Q74 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
If $\displaystyle\int_0^a \dfrac{dx}{1 + 4x^2} = \dfrac{\pi}{8}$, then $a =$
- A$\dfrac{1}{3}$
- Correct answerB$\dfrac{1}{2}$
- C$\dfrac{1}{4}$
- D$1$
Solution
Given $\displaystyle\int_0^a \dfrac{dx}{1 + 4x^2} = \dfrac{\pi}{8}$
Let $2x = t \Rightarrow 2dx = dt$
When $x = 0$, $t = 0$ and when $x = a$, $t = 2a$
The integral becomes $\dfrac{1}{2} \displaystyle\int_0^{2a} \dfrac{dt}{1 + t^2} = \dfrac{\pi}{8}$
$\dfrac{1}{2} [\tan^{-1} t]_0^{2a} = \dfrac{\pi}{8}$
$\dfrac{1}{2} \tan^{-1}(2a) = \dfrac{\pi}{8}$
$\tan^{-1}(2a) = \dfrac{\pi}{4}$
$2a = \tan\left(\dfrac{\pi}{4}\right) = 1$
$a = \dfrac{1}{2}$
Answer: $\dfrac{1}{2}$
Let $2x = t \Rightarrow 2dx = dt$
When $x = 0$, $t = 0$ and when $x = a$, $t = 2a$
The integral becomes $\dfrac{1}{2} \displaystyle\int_0^{2a} \dfrac{dt}{1 + t^2} = \dfrac{\pi}{8}$
$\dfrac{1}{2} [\tan^{-1} t]_0^{2a} = \dfrac{\pi}{8}$
$\dfrac{1}{2} \tan^{-1}(2a) = \dfrac{\pi}{8}$
$\tan^{-1}(2a) = \dfrac{\pi}{4}$
$2a = \tan\left(\dfrac{\pi}{4}\right) = 1$
$a = \dfrac{1}{2}$
Answer: $\dfrac{1}{2}$
Hence, the correct answer is option: 'B'.