What is the value of when ?

For , . By mathematical definition, .

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

[\cdot]

denote the greatest integer function. Then the value of

\int_{0}^{3} \left[ \frac{e^{x} + e^{-x}}{[x]!} \right] dx

is:

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}

B

\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)

C

e^{2} + e^{3} - \frac{1}{2e^{2}} - \frac{1}{2e^{3}}

D

\frac{1}{4} (e^{2} + e^{3}) - \frac{1}{e^{2}} - \frac{1}{e^{3}}

Step-by-step solutionView

Correct answer

Splitting the integral across integer intervals simplifies $[x]!$ to constants, leading to the result $\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)$ .

Option analysis

Why each option works or fails

A ·

e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}

Forgetting the factor of $2$ when evaluating the integral of hyperbolic cosine or failing to divide by $2! = 2$ in the interval where $[x] = 2$ . Ensure that the factor of $\frac{1}{[x]!} = \frac{1}{2}$ on the interval $[2, 3)$ distributes across the integrated antiderivative $e^x - e^{-x}$ .

B ·

\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)

None. The integral vanishes or simplifies consistently: on $[0,1)$ the integrand is handled via its components, and integrating $\frac{e^x + e^{-x}}{[x]!}$ yields $\frac{1}{2}\left(e^2 + e^3 - e^{-2} - e^{-3}\right)$ . This is the correct value.

C ·

e^{2} + e^{3} - \frac{1}{2e^{2}} - \frac{1}{2e^{3}}

Distributing the factorial divisor only to the negative exponential terms while leaving the positive exponential terms unscaled. Apply the constant denominator $[x]!$ to the entire antiderivative $e^x - e^{-x}$ uniformly.

D ·

\frac{1}{4} (e^{2} + e^{3}) - \frac{1}{e^{2}} - \frac{1}{e^{3}}

Introducing an extraneous factor of $2$ in the denominator for $e^x$ terms while incorrectly handling the signs and coefficients of $e^{-x}$ . Evaluate $\int (e^x + e^{-x}) dx = e^x - e^{-x}$ directly without altering the coefficients of individual terms.

Reviewed route

Solution

StepWorking

01given

We are given the integral $I = \int_{0}^{3} \frac{e^{x} + e^{-x}}{[x]!} \, dx$ , where $[\cdot]$ denotes the greatest integer function (interpreting the outer brackets as standard grouping).

02approach

Split the integration interval $[0, 3]$ into unit subintervals $[0, 1)$ , $[1, 2)$ , and $[2, 3)$ where $[x]$ is piecewise constant: $[x] = 0$ on $[0, 1)$ , $[x] = 1$ on $[1, 2)$ , and $[x] = 2$ on $[2, 3)$ .

03execute

Evaluate $[x]!$ on each interval: - For $x \in [0, 1)$ : $[x] = 0 \implies [x]! = 0! = 1$ - For $x \in [1, 2)$ : $[x] = 1 \implies [x]! = 1! = 1$ - For $x \in [2, 3)$ : $[x] = 2 \implies [x]! = 2! = 2$ Thus, $I = \int_{0}^{1} (e^x + e^{-x}) \, dx + \int_{1}^{2} (e^x + e^{-x}) \, dx + \int_{2}^{3} \frac{e^x + e^{-x}}{2} \, dx = \int_{0}^{2} (e^x + e^{-x}) \, dx + \frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) \, dx$

04execute

Compute each integral using $\int (e^x + e^{-x}) \, dx = e^x - e^{-x}$ : $\int_{0}^{2} (e^x + e^{-x}) \, dx = [e^x - e^{-x}]_{0}^{2} = (e^2 - e^{-2}) - (1 - 1) = e^2 - e^{-2}$ $\frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) \, dx = \frac{1}{2} [(e^3 - e^{-3}) - (e^2 - e^{-2})]$ Summing these gives: $I = e^2 - e^{-2} + \frac{1}{2}(e^3 - e^2 - e^{-3} + e^{-2}) = \frac{1}{2} \left( e^3 + e^2 - e^{-2} - e^{-3} \right)$

✓verify

Combine coefficients: $e^3$ has coefficient $1/2$ , $e^2$ has $1 - 1/2 = 1/2$ , $e^{-2}$ has $-1 + 1/2 = -1/2$ , and $e^{-3}$ has $-1/2$ . All match $\frac{1}{2}(e^2 + e^3 - e^{-2} - e^{-3})$ .

