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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. $\sin x$
in Calculus
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2 Answers

1 vote

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
0 votes

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

 

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