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If the value of the integral $$\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$$ is $$\frac{2}{\pi}$$.Then, a value of $$\alpha$$ is

JEE · Math · previous-year question

  1. A.$$\frac{\pi}{2}$$correct
  2. B.$$\frac{\pi}{4}$$
  3. C.$$\frac{\pi}{3}$$
  4. D.$$\frac{\pi}{6}$$

Answer

A. $$\frac{\pi}{2}$$

Explanation

$$\begin{aligned} & \text { Given, } \int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int\limits_0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int\limits_0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}$$

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