Q44 / 75 · JEE Main 2026 (02 April Shift 1) · Mathematics

Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :

  • A
    $e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}$
  • Correct answer
    B
    $\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)$
  • C
    $e^2 + e^3 - \dfrac{1}{2e^2} - \dfrac{1}{2e^3}$
  • D
    $\dfrac{1}{2}(e^2 + e^3) - \dfrac{1}{e^2} - \dfrac{1}{e^3}$

Solution

The given integral can be split at integer values of $x$ because of the greatest integer function $[x]$.

$\displaystyle I = \int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$

$\displaystyle I = \int_0^1 \dfrac{e^x + e^{-x}}{0!} dx + \int_1^2 \dfrac{e^x + e^{-x}}{1!} dx + \int_2^3 \dfrac{e^x + e^{-x}}{2!} dx$

Since $0! = 1$, $1! = 1$, and $2! = 2$, we get:

$\displaystyle I = \int_0^2 (e^x + e^{-x}) dx + \dfrac{1}{2} \int_2^3 (e^x + e^{-x}) dx$

Integrating the terms:

$\displaystyle I = \left[ e^x - e^{-x} \right]_0^2 + \dfrac{1}{2} \left[ e^x - e^{-x} \right]_2^3$

$\displaystyle I = (e^2 - e^{-2} - (1 - 1)) + \dfrac{1}{2} (e^3 - e^{-3} - (e^2 - e^{-2}))$

$\displaystyle I = e^2 - e^{-2} + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-3} - \dfrac{1}{2} e^2 + \dfrac{1}{2} e^{-2}$

$\displaystyle I = \dfrac{1}{2} e^2 + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-2} - \dfrac{1}{2} e^{-3}$

$\displaystyle I = \dfrac{1}{2} \left( e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3} \right)$

Answer: $\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)$

Hence, the correct answer is option: 'B'.

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