The definite integral has identical upper and lower limits, so its value is identically 0.

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why do we multiply by before differentiating instead of using the quotient rule directly?

Multiplying by simplifies the Leibniz rule application by avoiding quotient rule fractions, making the substitution at much cleaner.

\phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\mathrm{dt}, x > 0

, then

\phi'\left(\frac{\pi}{4}\right)

is equal to :

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What feels right?

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

\frac{8}{6+\sqrt{\pi}}

\frac{4}{6+\sqrt{\pi}}

\frac{8}{\sqrt{\pi}}

\frac{4}{6-\sqrt{\pi}}

Step-by-step solutionView

Correct answer

Rewriting the relation as $\sqrt{x}\,\phi(x) = \int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ and differentiating with respect to $x$ gives $\phi'(\pi/4) = \frac{8}{6+\sqrt{\pi}}$ .

Option analysis

Why each option works or fails

A ·

\frac{8}{6+\sqrt{\pi}}

None. This is the correct value obtained by using Leibniz's rule and evaluating at $x = \pi/4$ . Correctly applying the product rule to $\sqrt{x}\phi(x)$ gives $\frac{1}{2\sqrt{x}}\phi(x) + \sqrt{x}\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$ . Since $\phi(\pi/4) = 0$ , evaluating at $x = \pi/4$ yields $\phi'(\pi/4) = \frac{8}{6+\sqrt{\pi}}$ .

B ·

\frac{4}{6+\sqrt{\pi}}

The student forgets that $4\sqrt{2}\sin(\pi/4) = 4\sqrt{2}\cdot\frac{1}{\sqrt{2}} = 4$ , but then multiplies or divides incorrectly by an extra factor of $2$ , leading to a numerator of $4$ instead of $8$ . Compute $\left(3 + \sqrt{\frac{\pi}{4}}\right)\phi'\left(\frac{\pi}{4}\right) = 4$ , which means $\phi'\left(\frac{\pi}{4}\right) = \frac{4}{3 + \frac{\sqrt{\pi}}{2}} = \frac{8}{6+\sqrt{\pi}}$ .

C ·

\frac{8}{\sqrt{\pi}}

The student drops the $3\phi'(x)$ term when taking the derivative of the integral with respect to $x$ . The integrand contains $-3\phi'(t)$ , which under Leibniz's rule produces $-3\phi'(x)$ ; do not drop terms inside the integrand.

D ·

\frac{4}{6-\sqrt{\pi}}

The student makes a sign error when transposing $-3\phi'(x)$ to the left-hand side, obtaining $3 - \frac{\sqrt{\pi}}{2}$ instead of $3 + \frac{\sqrt{\pi}}{2}$ . Adding $3\phi'(x)$ to both sides results in $(3 + \sqrt{x})\phi'(x)$ , so the denominator must involve $6 + \sqrt{\pi}$ , not $6 - \sqrt{\pi}$ .

Reviewed route

Solution

StepWorking

01given

Given the integral equation: $\phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt, \quad x > 0$

02goal

Find the value of $\phi'\left(\frac{\pi}{4}\right)$ .

03approach

Clear the denominator by writing $\sqrt{x}\,\phi(x) = \int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ . Notice that $\phi(\pi/4) = 0$ because the upper and lower integration limits match. Then, differentiate both sides with respect to $x$ using the product rule and Leibniz's rule, and finally substitute $x = \pi/4$ .

04execute

Rewrite the expression: $\sqrt{x}\,\phi(x) = \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ Notice that evaluating at $x = \frac{\pi}{4}$ gives: $\sqrt{\frac{\pi}{4}}\,\phi\left(\frac{\pi}{4}\right) = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} (\dots)\,dt = 0 \implies \phi\left(\frac{\pi}{4}\right) = 0$ Differentiating both sides with respect to $x$ using the product rule on the LHS and Leibniz's rule on the RHS: $\frac{1}{2\sqrt{x}}\,\phi(x) + \sqrt{x}\,\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$ Group the $\phi'(x)$ terms: $(\sqrt{x} + 3)\,\phi'(x) + \frac{1}{2\sqrt{x}}\,\phi(x) = 4\sqrt{2}\sin x$

