JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 1)
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:
$$10 - {\log _e}(1 + e)$$
$$10 + {\log _e}2$$
$$10 + {\log _e}3$$
$$10 + {\log _e}(1 + e)$$
Explanation
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} = loge 2
So, $$a = [a] + \{ a\} = (10 + {\log _e}2)$$
$$\Rightarrow$$ Option (2) is correct.
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} = loge 2
So, $$a = [a] + \{ a\} = (10 + {\log _e}2)$$
$$\Rightarrow$$ Option (2) is correct.
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