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MHTCET 23 Apr 2025 Shift 1 Paper

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Question : 122 of 150

Marks: +1, -0

\int_{3}^{5} \frac{\sqrt{x} d x}{\sqrt{8-x}+\sqrt{x}}=

0
1
2
3

Solution:

Concept:

Use the standard property:

\int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)}\,dx = \frac{b-a}{2}

.

Explanation:

Here,

a = 3

and

b = 5

, so

a+b = 8

.

Let

f(x) = \sqrt{x}

. Then

f(a+b-x) = f(8-x) = \sqrt{8-x}

.

The given integrand

\frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}}

exactly matches

\frac{f(x)}{f(x)+f(a+b-x)}

.

Applying the property directly:

\int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}}\,dx = \frac{1}{2}(5-3)

.

= \frac{1}{2} \times 2 = 1

.

Answer:

The value of the integral is

1

, which corresponds to option B.

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