05execute

Substitute $x = \frac{\pi}{4}$ into the differentiated equation: $\left(\sqrt{\frac{\pi}{4}} + 3\right)\phi'\left(\frac{\pi}{4}\right) + \frac{1}{2\sqrt{\pi/4}}\,\phi\left(\frac{\pi}{4}\right) = 4\sqrt{2}\sin\left(\frac{\pi}{4}\right)$ Since $\phi\left(\frac{\pi}{4}\right) = 0$ and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ : $\left(\frac{\sqrt{\pi}}{2} + 3\right)\phi'\left(\frac{\pi}{4}\right) + 0 = 4\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 4$ $\left(\frac{\sqrt{\pi} + 6}{2}\right)\phi'\left(\frac{\pi}{4}\right) = 4$ $\phi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$

✓verify

Verify the boundary value: evaluating the original equation directly via L'Hopital's rule as $x \to \pi/4$ : $\lim_{x \to \pi/4} \frac{\int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt}{\sqrt{x}} = \frac{0}{\sqrt{\pi}/2} = 0 = \phi(\pi/4)$ . At $x = \pi/4$ , $\text{LHS} = (\sqrt{\pi}/2 + 3) \cdot \frac{8}{6+\sqrt{\pi}} = \frac{\sqrt{\pi}+6}{2} \cdot \frac{8}{\sqrt{\pi}+6} = 4$ , matching $\text{RHS} = 4$ .

Hints that build this answer step by step

To eliminate the fraction before differentiating, how can the given equation be rewritten?
$\sqrt{x}\,\phi(x) = \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,\mathrm{d}t$
What equation results from differentiating both sides of $\sqrt{x}\,\phi(x) = \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,\mathrm{d}t$ with respect to $x$ ?
$\frac{1}{2\sqrt{x}}\,\phi(x) + \sqrt{x}\,\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$
Evaluating at $x = \frac{\pi}{4}$ , what is $\phi\left(\frac{\pi}{4}\right)$ and the resulting equation for $\phi'\left(\frac{\pi}{4}\right)$ ?
$\phi\left(\frac{\pi}{4}\right) = 0$ , giving $\frac{\sqrt{\pi}}{2}\phi'\left(\frac{\pi}{4}\right) + 3\phi'\left(\frac{\pi}{4}\right) = 4$

✓ 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 do we multiply by

\sqrt{x}

before differentiating instead of using the quotient rule directly?

Multiplying by $\sqrt{x}$ simplifies the Leibniz rule application by avoiding quotient rule fractions, making the substitution at $x = \pi/4$ much cleaner.

Why is

\phi(\pi/4) = 0

The definite integral $\int_{\pi/4}^{\pi/4} f(t)\,dt$ has identical upper and lower limits, so its value is identically 0.

Answer

Rewriting the relation as $\sqrt{x}\,\phi(x) = \int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ and differentiating with respect to $x$ gives $\phi'(\pi/4) = \frac{8}{6+\sqrt{\pi}}$ .

