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Integral Calculus: JEE Main Mathematics Question with Solution

Let

\alpha>0

. If

\int\limits_0^a \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}

, then

\alpha

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

4

2 \sqrt{2}

\sqrt{2}

2

Step-by-step solutionView

Correct answer

Rationalizing the integrand or substituting $x = \alpha t$ yields the integral as $\frac{\alpha^{3/2}}{15}(16\sqrt{2} + 4\sqrt{2}) = \frac{20\sqrt{2} - 8 + 24}{15}$ , solving to $\alpha = 2$ .

Option analysis

Why each option works or fails

A ·

4

Solving $\alpha^{3/2} = 2\sqrt{2} = 2^{3/2}$ incorrectly as $\alpha = 2^2 = 4$ due to squaring the base rather than equating exponents. Equate the fractional exponents directly: $\alpha^{3/2} = 2^{3/2} \implies \alpha = 2$ .

B ·

2 \sqrt{2}

Confusing the value of $\alpha^{3/2}$ ( $2\sqrt{2}$ ) with the final value of $\alpha$ itself. Ensure you solve fully for $\alpha$ rather than stopping at $\alpha^{3/2} = 2\sqrt{2}$ .

C ·

\sqrt{2}

Making an arithmetic error in simplifying the integration coefficients, arriving at $\alpha^{3/2} = 2^{3/4}$ or misidentifying $\alpha = \sqrt{2}$ . Carefully track the powers when evaluating $\frac{2}{5}((1+\alpha)^{5/2}-\dots)$ and equating with $\frac{16+20\sqrt{2}}{15}$ .

D ·

2

None. Rationalizing the denominator and evaluating the integral from $0$ to $\alpha$ gives $\frac{\alpha^{3/2}}{15}(16+20\sqrt{2})$ , which directly gives $\alpha = 2$ . This is the correct answer.

Reviewed route

Solution

StepWorking

01given

Given $\alpha > 0$ and the definite integral $I = \int_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \, dx = \frac{16+20\sqrt{2}}{15}$ . Note: the upper limit of integration in the printed question is $\alpha$ .

02approach

Rationalize the denominator by multiplying numerator and denominator by $\sqrt{x+\alpha}+\sqrt{x}$ . Then split the integrand into standard powers of $(x+\alpha)$ and $x$ .

03execute

Rationalizing the denominator gives $\frac{x(\sqrt{x+\alpha}+\sqrt{x})}{(x+\alpha)-x} = \frac{x(\sqrt{x+\alpha}+\sqrt{x})}{\alpha}$ . Thus, $I = \frac{1}{\alpha} \int_0^\alpha x\sqrt{x+\alpha}\,dx + \frac{1}{\alpha}\int_0^\alpha x^{3/2}\,dx$ .

04execute

Evaluate $I_2 = \int_0^\alpha x^{3/2}\,dx = \left[\frac{2}{5}x^{5/2}\right]_0^\alpha = \frac{2}{5}\alpha^{5/2}$ . Evaluate $I_1 = \int_0^\alpha x\sqrt{x+\alpha}\,dx = \int_0^\alpha ((x+\alpha)-\alpha)\sqrt{x+\alpha}\,dx = \int_0^\alpha (x+\alpha)^{3/2}\,dx - \alpha\int_0^\alpha (x+\alpha)^{1/2}\,dx$ . $= \left[\frac{2}{5}(x+\alpha)^{5/2} - \frac{2\alpha}{3}(x+\alpha)^{3/2}\right]_0^\alpha$ $= \left(\frac{2}{5}(2\alpha)^{5/2} - \frac{2\alpha}{3}(2\alpha)^{3/2}\right) - \left(\frac{2}{5}\alpha^{5/2} - \frac{2\alpha}{3}\alpha^{3/2}\right)$ $= \alpha^{5/2}\left(\frac{8\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} - \frac{2}{5} + \frac{2}{3}\right) = \alpha^{5/2}\left(\frac{4\sqrt{2}}{15} + \frac{4}{15}\right) = \frac{4(\sqrt{2}+1)}{15}\alpha^{5/2}$ .

