Substituting in the first integral and in the second gives in both cases.

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why is the minimum guaranteed to lie in ?

Because outside , the integrand grows strictly as moves away from the interval . Moving closer to the interval always decreases the integrand pointwise.

The minimum value of the function

f(x) = \int_0^2 e^{|x-t|} dt

is :

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What feels right?

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e(e-1)

2(e-1)

2

2e-1

Step-by-step solutionView

Correct answer

The minimum value of $f(x) = \int_0^2 e^{|x-t|} dt$ is $2(e-1)$ , achieved at $x = 1$ .

Option analysis

Why each option works or fails

A ·

e(e-1)

Evaluating the function at the boundaries $x=0$ or $x=2$ where $f(0) = f(2) = e^2 - 1 = (e-1)(e+1)$ , or mistakenly factoring $e^2 - e$ as the minimum instead of locating the interior stationary point. Differentiate $f(x)$ or inspect symmetry to see that the global minimum lies at the midpoint $x = 1$ , not at the interval boundaries.

B ·

2(e-1)

None. By symmetry or differentiation, the critical point is $x = 1$ , giving $f(1) = \int_0^1 e^{1-t} dt + \int_1^2 e^{t-1} dt = (e - 1) + (e - 1) = 2(e - 1)$ . Correctly identified $x=1$ as the minimizing point and integrated properly across the split domain.

C ·

2

Integrating $e^{|x-t|}$ as if the exponential and absolute value cancel out to a constant, or setting $e^0 = 1$ and integrating $1$ over $[0, 2]$ . Keep the exponential term: $\int_0^1 e^{1-t} dt = e - 1$ , which does not simplify to $1$ .

D ·

2e-1

Evaluating only one half of the symmetric integral at $x=1$ , calculating $\int_0^1 e^{1-t} dt = e - 1$ and adding $e$ instead of $e - 1$ . Ensure both sub-integrals $\int_0^1 e^{1-t}dt$ and $\int_1^2 e^{t-1}dt$ are computed and summed: $(e-1) + (e-1) = 2(e-1)$ .

Reviewed route

Solution

StepWorking

01given

$f(x) = \int_0^2 e^{|x-t|} dt$ , where $t \in [0, 2]$ and $x \in \mathbb{R}$ .

02goal

Find the global minimum value of $f(x)$ over $x \in \mathbb{R}$ .

03approach

Partition the real line into $x \le 0$ , $0 < x < 2$ , and $x \ge 2$ to split the absolute value $|x-t|$ . In the middle interval, find the critical point using calculus or AM-GM inequality, and verify that the endpoints match the outer monotonic pieces.

04execute

For $x \in [0, 2]$ , split at $t = x$ : $f(x) = \int_0^x e^{x-t} dt + \int_x^2 e^{t-x} dt = e^x \left[-e^{-t}\right]_0^x + e^{-x} \left[e^t\right]_x^2 = (e^x - 1) + (e^{2-x} - 1) = e^x + e^{2-x} - 2$

05execute

For $0 \le x \le 2$ , apply AM-GM to $e^x$ and $e^{2-x}$ : $e^x + e^{2-x} \ge 2\sqrt{e^x \cdot e^{2-x}} = 2\sqrt{e^2} = 2e$ Equality holds when $e^x = e^{2-x} \implies x = 1$ . Thus, the minimum value on $[0, 2]$ is $2e - 2 = 2(e - 1)$ .

06execute

For $x < 0$ , $|x-t| = t-x$ , so $f(x) = e^{-x}(e^2 - 1)$ , which is strictly decreasing with infimum at $x \to 0^-$ equal to $e^2 - 1$ . For $x > 2$ , $|x-t| = x-t$ , so $f(x) = e^x(1 - e^{-2})$ , which is strictly increasing with infimum at $x \to 2^+$ equal to $e^2 - 1$ . Since $e^2 - 1 = (e-1)(e+1) > 2(e-1)$ (as $e+1 > 2$ ), the global minimum is indeed at $x = 1$ .

✓verify

Differentiating using Leibniz rule: $f'(x) = \int_0^x e^{x-t} dt - \int_x^2 e^{t-x} dt = (e^x - 1) - (e^{2-x} - 1) = e^x - e^{2-x}$ . Setting $f'(x) = 0$ gives $x = 1$ , with $f''(1) = e + e > 0$ , confirming a unique global minimum.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the minimum guaranteed to lie in

[0, 2]

Because outside $[0, 2]$ , the integrand $e^{|x-t|}$ grows strictly as $x$ moves away from the interval $[0, 2]$ . Moving $x$ closer to the interval always decreases the integrand pointwise.

Why is

\int_0^1 e^{1-t} dt

equal to

\int_1^2 e^{t-1} dt

Substituting $u = 1-t$ in the first integral and $u = t-1$ in the second gives $\int_0^1 e^u du$ in both cases.

Answer

The minimum value of $f(x) = \int_0^2 e^{|x-t|} dt$ is $2(e-1)$ , achieved at $x = 1$ .

