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# Problem on Function | TOMATO BStat Objective 720

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Function.

## Problem on Function (B.Stat Objective Question )

Consider the function f(x)=$tan^{-1}(2tan(\frac{x}{2}))$, where $\frac{-\pi}{2} \leq f(x) \leq \frac{\pi}{2}$ Then

• $\lim\limits_{x \to \pi-0}f(x)=\frac{\pi}{2}$, $\lim\limits_{x \to \pi+0}f(x)=\frac{-\pi}{2}$
• $\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$
• $\lim\limits_{x \to \pi-0}f(x)=\frac{-\pi}{2}$, $\lim\limits_{x \to \pi+0}f(x)=\frac{\pi}{2}$
• $\lim\limits_{x \to \pi}f(x)=\frac{-\pi}{2}$

### Key Concepts

Equation

Roots

Algebra

Answer:$\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$

B.Stat Objective Problem 720

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

f(x)=$tan^{-1}(2tan{\frac{x}{2}})$

Second Hint

$\lim\limits_{x \to \pi}f(x)$

$=\lim\limits_{x \to \pi}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

$\lim\limits_{x \to \pi-0}f(x)$

$=\lim\limits_{x \to \pi-0}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

Final Step

$\lim\limits_{x \to \pi+0}f(x)$

$=\lim\limits_{x \to \pi+0}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

So $\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$