ISI2014-DCG-37

Question

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 $f(x)$?

• A. $\cos x$
• B. $\tan x$
• C. $\tan^{-1} x$
• D. $\sin x$

Answer 1

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.

Answer 2

Just observe the graph of tanx and you’ll get the idea