Hints that build this answer step by step

How should the domain of integration $[0, 3]$ be partitioned to evaluate $[x]$ and $[x]!$ ?
Partition into unit intervals: $[0, 1)$ , $[1, 2)$ , and $[2, 3)$ .
What does the sum of integrals simplify to across these three sub-intervals?
$\int_{0}^{1} (e^x + e^{-x})\,dx + \int_{1}^{2} (e^x + e^{-x})\,dx + \frac{1}{2}\int_{2}^{3} (e^x + e^{-x})\,dx$
What is the result of evaluating the combined integral $\int_{0}^{2} (e^x + e^{-x})\,dx + \frac{1}{2}\int_{2}^{3} (e^x + e^{-x})\,dx$ ?
$\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)$

✓ 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

What is the value of

[x]!

when

x \in [0, 1)

?

For $x \in [0, 1)$ , $[x] = 0$ . By mathematical definition, $0! = 1$ .

Answer

Splitting the integral across integer intervals simplifies $[x]!$ to constants, leading to the result $\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)$ .

Why each option works or fails

A: $e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}$ - Forgetting the factor of $2$ when evaluating the integral of hyperbolic cosine or failing to divide by $2! = 2$ in the interval where $[x] = 2$ . Ensure that the factor of $\frac{1}{[x]!} = \frac{1}{2}$ on the interval $[2, 3)$ distributes across the integrated antiderivative $e^x - e^{-x}$ .
B · correct: $\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)$ - None. The integral vanishes or simplifies consistently: on $[0,1)$ the integrand is handled via its components, and integrating $\frac{e^x + e^{-x}}{[x]!}$ yields $\frac{1}{2}\left(e^2 + e^3 - e^{-2} - e^{-3}\right)$ . This is the correct value.
C: $e^{2} + e^{3} - \frac{1}{2e^{2}} - \frac{1}{2e^{3}}$ - Distributing the factorial divisor only to the negative exponential terms while leaving the positive exponential terms unscaled. Apply the constant denominator $[x]!$ to the entire antiderivative $e^x - e^{-x}$ uniformly.
D: $\frac{1}{4} (e^{2} + e^{3}) - \frac{1}{e^{2}} - \frac{1}{e^{3}}$ - Introducing an extraneous factor of $2$ in the denominator for $e^x$ terms while incorrectly handling the signs and coefficients of $e^{-x}$ . Evaluate $\int (e^x + e^{-x}) dx = e^x - e^{-x}$ directly without altering the coefficients of individual terms.

Step-by-step solution

given: We are given the integral $I = \int_{0}^{3} \frac{e^{x} + e^{-x}}{[x]!} \, dx$ , where $[\cdot]$ denotes the greatest integer function (interpreting the outer brackets as standard grouping).
approach: Split the integration interval $[0, 3]$ into unit subintervals $[0, 1)$ , $[1, 2)$ , and $[2, 3)$ where $[x]$ is piecewise constant: $[x] = 0$ on $[0, 1)$ , $[x] = 1$ on $[1, 2)$ , and $[x] = 2$ on $[2, 3)$ .
execute: Evaluate $[x]!$ on each interval: - For $x \in [0, 1)$ : $[x] = 0 \implies [x]! = 0! = 1$ - For $x \in [1, 2)$ : $[x] = 1 \implies [x]! = 1! = 1$ - For $x \in [2, 3)$ : $[x] = 2 \implies [x]! = 2! = 2$ Thus, $I = \int_{0}^{1} (e^x + e^{-x}) \, dx + \int_{1}^{2} (e^x + e^{-x}) \, dx + \int_{2}^{3} \frac{e^x + e^{-x}}{2} \, dx = \int_{0}^{2} (e^x + e^{-x}) \, dx + \frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) \, dx$
execute: Compute each integral using $\int (e^x + e^{-x}) \, dx = e^x - e^{-x}$ : $\int_{0}^{2} (e^x + e^{-x}) \, dx = [e^x - e^{-x}]_{0}^{2} = (e^2 - e^{-2}) - (1 - 1) = e^2 - e^{-2}$ $\frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) \, dx = \frac{1}{2} [(e^3 - e^{-3}) - (e^2 - e^{-2})]$ Summing these gives: $I = e^2 - e^{-2} + \frac{1}{2}(e^3 - e^2 - e^{-3} + e^{-2}) = \frac{1}{2} \left( e^3 + e^2 - e^{-2} - e^{-3} \right)$
verify: Combine coefficients: $e^3$ has coefficient $1/2$ , $e^2$ has $1 - 1/2 = 1/2$ , $e^{-2}$ has $-1 + 1/2 = -1/2$ , and $e^{-3}$ has $-1/2$ . All match $\frac{1}{2}(e^2 + e^3 - e^{-2} - e^{-3})$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Splitting the integral across integer intervals simplifies $[x]!$ to constants, leading to the result $\frac{1}{2} \left( e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}} \right)$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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