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QJEEM22Mac17b: JEE Main 2022 Mathematics - Class 12 JEEM22Mac17b Solution: Let denote the greatest integer less than or equal to . Then…

Let [t][t] denote the greatest integer less than or equal to tt . Then the value of the integral ∫−3101([sin⁡(πx)]+e[cos⁡(2πx)])dx\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x is equal to

Multiple Choice Options

  • A 52(1−e)e\frac{52(1-e)}{e}
  • B 52e\frac{52}{e}
  • C 52(2+e)e\frac{52(2+e)}{e}
  • D 104e\frac{104}{e}

Step-by-Step Verified Solution

I=∫−3101([sin⁡(πx)]+e[cos⁡(2πx)])dxI = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx}

[sin⁡πx][\sin \pi x] is periodic with period 2 and e[cos⁡(2πx)]{{e^{[\cos (2\pi x)]}}} is periodic with period 1.

So,

I=52∫02([sin⁡πx]+e[cos⁡2πx])dxI = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx}

=52{∫12−1 dx+∫1434e−1 dx+∫5474e−1 dx+∫014e0 dx+∫3454e0 dx+∫742e0 dx}= 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}

=52e= {{52} \over e}

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