lim _{x \rightarrow(\pi / 2)^{+}} f(x)=\lim _{h \rightarrow 0^{+}} f\left(\frac{\pi}{2}+h\right) \\ & =\lim _{h \rightarrow 0} \frac{\pi}{2} \sin \left( \frac{3 \pi}{2} + h \right)=\frac{\pi}{2} \end{aligned} $$
So, $f(x)$ is not continuous at $x=\pi / 2$.
BETA
