Register

ISI2014-DCG-37

• recategorized by
580 views
2 votes
2 votes

Let $f: \bigg( – \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg) \to \mathbb{R}$ be a continuous function, $f(x) \to +\infty$ as $x \to \dfrac{\pi^-}{2}$ and $f(x) \to – \infty$ as $x \to -\dfrac{\pi^+}{2}$. Which one of the following functions satisfies the above properties of $f(x)$?

  1. $\cos x$
  2. $\tan x$
  3. $\tan^{-1} x$
  4. $\sin x$
• recategorized by
User Avatar

2 Answers

1 votes
1 votes

Here, $\cos (-\frac{\pi}{2})=\cos (\frac{\pi}{2})=0$ and $\sin (-\frac{\pi}{2})=-1,~ \sin (\frac{\pi}{2})=1$

$\displaystyle \lim_{x \to \frac{\pi^-}{2}} \tan x = \lim_{x \to \frac{\pi^-}{2}} \frac{\sin x}{\cos x}=\frac{+1}{0^+}\to +\infty$

Again

$\displaystyle \lim_{x \to -\frac{\pi^+}{2}} \tan x = \lim_{x \to -\frac{\pi^+}{2}} \frac{\sin x}{\cos x}=\frac{-1}{0^+}\to -\infty$

$$\therefore f(x)=\tan x$$

So the correct answer is B.

• edited by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

1 votes
1 votes
0 answers
1
Arjun asked Sep 23, 2019
478 views
ISI2014-DCG-33
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties.$f(0)=0$,$f(1)=1$, and$f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_...
User Avatar
0 votes
0 votes
0 answers
2
Arjun asked Sep 23, 2019
361 views
ISI2014-DCG-43
Let $f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$ Then $\underset{x \to a}{\lim} f(x)$ exists if $a=0$ for all $a \in R$ for all $a \neq 0$ only if $a=1$
Let $$f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$$ Then $\underset{x \to a}{\lim}...
User Avatar
0 votes
0 votes
2 answers
3
Arjun asked Sep 23, 2019
721 views
ISI2014-DCG-29
If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then $f(x)$ is continuous at $x=0$, but not differentiable at $x=0$ $f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$ $f(x)$ is differentiable at $x=0$, and $f’(0) = 0$ None of the above
If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then$f(x)$ is continuous at $x=0$, but not differentiable at $x=0$$f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$...
User Avatar
3 votes
3 votes
1 answer
4
Arjun asked Sep 23, 2019
829 views
ISI2014-DCG-2
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$equals $1$does not...
User Avatar
Link Copied!