05execute

Sum the components: $I = \frac{1}{\alpha}(I_1 + I_2) = \frac{1}{\alpha}\left(\frac{4\sqrt{2}+4}{15}\alpha^{5/2} + \frac{6}{15}\alpha^{5/2}\right) = \frac{4\sqrt{2}+10}{15}\alpha^{3/2} = \frac{2(2\sqrt{2}+5)}{15}\alpha^{3/2}$ . Set equal to the given value: $\frac{2(2\sqrt{2}+5)}{15}\alpha^{3/2} = \frac{16+20\sqrt{2}}{15} = \frac{4(5\sqrt{2}+4)}{15} = \frac{4\sqrt{2}(5+2\sqrt{2})}{15}$ . Dividing both sides by $\frac{2(2\sqrt{2}+5)}{15}$ gives $\alpha^{3/2} = 2\sqrt{2} = 2^{3/2}$ , hence $\alpha = 2$ .

✓verify

Checking scaling: substituting $\alpha=2$ , $\alpha^{3/2} = 2\sqrt{2}$ , $(4\sqrt{2}+10)(2\sqrt{2})/15 = (16+20\sqrt{2})/15$ , which matches the given RHS exactly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why didn't we use the substitution

x = \alpha \tan^2\theta

Rationalization immediately simplifies the denominator because $(\sqrt{x+\alpha}-\sqrt{x})(\sqrt{x+\alpha}+\sqrt{x}) = \alpha$ , a constant, avoiding trigonometric powers and substitutions.

Answer

Rationalizing the integrand or substituting $x = \alpha t$ yields the integral as $\frac{\alpha^{3/2}}{15}(16\sqrt{2} + 4\sqrt{2}) = \frac{20\sqrt{2} - 8 + 24}{15}$ , solving to $\alpha = 2$ .

Why each option works or fails

A: $4$ - Solving $\alpha^{3/2} = 2\sqrt{2} = 2^{3/2}$ incorrectly as $\alpha = 2^2 = 4$ due to squaring the base rather than equating exponents. Equate the fractional exponents directly: $\alpha^{3/2} = 2^{3/2} \implies \alpha = 2$ .
B: $2 \sqrt{2}$ - Confusing the value of $\alpha^{3/2}$ ( $2\sqrt{2}$ ) with the final value of $\alpha$ itself. Ensure you solve fully for $\alpha$ rather than stopping at $\alpha^{3/2} = 2\sqrt{2}$ .
C: $\sqrt{2}$ - Making an arithmetic error in simplifying the integration coefficients, arriving at $\alpha^{3/2} = 2^{3/4}$ or misidentifying $\alpha = \sqrt{2}$ . Carefully track the powers when evaluating $\frac{2}{5}((1+\alpha)^{5/2}-\dots)$ and equating with $\frac{16+20\sqrt{2}}{15}$ .
D · correct: $2$ - None. Rationalizing the denominator and evaluating the integral from $0$ to $\alpha$ gives $\frac{\alpha^{3/2}}{15}(16+20\sqrt{2})$ , which directly gives $\alpha = 2$ . This is the correct answer.

