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MHTCET 20 Apr 2026 Shift 1 Paper

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Question : 132 of 150

Marks: +1, -0

If

\int_{0}^{1} \frac{d x}{\sqrt{x+1}-\sqrt{x}}=\sqrt{2} k

, then the value of

k

is...

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

Solution:

Concept:

Rationalising the denominator converts the integrand into a sum of standard power functions that are easy to integrate.

Explanation:

Given integral is

\int_{0}^{1} \frac{dx}{\sqrt{x+1}-\sqrt{x}}=\sqrt{2}\,k

.

Rationalise:

\frac{1}{\sqrt{x+1}-\sqrt{x}}\times \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}} = \sqrt{x+1}+\sqrt{x}

.

Thus the integral becomes

\int_{0}^{1}\left((x+1)^{1/2}+x^{1/2}\right)dx

.

Evaluate:

\int_{0}^{1}(x+1)^{1/2}dx=\frac{2}{3}\left[(x+1)^{3/2}\right]_{0}^{1}=\frac{2}{3}(2^{3/2}-1)

.

Also

\int_{0}^{1}x^{1/2}dx=\frac{2}{3}\left[x^{3/2}\right]_{0}^{1}=\frac{2}{3}

.

Adding:

\frac{2}{3}(2^{3/2}-1)+\frac{2}{3}=\frac{2}{3}\cdot 2^{3/2}=\frac{4\sqrt{2}}{3}

.

So

\sqrt{2}\,k=\frac{4\sqrt{2}}{3}

, giving

k=\frac{4}{3}

.

Answer:

k=\frac{4}{3}

, which is Option B.

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