QJEEM21Ma0542: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0542 Solution: Let a be a positive real number such that where [ x ] is the…

Question

Verified Solution • 2025–26 Session moderate

Let a be a positive real number such that

\int_0^a {{e^{x - [x]}}} dx = 10e - 9

where [ x ] is the greatest integer less than or equal to x. Then a is equal to:

Accepted Answer

a > 0

Let

n \le a &lt; n + 1,n \in W

a = \begin{matrix} [a] & + & {a} \\ ⇓ & ⇓ \\ G . I . F . & F r a c t i o n a l p a r t \end{matrix}

Here [ a ] = n

Now,

\int_0^a {{e^{x - [x]}}} dx = 10e - 9

\Rightarrow \int\limits_0^n {{e^{\{ x\} }}dx} + \int\limits_n^a {{e^{x - [x]}}dx} = 10e - 9

\therefore

n\int\limits_0^1 {{e^x}dx} + \int\limits_n^a {{e^{x - n}}dx} = 10e - 9

\Rightarrow n(e - 1) + ({e^{a - n}} - 1) = 10e - 9

\therefore

n = 0 and {a} = log_e 2

So,

a = [a] + \{ a\} = (10 + {\log _e}2)

\Rightarrow

Option (2) is correct.

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