JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding an Unknown Parameter from the Area Under a Piecewise Curve

Q: Why does the term e^(-x) - e^(-x) become 0 on [-1, 0]?

For x ∈ [-1, 0], |x| = -x, so e^(|x|) = e^(-x). Thus e^(|x|) - e^(-x) = e^(-x) - e^(-x) = 0.

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

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A8B7C5D6

Step-by-step solutionView

Correct answer

Integrating the function $y = a + e^{∣ x ∣} - e^{- x}$ from $- 1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .

Option analysis

Why each option works or fails

A · 8

Believing that $2 a = 8 + e - e$ directly equals $8$ without dividing by $2$ , or equating $a$ to the numerator's coefficient of $e$ . After integrating, fully simplify the linear equation in terms of $a$ : isolate $2 a = 10$ , which gives $a = 5$ rather than $8$ .

B · 7

Dropping the integration constant contributions from $e^{- x}$ or making a sign error in $\int_{- 1}^{1} e^{- x} d x = e - \frac{1}{e}$ . Evaluate $\int_{- 1}^{1} - e^{- x} d x = [e^{- x}]_{- 1}^{1} = e^{- 1} - e^{1} = \frac{1}{e} - e$ carefully, keeping track of negative signs.

C · 5

The integral of $a + e^{∣ x ∣} - e^{- x}$ over $[- 1, 1]$ correctly simplifies to $2 a + e - 1 - \frac{1}{e} + (e - 1) + 1 - e^{- 1} = 2 a + 2 e - 2 - (e - e^{- 1})$ , which equates to the given area to yield $a = 5$ . This is the correct result.

D · 6

Making an arithmetic error when combining the constant terms from $[e^{x}]$ and $[e^{- x}]$ limits, leading to $2 a = 12$ instead of $2 a = 10$ . Ensure that the constant terms from $(e^{0} - e^{- 1}) + (e^{1} - e^{0}) = e - e^{- 1}$ and other bounds are added precisely.

Reviewed route

Solution

StepWorking

01Given

Region is bounded by $- 1 \leq x \leq 1$ and $0 \leq y \leq a + e^{∣ x ∣} - e^{- x}$ , where $a > 0$ , and the area is $\frac{e^{2} + 8 e + 1}{e}$ .

02Approach

Set up the definite integral $Area = \int_{- 1}^{1} (a + e^{∣ x ∣} - e^{- x}) d x$ . Split the integral at $x = 0$ to handle $∣ x ∣$ .

03Execute

Split into $[- 1, 0]$ and $[0, 1]$ : $Area = \int_{- 1}^{0} (a + e^{- x} - e^{- x}) d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$ $= \int_{- 1}^{0} a d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$ $= a (0 - (- 1)) + [a x + e^{x} + e^{- x}]_{0}^{1}$ $= a + (a (1) + e^{1} + e^{- 1}) - (0 + e^{0} + e^{0})$ $= a + a + e + \frac{1}{e} - 2 = 2 a + e + \frac{1}{e} - 2 = 2 a - 2 + \frac{e^{2} + 1}{e}$ .

04Execute

Equate the computed area to the given value: $2 a - 2 + \frac{e^{2} + 1}{e} = \frac{e^{2} + 8 e + 1}{e} = \frac{e^{2} + 1}{e} + 8$ $2 a - 2 = 8 ⟹ 2 a = 10 ⟹ a = 5$ .

✓Verify

For $a = 5$ , check integrand positivity: $5 + e^{∣ x ∣} - e^{- x} \geq 5 + 1 - e > 0$ for all $x \in [- 1, 1]$ . Area $= 2 (5) - 2 + \frac{e^{2} + 1}{e} = 8 + \frac{e^{2} + 1}{e} = \frac{e^{2} + 8 e + 1}{e}$ , which matches exactly.

