"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:"

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Accepted Answer

"a > 0

Let $n \le a < n + 1,n \in W$

$$a=\matrix{ {[a]} & + & {\{ a\} } \cr \Downarrow & {} & \Downarrow \cr {G.I.F.} & {} & {Fractional\,part} \cr } $$

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."

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:

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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:

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