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Yukti Vault

Application of Integrals - JEE Mathematics question (JEE Advanced 2026 Paper 2)

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Consider the curve $C_1$ given by $$y=e^{-x}\quad\text{for }x\in[0,10\pi],$$ and the curve $C_2$ given by $$y=e^{-x}(\sin x+\cos x)\quad\text{for }x\in[0,10\pi].$$ Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$. Suppose that $\alpha_1,\alpha_2,\ldots,\alpha_n\in[0,10\pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that $$\alpha_1<\alpha_2<\cdots<\alpha_n.$$
JEE Advanced 2026 Paper 2Question 1
Let $\beta$ be the area of the region enclosed between the curves $C_1$, $C_2$, and the lines $x=\alpha_1$ and $x=\alpha_4$. Then the value of $$-\frac{1}{\pi}\log_e\left(\beta-2e^{-\frac{\pi}{2}}\right)$$ is ____________.
Application of Integrals

Answer$2.4 to 2.6$

Solution

Intersections: $\sin x+\cos x=1\Rightarrow x=0,\frac\pi2,2\pi,\frac{5\pi}2,\dots$. With $H=e^{-x}(1-\cos x)$ an antiderivative of $e^{-x}(\sin x+\cos x-1)$: $\beta=e^{-\pi/2}+e^{-\pi/2}+e^{-5\pi/2}$. So $\beta-2e^{-\pi/2}=e^{-5\pi/2}$ and the value is $\frac52$.

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