If ∫_0^kdisplaystyle1/(2+8x^2) dx=(π)/16 then the value of k is:

The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)

Taking 8 out of the denominator:

    \( \displaystyle \int_0^k\frac{dx}{2+8x^2} = \frac{1}{8}\int_0^k\frac{dx}{x^2+\left(\frac{1}{2}\right)^2} \)

Using the standard integral \(\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}+C\) with \(a=\frac{1}{2}\):

    \( \displaystyle = \frac{1}{8}\times 2\Big[\tan^{-1}2x\Big]_0^k \)

    \( \displaystyle = \frac{1}{4}\tan^{-1}2k \)

Equating to the given value:

    \( \displaystyle \frac{1}{4}\tan^{-1}2k = \frac{\pi}{16} \)

    \( \displaystyle \tan^{-1}2k = \frac{\pi}{4} \)

    \( \displaystyle 2k = 1,\quad k = \frac{1}{2} \)