Why each option works or fails

A: $e(e-1)$ - Evaluating the function at the boundaries $x=0$ or $x=2$ where $f(0) = f(2) = e^2 - 1 = (e-1)(e+1)$ , or mistakenly factoring $e^2 - e$ as the minimum instead of locating the interior stationary point. Differentiate $f(x)$ or inspect symmetry to see that the global minimum lies at the midpoint $x = 1$ , not at the interval boundaries.
B · correct: $2(e-1)$ - None. By symmetry or differentiation, the critical point is $x = 1$ , giving $f(1) = \int_0^1 e^{1-t} dt + \int_1^2 e^{t-1} dt = (e - 1) + (e - 1) = 2(e - 1)$ . Correctly identified $x=1$ as the minimizing point and integrated properly across the split domain.
C: $2$ - Integrating $e^{|x-t|}$ as if the exponential and absolute value cancel out to a constant, or setting $e^0 = 1$ and integrating $1$ over $[0, 2]$ . Keep the exponential term: $\int_0^1 e^{1-t} dt = e - 1$ , which does not simplify to $1$ .
D: $2e-1$ - Evaluating only one half of the symmetric integral at $x=1$ , calculating $\int_0^1 e^{1-t} dt = e - 1$ and adding $e$ instead of $e - 1$ . Ensure both sub-integrals $\int_0^1 e^{1-t}dt$ and $\int_1^2 e^{t-1}dt$ are computed and summed: $(e-1) + (e-1) = 2(e-1)$ .

Step-by-step solution

given: $f(x) = \int_0^2 e^{|x-t|} dt$ , where $t \in [0, 2]$ and $x \in \mathbb{R}$ .
goal: Find the global minimum value of $f(x)$ over $x \in \mathbb{R}$ .
approach: Partition the real line into $x \le 0$ , $\int_1^2 e^{t-1}dt$ , and $x \ge 2$ to split the absolute value $|x-t|$ . In the middle interval, find the critical point using calculus or AM-GM inequality, and verify that the endpoints match the outer monotonic pieces.
execute: For $x \in [0, 2]$ , split at $t = x$ : $f(x) = \int_0^x e^{x-t} dt + \int_x^2 e^{t-x} dt = e^x \left[-e^{-t}\right]_0^x + e^{-x} \left[e^t\right]_x^2 = (e^x - 1) + (e^{2-x} - 1) = e^x + e^{2-x} - 2$
execute: For $0 \le x \le 2$ , apply AM-GM to $e^x$ and $e^{2-x}$ : $e^x + e^{2-x} \ge 2\sqrt{e^x \cdot e^{2-x}} = 2\sqrt{e^2} = 2e$ Equality holds when $e^x = e^{2-x} \implies x = 1$ . Thus, the minimum value on $[0, 2]$ is $2e - 2 = 2(e - 1)$ .
execute: For $x < 0$ , $|x-t| = t-x$ , so $e^x + e^{2-x} \ge 2\sqrt{e^x \cdot e^{2-x}} = 2\sqrt{e^2} = 2e$ , which is strictly decreasing with infimum at $x \to 0^-$ equal to $e^2 - 1$ . For $x > 2$ , $|x-t| = x-t$ , so $f(x) = e^x(1 - e^{-2})$ , which is strictly increasing with infimum at $x \to 2^+$ equal to $e^2 - 1$ . Since $|x-t| = t-x$ (as $e+1 > 2$ ), the global minimum is indeed at $x = 1$ .
verify: Differentiating using Leibniz rule: $f'(x) = \int_0^x e^{x-t} dt - \int_x^2 e^{t-x} dt = (e^x - 1) - (e^{2-x} - 1) = e^x - e^{2-x}$ . Setting $f'(x) = 0$ gives $x = 1$ , with $f''(1) = e + e > 0$ , confirming a unique global minimum.

Shortcut:

When to use it: Fastest route: differentiate $f(x)$ directly to find the critical point without computing the full integral beforehand.

given: $f(x) = \int_0^2 e^{|x-t|} dt$ . The integrand is convex in $x$ , so $f(x)$ is convex.

goal: Find the stationary point $x^*$ where $f'(x^*) = 0$ , then evaluate $f(x^*)$ .

approach: Assume the minimum occurs at $x \in (0, 2)$ . Split the integral: $f(x) = \int_0^x e^{x-t} dt + \int_x^2 e^{t-x} dt$ . Differentiate with respect to $x$ via Leibniz rule.

execute: $f'(x) = e^{x-x}(1) + \int_0^x e^{x-t} dt - e^{x-x}(1) - \int_x^2 e^{t-x} dt = (e^x - 1) - (e^{2-x} - 1) = e^x - e^{2-x}$ . Setting $f'(x) = 0 \implies e^x = e^{2-x} \implies x = 1$ .

execute: Calculate $f(1)$ : $f(1) = \int_0^2 e^{|1-t|} dt = \int_0^1 e^{1-t} dt + \int_1^2 e^{t-1} dt = 2 \int_0^1 e^u du = 2(e - 1)$

verify: Symmetry check: $|1 - t|$ on $t \in [0, 2]$ is symmetric around $t = 1$ , giving $2 \int_0^1 e^u du = 2(e - 1)$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The minimum value of $f(x) = \int_0^2 e^{|x-t|} dt$ is $2(e-1)$ , achieved at $x = 1$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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