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Area Under the Curves question

2025 · 23 Jan · Shift 2 · Q33
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  5. /2025 · 23 Jan · Shift 2 · Q33

Area Under the Curves question

2025 · 23 Jan · Shift 2 · Q33

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
If the area of the region {(x,y):−1≤x≤1,0≤y≤a+e∣x∣−e−x,a>0}\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}\gt 0\right\}{(x,y):−1≤x≤1,0≤y≤a+e∣x∣−e−x,a>0} is e2+8e+1e\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}ee2+8e+1​, then the value of aaa is :
  1. A
    7
  2. B
    5
  3. C
    6
  4. D
    8
View written solutionFree

Correct answer: B

  1. Given region and required area

The area is

A=∫−11(a+e∣x∣−e−x)dxA=\int_{-1}^{1}\left(a+e^{|x|}-e^{-x}\right)dxA=∫−11​(a+e∣x∣−e−x)dx

and it is given that

A=e2+8e+1e=e+8+1e.A=\frac{e^2+8e+1}{e}=e+8+\frac1e.A=ee2+8e+1​=e+8+e1​.

We must find a>0a>0a>0.

  1. Handle the absolute value

Since ∣x∣|x|∣x∣ changes at x=0x=0x=0, split the integral:

  • For −1≤x≤0-1\le x\le 0−1≤x≤0, ∣x∣=−x|x|=-x∣x∣=−x, so e∣x∣=e−xe^{|x|}=e^{-x}e∣x∣=e−x. Hence the integrand becomes a+e−x−e−x=a.a+e^{-x}-e^{-x}=a.a+e−x−e−x=a.

  • For 0≤x≤10\le x\le 10≤x≤1, ∣x∣=x|x|=x∣x∣=x, so e∣x∣=exe^{|x|}=e^xe∣x∣=ex. Hence the integrand becomes a+ex−e−x.a+e^x-e^{-x}.a+ex−e−x.

Therefore,

A=∫−10a dx+∫01(a+ex−e−x) dx.A=\int_{-1}^{0} a\,dx+\int_{0}^{1}(a+e^x-e^{-x})\,dx.A=∫−10​adx+∫01​(a+ex−e−x)dx.
  1. Evaluate each integral

First part:

∫−10a dx=a(0−(−1))=a.\int_{-1}^{0} a\,dx=a(0-(-1))=a.∫−10​adx=a(0−(−1))=a.

Second part:

∫01(a+ex−e−x)dx=∫01a dx+∫01ex dx−∫01e−x dx.\int_{0}^{1}(a+e^x-e^{-x})dx =\int_0^1 a\,dx+\int_0^1 e^x\,dx-\int_0^1 e^{-x}\,dx.∫01​(a+ex−e−x)dx=∫01​adx+∫01​exdx−∫01​e−xdx.

Now,

∫01a dx=a,\int_0^1 a\,dx=a,∫01​adx=a, ∫01ex dx=e−1,\int_0^1 e^x\,dx=e-1,∫01​exdx=e−1, ∫01e−x dx=[−e−x]01=1−1e.\int_0^1 e^{-x}\,dx=\left[-e^{-x}\right]_0^1=1-\frac1e.∫01​e−xdx=[−e−x]01​=1−e1​.

So the second part is

a+(e−1)−(1−1e)=a+e−2+1e.a+(e-1)-\left(1-\frac1e\right)=a+e-2+\frac1e.a+(e−1)−(1−e1​)=a+e−2+e1​.

Hence total area:

A=a+(a+e−2+1e)=2a+e−2+1e.A=a+\left(a+e-2+\frac1e\right)=2a+e-2+\frac1e.A=a+(a+e−2+e1​)=2a+e−2+e1​.
  1. Use the given area

Given

2a+e−2+1e=e+8+1e.2a+e-2+\frac1e=e+8+\frac1e.2a+e−2+e1​=e+8+e1​.

Cancel eee and 1e\frac1ee1​ from both sides:

2a−2=8.2a-2=8.2a−2=8.

So,

2a=10  ⟹  a=5.2a=10 \implies a=5.2a=10⟹a=5.
  1. Check with options

The correct option is:

5\boxed{5}5​

which is Option B.

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