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JEE Mathematics 2025 Question with Solution

Let gg be a differentiable function such that ∫0xg(t) dt=x−∫0xtg(t) dt\int_0^x g(t) \, dt = x - \int_0^x t g(t) \, dt, x≥0x \ge 0 and let y=y(x)y = y(x) satisfy the differential equation dydx−ytan⁡x=2(x+1)sec⁡xg(x)\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x), x∈[0,π2)x \in \left[ 0, \frac{\pi}{2} \right). If y(0)=0y(0) = 0, then y(π3)y\left( \frac{\pi}{3} \right) is equal to](streamdown:incomplete-link)

  • A

    2π33\frac{2\pi}{3\sqrt{3}}

  • B

    4π3\frac{4\pi}{3}

  • C

    2π3\frac{2\pi}{3}

  • D

    4π33\frac{4\pi}{3\sqrt{3}}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

∫0xg(t) dt=x−∫0xtg(t) dt\int_0^x g(t) \, dt = x - \int_0^x t g(t) \, dt

and

dydx−ytan⁡x=2(x+1)sec⁡xg(x),y(0)=0\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x), \qquad y(0)=0

Find: y(π3)y\left( \frac{\pi}{3} \right)

Differentiate the integral relation with respect to xx:

g(x)=1−xg(x)g(x) = 1 - xg(x)

So,

g(x)(1+x)=1g(x)(1+x)=1

and hence

g(x)=11+xg(x)=\frac{1}{1+x}

Substitute this in the differential equation:

dydx−ytan⁡x=2(x+1)sec⁡x⋅11+x=2sec⁡x\frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \frac{1}{1+x} = 2\sec x

Thus the equation becomes

dydx−ytan⁡x=2sec⁡x\frac{dy}{dx} - y \tan x = 2\sec x

This is a linear differential equation. Its integrating factor is

IF=e∫−tan⁡x dx=eln⁡∣cos⁡x∣=cos⁡x\text{IF} = e^{\int -\tan x \, dx} = e^{\ln|\cos x|} = \cos x

Multiplying throughout by cos⁡x\cos x,

cos⁡x⋅dydx−ycos⁡xtan⁡x=2\cos x \cdot \frac{dy}{dx} - y \cos x \tan x = 2

Therefore,

ddx(ycos⁡x)=2\frac{d}{dx}(y\cos x)=2

Integrate both sides:

ycos⁡x=∫2 dx=2x+Cy\cos x = \int 2 \, dx = 2x + C

Using the initial condition y(0)=0y(0)=0,

0⋅cos⁡0=2⋅0+C0 \cdot \cos 0 = 2\cdot 0 + C

so C=0C=0.

Hence,

ycos⁡x=2xy\cos x = 2x

which gives

y=2xcos⁡x=2xsec⁡xy = \frac{2x}{\cos x} = 2x\sec x

Now evaluate at x=π3x=\frac{\pi}{3}:

y(π3)=2⋅π3⋅sec⁡(π3)y\left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3} \cdot \sec\left( \frac{\pi}{3} \right)

Since sec⁡(π3)=2\sec\left( \frac{\pi}{3} \right)=2,

y(π3)=4π3y\left( \frac{\pi}{3} \right)=\frac{4\pi}{3}

Therefore, the correct option is B.

The solution also states explicitly: The Correct Option is B.

Stepwise Derivation

Start from

∫0xg(t) dt=x−∫0xtg(t) dt\int_0^x g(t) \, dt = x - \int_0^x t g(t) \, dt

By differentiating both sides using the Fundamental Theorem of Calculus,

g(x)=1−xg(x)g(x)=1-xg(x)

so

g(x)=1x+1g(x)=\frac{1}{x+1}

Now the given equation becomes

dydx−ytan⁡x=2(x+1)sec⁡x⋅1x+1=2sec⁡x\frac{dy}{dx} - y\tan x = 2(x+1)\sec x \cdot \frac{1}{x+1}=2\sec x

For the linear equation

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

we have

P(x)=−tan⁡x,Q(x)=2sec⁡xP(x)=-\tan x, \qquad Q(x)=2\sec x

So the integrating factor is

μ(x)=e∫P(x) dx=e∫−tan⁡x dx=cos⁡x\mu(x)=e^{\int P(x) \, dx}=e^{\int -\tan x \, dx}=\cos x

Multiplying the equation by cos⁡x\cos x gives

cos⁡xdydx−ysin⁡x=2\cos x \frac{dy}{dx} - y\sin x = 2

which is exactly

ddx(ycos⁡x)=2\frac{d}{dx}(y\cos x)=2

Integrating,

ycos⁡x=2x+Cy\cos x = 2x + C

At x=0x=0, we use y(0)=0y(0)=0:

0⋅1=0+C0 \cdot 1 = 0 + C

Therefore,

C=0C=0

and hence

y=2xsec⁡xy=2x\sec x

Now,

y(π3)=2⋅π3⋅2=4π3y\left( \frac{\pi}{3} \right)=2\cdot \frac{\pi}{3} \cdot 2 = \frac{4\pi}{3}

Common mistakes

  • Differentiating ∫0xtg(t) dt\int_0^x t g(t) \, dt incorrectly as xg(x)xg(x) without the factor xx coming from the integrand. By the Fundamental Theorem of Calculus, it becomes xg(x)xg(x) because the integrand is tg(t)t g(t) evaluated at t=xt=x. Always substitute the upper limit into the entire integrand.

  • Using the wrong integrating factor for dydx−ytan⁡x=2sec⁡x\frac{dy}{dx} - y\tan x = 2\sec x. Here P(x)=−tan⁡xP(x)=-\tan x, so the integrating factor is e∫−tan⁡x dx=cos⁡xe^{\int -\tan x \, dx}=\cos x, not sec⁡x\sec x. Always identify the coefficient of yy carefully before forming the integrating factor.

  • After obtaining ddx(ycos⁡x)=2\frac{d}{dx}(y\cos x)=2, forgetting to apply the initial condition y(0)=0y(0)=0 and leaving the constant undetermined. The constant must be found before evaluating the required value.

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