If the area of the region \ (x, y):-1 ≤ x ≤ 1,0 ≤ y ≤ a+e^ |x| -e^ -x , a 0\ is e^2+8 e+1 e , then the value of a is :

Correct answer: C. 5 aligned & y ∈[0, a+e^ |x| -e^ -x ] & (i) If x ≥ 0 y ∈(0, a+e^x-(1)/(e^x)) & if x < 0 y ∈(0, a+e^ -x -e^ -x ) & y ∈(0, a) aligned aligned & Area =(a)+∫_0^1(a+e^x-e^ -x ) d x=(e^2+8 e+1)/(e) & =a+.(a x+e^x+e^ -x )|_0 ^1=e+8+(1)/(e) & =a+(a+e+(1)/(e)-2)=e+(1)/(e)+8 & 2 a-2=8 a=5 aligned

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Area Under Curves

If the area of the region \ (x, y):-1 ≤ x ≤ 1,0 ≤ y ≤ a+e^ |x| -e^...

If the area of the region

\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq a+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}\gt0\right\}

\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}

, then the value of

a

is :

Options

Correct answer

C. 5

Step-by-step solution

\begin{aligned} & y \in\left[0, a+e^{|x|}-e^{-x}\right] \\ & \text{ (i) If } x \geq 0 \Rightarrow y \in\left(0, a+e^x-\frac{1}{e^x}\right) \\ & \text{ if } x < 0 \Rightarrow y \in\left(0, a+e^{-x}-e^{-x}\right) \\ & \Rightarrow y \in(0, a)\end{aligned}

\begin{aligned} & \text{ Area }=(a)+\int_0^1\left(a+e^x-e^{-x}\right) d x=\frac{e^2+8 e+1}{e} \\ & =a+\left.\left(a x+e^x+e^{-x}\right)\right|_0 ^1=e+8+\frac{1}{e} \\ & =a+\left(a+e+\frac{1}{e}-2\right)=e+\frac{1}{e}+8 \\ & \Rightarrow 2 a-2=8 \Rightarrow a=5\end{aligned}

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