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
- \(\lim\limits_{x \to \pi-0}f(x)=\frac{\pi}{2}\), \(\lim\limits_{x \to \pi+0}f(x)=\frac{-\pi}{2}\)
- \(\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}\)
- \(\lim\limits_{x \to \pi-0}f(x)=\frac{-\pi}{2}\), \(\lim\limits_{x \to \pi+0}f(x)=\frac{\pi}{2}\)
- \(\lim\limits_{x \to \pi}f(x)=\frac{-\pi}{2}\)
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Equation
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Algebra
Check the Answer
But try the problem first…
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 \(\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}\)
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