Step-by-step solution

given: Given $\alpha > 0$ and the definite integral $I = \int_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \, dx = \frac{16+20\sqrt{2}}{15}$ . Note: the upper limit of integration in the printed question is $\alpha$ .
approach: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{x+\alpha}+\sqrt{x}$ . Then split the integrand into standard powers of $(x+\alpha)$ and $x$ .
execute: Rationalizing the denominator gives $\frac{x(\sqrt{x+\alpha}+\sqrt{x})}{(x+\alpha)-x} = \frac{x(\sqrt{x+\alpha}+\sqrt{x})}{\alpha}$ . Thus, $I = \frac{1}{\alpha} \int_0^\alpha x\sqrt{x+\alpha}\,dx + \frac{1}{\alpha}\int_0^\alpha x^{3/2}\,dx$ .
execute: Evaluate $I_2 = \int_0^\alpha x^{3/2}\,dx = \left[\frac{2}{5}x^{5/2}\right]_0^\alpha = \frac{2}{5}\alpha^{5/2}$ . Evaluate $I_1 = \int_0^\alpha x\sqrt{x+\alpha}\,dx = \int_0^\alpha ((x+\alpha)-\alpha)\sqrt{x+\alpha}\,dx = \int_0^\alpha (x+\alpha)^{3/2}\,dx - \alpha\int_0^\alpha (x+\alpha)^{1/2}\,dx$ . $= \left[\frac{2}{5}(x+\alpha)^{5/2} - \frac{2\alpha}{3}(x+\alpha)^{3/2}\right]_0^\alpha$ $= \left(\frac{2}{5}(2\alpha)^{5/2} - \frac{2\alpha}{3}(2\alpha)^{3/2}\right) - \left(\frac{2}{5}\alpha^{5/2} - \frac{2\alpha}{3}\alpha^{3/2}\right)$ $= \alpha^{5/2}\left(\frac{8\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} - \frac{2}{5} + \frac{2}{3}\right) = \alpha^{5/2}\left(\frac{4\sqrt{2}}{15} + \frac{4}{15}\right) = \frac{4(\sqrt{2}+1)}{15}\alpha^{5/2}$ .
execute: Sum the components: $I = \frac{1}{\alpha}(I_1 + I_2) = \frac{1}{\alpha}\left(\frac{4\sqrt{2}+4}{15}\alpha^{5/2} + \frac{6}{15}\alpha^{5/2}\right) = \frac{4\sqrt{2}+10}{15}\alpha^{3/2} = \frac{2(2\sqrt{2}+5)}{15}\alpha^{3/2}$ . Set equal to the given value: $\frac{2(2\sqrt{2}+5)}{15}\alpha^{3/2} = \frac{16+20\sqrt{2}}{15} = \frac{4(5\sqrt{2}+4)}{15} = \frac{4\sqrt{2}(5+2\sqrt{2})}{15}$ . Dividing both sides by $\frac{2(2\sqrt{2}+5)}{15}$ gives $\alpha^{3/2} = 2\sqrt{2} = 2^{3/2}$ , hence $\alpha = 2$ .
verify: Checking scaling: substituting $\alpha=2$ , $\alpha^{3/2} = 2\sqrt{2}$ , $(4\sqrt{2}+10)(2\sqrt{2})/15 = (16+20\sqrt{2})/15$ , which matches the given RHS exactly.

Shortcut:

When to use it: Use dimensional scaling substitution $x = \alpha t$ to pull out the parameter $\alpha$ completely before integrating.

given: $I(\alpha) = \int_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \, dx = \frac{16+20\sqrt{2}}{15}$ .

approach: Substitute $x = \alpha t$ , so $dx = \alpha dt$ , with limits $t \in [0, 1]$ . This factors out all powers of $\alpha$ as a single pre-factor $\alpha^{3/2}$ .

execute: With $(4\sqrt{2}+10)(2\sqrt{2})/15 = (16+20\sqrt{2})/15$ : $\frac{x}{\sqrt{x+\alpha}-\sqrt{x}} dx = \frac{\alpha t}{\sqrt{\alpha}(\sqrt{t+1}-\sqrt{t})} \alpha dt = \alpha^{3/2} t (\sqrt{t+1}+\sqrt{t}) dt$ . Thus $I(\alpha) = \alpha^{3/2} \int_0^1 (t\sqrt{t+1} + t^{3/2}) dt$ .

execute: Compute the numerical definite integral: $\int_0^1 t^{3/2} dt = \frac{2}{5}$ . $\int_0^1 t\sqrt{t+1} dt = \int_0^1 ((t+1)-1)\sqrt{t+1} dt = \left[\frac{2}{5}(t+1)^{5/2} - \frac{2}{3}(t+1)^{3/2}\right]_0^1 = \left(\frac{2}{5} \cdot 4\sqrt{2} - \frac{2}{3} \cdot 2\sqrt{2}\right) - \left(\frac{2}{5} - \frac{2}{3}\right) = \frac{4\sqrt{2}+4}{15}$ . Summing: $\frac{4\sqrt{2}+4}{15} + \frac{6}{15} = \frac{4\sqrt{2}+10}{15}$ . Equating to the given RHS: $\alpha^{3/2} \frac{4\sqrt{2}+10}{15} = \frac{20\sqrt{2}+16}{15} \implies \alpha^{3/2} = \frac{20\sqrt{2}+16}{4\sqrt{2}+10} = 2\sqrt{2} = 2^{3/2} \implies \alpha = 2$ .

verify: $2^{3/2} = 2\sqrt{2}$ , matching option (3).

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Rationalizing the integrand or substituting $x = \alpha t$ yields the integral as $\frac{\alpha^{3/2}}{15}(16\sqrt{2} + 4\sqrt{2}) = \frac{20\sqrt{2} - 8 + 24}{15}$ , solving to $\alpha = 2$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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