Why each option works or fails

A · correct: $\frac{8}{6+\sqrt{\pi}}$ - None. This is the correct value obtained by using Leibniz's rule and evaluating at $x = \pi/4$ . Correctly applying the product rule to $\sqrt{x}\phi(x)$ gives $\frac{1}{2\sqrt{x}}\phi(x) + \sqrt{x}\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$ . Since $\phi(\pi/4) = 0$ , evaluating at $x = \pi/4$ yields $\phi'(\pi/4) = \frac{8}{6+\sqrt{\pi}}$ .
B: $\frac{4}{6+\sqrt{\pi}}$ - The student forgets that $4\sqrt{2}\sin(\pi/4) = 4\sqrt{2}\cdot\frac{1}{\sqrt{2}} = 4$ , but then multiplies or divides incorrectly by an extra factor of $2$ , leading to a numerator of $4$ instead of $8$ . Compute $\left(3 + \sqrt{\frac{\pi}{4}}\right)\phi'\left(\frac{\pi}{4}\right) = 4$ , which means $\phi'\left(\frac{\pi}{4}\right) = \frac{4}{3 + \frac{\sqrt{\pi}}{2}} = \frac{8}{6+\sqrt{\pi}}$ .
C: $\frac{8}{\sqrt{\pi}}$ - The student drops the $3\phi'(x)$ term when taking the derivative of the integral with respect to $x$ . The integrand contains $-3\phi'(t)$ , which under Leibniz's rule produces $-3\phi'(x)$ ; do not drop terms inside the integrand.
D: $\frac{4}{6-\sqrt{\pi}}$ - The student makes a sign error when transposing $-3\phi'(x)$ to the left-hand side, obtaining $3 - \frac{\sqrt{\pi}}{2}$ instead of $3 + \frac{\sqrt{\pi}}{2}$ . Adding $3\phi'(x)$ to both sides results in $(3 + \sqrt{x})\phi'(x)$ , so the denominator must involve $6 + \sqrt{\pi}$ , not $6 - \sqrt{\pi}$ .

Step-by-step solution

given: Given the integral equation: $\phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt, \quad x > 0$
goal: Find the value of $\phi'\left(\frac{\pi}{4}\right)$ .
approach: Clear the denominator by writing $\sqrt{x}\,\phi(x) = \int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ . Notice that $\phi(\pi/4) = 0$ because the upper and lower integration limits match. Then, differentiate both sides with respect to $x$ using the product rule and Leibniz's rule, and finally substitute $x = \pi/4$ .
execute: Rewrite the expression: $\sqrt{x}\,\phi(x) = \int_{\frac{\pi}{4}}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ Notice that evaluating at $x = \frac{\pi}{4}$ gives: $\sqrt{\frac{\pi}{4}}\,\phi\left(\frac{\pi}{4}\right) = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} (\dots)\,dt = 0 \implies \phi\left(\frac{\pi}{4}\right) = 0$ Differentiating both sides with respect to $x$ using the product rule on the LHS and Leibniz's rule on the RHS: $\frac{1}{2\sqrt{x}}\,\phi(x) + \sqrt{x}\,\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$ Group the $\phi'(x)$ terms: $(\sqrt{x} + 3)\,\phi'(x) + \frac{1}{2\sqrt{x}}\,\phi(x) = 4\sqrt{2}\sin x$
execute: Substitute $x = \frac{\pi}{4}$ into the differentiated equation: $\left(\sqrt{\frac{\pi}{4}} + 3\right)\phi'\left(\frac{\pi}{4}\right) + \frac{1}{2\sqrt{\pi/4}}\,\phi\left(\frac{\pi}{4}\right) = 4\sqrt{2}\sin\left(\frac{\pi}{4}\right)$ Since $\phi\left(\frac{\pi}{4}\right) = 0$ and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ : $\left(\frac{\sqrt{\pi}}{2} + 3\right)\phi'\left(\frac{\pi}{4}\right) + 0 = 4\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 4$ $\left(\frac{\sqrt{\pi} + 6}{2}\right)\phi'\left(\frac{\pi}{4}\right) = 4$ $\phi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$
verify: Verify the boundary value: evaluating the original equation directly via L'Hopital's rule as $x \to \pi/4$ : $\lim_{x \to \pi/4} \frac{\int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt}{\sqrt{x}} = \frac{0}{\sqrt{\pi}/2} = 0 = \phi(\pi/4)$ . At $x = \pi/4$ , $\text{LHS} = (\sqrt{\pi}/2 + 3) \cdot \frac{8}{6+\sqrt{\pi}} = \frac{\sqrt{\pi}+6}{2} \cdot \frac{8}{\sqrt{\pi}+6} = 4$ , matching $\text{RHS} = 4$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Rewriting the relation as $\sqrt{x}\,\phi(x) = \int_{\pi/4}^x (4\sqrt{2}\sin t - 3\phi'(t))\,dt$ and differentiating with respect to $x$ gives $\phi'(\pi/4) = \frac{8}{6+\sqrt{\pi}}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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