Problem on Function | TOMATO BStat Objective 720

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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

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

Equation

Roots

Algebra

But try the problem first…

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

Source

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