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MCQMediumJEE 2025Linear Differential Equations

JEE Mathematics 2025 Question with Solution

Let for some function y=f(x)y = f(x), ∫0xtf(t) dt=x2f(x),x>0\int_0^x t f(t) \, dt = x^2 f(x), x > 0 and f(2)=3f(2) = 3. Then f(6)f(6) is equal to:

  • A

    33

  • B

    11

  • C

    66

  • D

    22

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

∫0xtf(t) dt=x2f(x),x>0\int_0^x t f(t) \, dt = x^2 f(x), \quad x>0

and f(2)=3f(2)=3.

Find: f(6)f(6).

Differentiate both sides with respect to xx using the fundamental theorem of calculus:

ddx(∫0xtf(t) dt)=xf(x)\frac{d}{dx}\left(\int_0^x t f(t) \, dt\right)=x f(x)

Also,

ddx(x2f(x))=2xf(x)+x2f′(x)\frac{d}{dx}\left(x^2 f(x)\right)=2x f(x)+x^2 f'(x)

Equating derivatives,

xf(x)=2xf(x)+x2f′(x)x f(x)=2x f(x)+x^2 f'(x)

So,

0=xf(x)+x2f′(x)0=x f(x)+x^2 f'(x)

which gives

f(x)+xf′(x)=0f(x)+x f'(x)=0

Now separate variables:

f′(x)f(x)=−1x\frac{f'(x)}{f(x)}=-\frac{1}{x}

Integrating,

ln⁡∣f(x)∣=−ln⁡∣x∣+C\ln|f(x)|=-\ln|x|+C

Hence,

f(x)=kxf(x)=\frac{k}{x}

for some constant kk.

Use the condition f(2)=3f(2)=3:

k2=3\frac{k}{2}=3

Therefore,

k=6k=6

and hence

f(x)=6xf(x)=\frac{6}{x}

Now evaluate at x=6x=6:

f(6)=66=1f(6)=\frac{6}{6}=1

Therefore, the correct option is B.

Using the correct differentiation rule

Given:

∫0xtf(t) dt=x2f(x)\int_0^x t f(t) \, dt = x^2 f(x)

and f(2)=3f(2)=3.

Find: f(6)f(6).

The key step is that when differentiating an integral of the form

∫0xg(t) dt\int_0^x g(t) \, dt

with respect to xx, the derivative is

g(x)g(x)

not an integral involving f′(t)f'(t). Here,

g(t)=tf(t)g(t)=t f(t)

so

ddx(∫0xtf(t) dt)=xf(x)\frac{d}{dx}\left(\int_0^x t f(t) \, dt\right)=x f(x)

Then differentiate the right-hand side:

ddx(x2f(x))=2xf(x)+x2f′(x)\frac{d}{dx}(x^2 f(x))=2x f(x)+x^2 f'(x)

Thus,

xf(x)=2xf(x)+x2f′(x)x f(x)=2x f(x)+x^2 f'(x)

which reduces to

f(x)+xf′(x)=0f(x)+x f'(x)=0

Solve the differential equation:

f′(x)f(x)=−1x\frac{f'(x)}{f(x)}=-\frac{1}{x}

Integrate both sides:

ln⁡∣f(x)∣=−ln⁡x+C\ln|f(x)|=-\ln x + C

since x>0x>0. Therefore,

f(x)=kxf(x)=\frac{k}{x}

Apply the initial value:

f(2)=k2=3⇒k=6f(2)=\frac{k}{2}=3 \Rightarrow k=6

So,

f(x)=6xf(x)=\frac{6}{x}

and

f(6)=1f(6)=1

Therefore, the answer is 11.

Common mistakes

  • Differentiating ∫0xtf(t) dt\int_0^x t f(t) \, dt incorrectly using a Leibniz-rule expression with f′(t)f'(t). Here the integrand depends only on tt, so by the fundamental theorem of calculus the derivative is directly xf(x)x f(x). Use ddx∫0xg(t) dt=g(x)\frac{d}{dx}\int_0^x g(t) \, dt=g(x).

  • After obtaining xf(x)=2xf(x)+x2f′(x)x f(x)=2x f(x)+x^2 f'(x), forgetting to divide by xx for x>0x>0. Since the question states x>0x>0, reducing to f(x)+xf′(x)=0f(x)+x f'(x)=0 is valid and necessary.

  • Solving f′(x)f(x)=−1x\frac{f'(x)}{f(x)}=-\frac{1}{x} but missing the constant of integration or writing the final form incorrectly. After integration, rewrite carefully as f(x)=kxf(x)=\frac{k}{x} before using f(2)=3f(2)=3.

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