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MHTCET 13 Apr 2026 Shift 1 Paper

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Question : 130 of 150

Marks: +1, -0

If

\int_{0}^{a} \frac{d x}{1+4 x^{2}}=\frac{\pi}{8}

a=

$\frac{1}{3}$
$\frac{1}{2}$
$\frac{1}{4}$
1

Solution:

Concept:

Use the standard integral

\int \frac{dx}{1+(2x)^2} = \frac{1}{2}\tan^{-1}(2x)

and apply the limits.

Explanation:

Given:

\int_{0}^{a} \frac{dx}{1+4x^{2}} = \frac{\pi}{8}

Rewrite the integrand as

\frac{1}{1+(2x)^2}

.

So,

\int_{0}^{a} \frac{dx}{1+4x^{2}} = \left[\frac{1}{2}\tan^{-1}(2x)\right]_{0}^{a}

= \frac{1}{2}\tan^{-1}(2a) - \frac{1}{2}\tan^{-1}(0)

Since

\tan^{-1}(0)=0

, we get:

\frac{1}{2}\tan^{-1}(2a) = \frac{\pi}{8}

Multiply by 2:

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

Taking tangent on both sides:

2a = \tan\frac{\pi}{4} = 1

Therefore,

a = \frac{1}{2}

Answer:

a = \frac{1}{2}

, which is option B.

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