Hints that build this answer step by step

How should the area integral $\int_{- 1}^{1} (a + e^{∣ x ∣} - e^{- x}) d x$ be split to handle the absolute value $∣ x ∣$ ?
$\int_{- 1}^{0} (a + e^{- x} - e^{- x}) d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$
What is the evaluated result of the total area integral in terms of $a$ and $e$ ?
$2 a + e - 1 + \frac{1}{e}$
Equating the computed area $2 a + e - 2 + \frac{1}{e}$ or $2 a + e + \frac{1}{e} - 2$ to $\frac{e^{2} + 8 e + 1}{e} = e + 8 + \frac{1}{e}$ , what is the value of $a$ ?
$a = 5$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does the term

e^{- x} - e^{- x}

become 0 on

[- 1, 0]

For $x \in [- 1, 0]$ , $∣ x ∣ = - x$ , so $e^{∣ x ∣} = e^{- x}$ . Thus $e^{∣ x ∣} - e^{- x} = e^{- x} - e^{- x} = 0$ .

Answer

Integrating the function $y = a + e^{∣ x ∣} - e^{- x}$ from $- 1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .

Why each option works or fails

A: 8 - Believing that $2 a = 8 + e - e$ directly equals $8$ without dividing by $2$ , or equating $a$ to the numerator's coefficient of $e$ . After integrating, fully simplify the linear equation in terms of $a$ : isolate $2 a = 10$ , which gives $a = 5$ rather than $8$ .
B: 7 - Dropping the integration constant contributions from $e^{- x}$ or making a sign error in $\int_{- 1}^{1} e^{- x} d x = e - \frac{1}{e}$ . Evaluate $\int_{- 1}^{1} - e^{- x} d x = [e^{- x}]_{- 1}^{1} = e^{- 1} - e^{1} = \frac{1}{e} - e$ carefully, keeping track of negative signs.
C · correct: 5 - The integral of $a + e^{∣ x ∣} - e^{- x}$ over $[- 1, 1]$ correctly simplifies to $2 a + e - 1 - \frac{1}{e} + (e - 1) + 1 - e^{- 1} = 2 a + 2 e - 2 - (e - e^{- 1})$ , which equates to the given area to yield $a = 5$ . This is the correct result.
D: 6 - Making an arithmetic error when combining the constant terms from $[e^{x}]$ and $[e^{- x}]$ limits, leading to $2 a = 12$ instead of $2 a = 10$ . Ensure that the constant terms from $(e^{0} - e^{- 1}) + (e^{1} - e^{0}) = e - e^{- 1}$ and other bounds are added precisely.

Step-by-step solution

Given: Region is bounded by $- 1 \leq x \leq 1$ and $0 \leq y \leq a + e^{∣ x ∣} - e^{- x}$ , where $a > 0$ , and the area is $\frac{e^{2} + 8 e + 1}{e}$ .
Approach: Set up the definite integral $Area = \int_{- 1}^{1} (a + e^{∣ x ∣} - e^{- x}) d x$ . Split the integral at $x = 0$ to handle $∣ x ∣$ .
Execute: Split into $[- 1, 0]$ and $[0, 1]$ : $Area = \int_{- 1}^{0} (a + e^{- x} - e^{- x}) d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$ $= \int_{- 1}^{0} a d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$ $= a (0 - (- 1)) + [a x + e^{x} + e^{- x}]_{0}^{1}$ $= a + (a (1) + e^{1} + e^{- 1}) - (0 + e^{0} + e^{0})$ $= a + a + e + \frac{1}{e} - 2 = 2 a + e + \frac{1}{e} - 2 = 2 a - 2 + \frac{e^{2} + 1}{e}$ .
Execute: Equate the computed area to the given value: $2 a - 2 + \frac{e^{2} + 1}{e} = \frac{e^{2} + 8 e + 1}{e} = \frac{e^{2} + 1}{e} + 8$ $2 a - 2 = 8 ⟹ 2 a = 10 ⟹ a = 5$ .
Verify: For $a = 5$ , check integrand positivity: $5 + e^{∣ x ∣} - e^{- x} \geq 5 + 1 - e > 0$ for all $x \in [- 1, 1]$ . Area $= 2 (5) - 2 + \frac{e^{2} + 1}{e} = 8 + \frac{e^{2} + 1}{e} = \frac{e^{2} + 8 e + 1}{e}$ , which matches exactly.

Finding an Unknown Parameter from the Area Under a Piecewise Curve

What feels right?

Integrating the function $y = a + e^{∣ x ∣} - e^{- x}$ from $- 1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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