JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 8)
If the integral
$$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are integers and [x] denotes the greatest integer less than or equal to x, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :
$$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $$, where $$\alpha$$, $$\beta$$, $$\gamma$$ are integers and [x] denotes the greatest integer less than or equal to x, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is equal to :
0
10
20
25
Explanation
Given integral
$$\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } $$ (using property of definite in.)
$$ = 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]$$
= $$ - 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]$$
$$ = 10{e^{ - 1}} - 10{e^{ - 1/2}}$$
comparing with the given relation,
$$\alpha$$ = 10, $$\beta$$ = $$-$$10, $$\gamma$$ = 0
$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 0
Therefore, the correct answer is (A).
$$\int\limits_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}dx = 10\int\limits_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}dx} } $$ (using property of definite in.)
$$ = 10\left[ {\int\limits_0^{1/2} {0.dx} + \int\limits_{1/2}^1 {{{ - 1} \over {{e^x}}}dx} } \right]$$
= $$ - 10\left[ {{{{e^{ - x}}} \over { - 1}}} \right]_{1/2}^1 = 10\left[ {{e^{ - 1}} - {e^{ - 1/2}}} \right]$$
$$ = 10{e^{ - 1}} - 10{e^{ - 1/2}}$$
comparing with the given relation,
$$\alpha$$ = 10, $$\beta$$ = $$-$$10, $$\gamma$$ = 0
$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 0
Therefore, the correct answer is (A).
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