Q116 / 160 · TS EAMCET 2013 · Mathematics

If $\int_0^b \frac{d x}{1+x^2}=\int_b^{\infty} \frac{d x}{1+x^2}$, then $b$ is equal to

  • A
    $\tan ^{-1}\left(\frac{1}{3}\right)$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\sqrt{2}$
  • Correct answer
    D
    $1$

Solution

We have, $$ \begin{aligned} & \int_0^b \frac{d x}{1+x^2}=\int_b^{\infty} \frac{d x}{1+x^2} \\ \Rightarrow \quad & {\left[\tan ^{-1} x\right]_0^b=\left[\tan ^{-1} x\right]_b^{\infty} } \\ \Rightarrow & \tan ^{-1}(b)-\tan ^{-1}(0)=\tan ^{-1}(\infty)-\tan ^{-1}(b) \\ \Rightarrow \quad & 2 \tan ^{-1}(b)-0=\frac{\pi}{2} \\ \Rightarrow \quad & \tan ^{-1}(b)=\frac{\pi}{4} \Rightarrow b=\tan \left(\frac{\pi}{4}\right) \\ \therefore & b=1 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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