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← JEE Advanced 2026 · Paper 2
MathematicsNumerical+2Question 16 of 54

JEE Advanced 2026 · Paper 2, Mathematics question 16

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Question Stem for Question Nos. 15 and 16

Consider the curve C1C_1 given by
y=e−xfor x∈[0,10π],y=e^{-x}\qquad\text{for }x\in[0,10\pi],
and the curve C2C_2 given by
y=e−x(sin⁡x+cos⁡x)for x∈[0,10π].y=e^{-x}(\sin x+\cos x)\qquad\text{for }x\in[0,10\pi].
Let nn be the total number of points of intersection of the curves C1C_1 and C2C_2.

Suppose that α1,α2,…,αn∈[0,10π]\alpha_1,\alpha_2,\dots,\alpha_n\in[0,10\pi] are the xx-coordinates of the points of intersection of the curves C1C_1 and C2C_2 such that
α1<α2<⋯<αn.\alpha_1<\alpha_2<\dots<\alpha_n.

Let β\beta be the area of the region enclosed between the curves C1C_1, C2C_2, and the lines x=α1x=\alpha_1 and x=α4x=\alpha_4. Then the value of
−1πlog⁡e(β−2 e−π2)-\frac1\pi\log_e\left(\beta-2\,e^{-\frac\pi2}\right)
is ________.

Answer and worked solution

Official answer: 2.4

In order, α1=0\alpha_1=0, α2=π2\alpha_2=\frac\pi2, α3=2π\alpha_3=2\pi, α4=5π2\alpha_4=\frac{5\pi}2.

Antiderivative. ddx(−e−xcos⁡x)=e−x(sin⁡x+cos⁡x)\frac{d}{dx}\left(-e^{-x}\cos x\right)=e^{-x}(\sin x+\cos x), so
F(x)=∫e−x(sin⁡x+cos⁡x−1) dx=e−x(1−cos⁡x).F(x)=\int e^{-x}(\sin x+\cos x-1)\,dx=e^{-x}(1-\cos x).
F(0)=0F(0)=0, F(π2)=e−π/2F(\frac\pi2)=e^{-\pi/2}, F(2π)=0F(2\pi)=0, F(5π2)=e−5π/2F(\frac{5\pi}2)=e^{-5\pi/2}.

Signs. sin⁡x+cos⁡x−1\sin x+\cos x-1 is positive on (0,π2)(0,\frac\pi2), negative on (π2,2π)(\frac\pi2,2\pi) and positive on (2π,5π2)(2\pi,\frac{5\pi}2). So
β=e−π/2+e−π/2+e−5π/2.\beta=e^{-\pi/2}+e^{-\pi/2}+e^{-5\pi/2}.
Then β−2e−π/2=e−5π/2\beta-2e^{-\pi/2}=e^{-5\pi/2} and
−1πln⁡(e−5π/2)=52=2.5.-\frac1\pi\ln\left(e^{-5\pi/2}\right)=\frac52=2.5.

Topics:Area between curvesIntegration by parts
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