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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
Let $g(t)=\int \limits_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} t+f(x)\right) \,d x$, where $f(x)=\log _{e}\left(x+\sqrt{x^{2}+1}\right), x \in R$. Then which one of the following is correct?
  • A
    $g(1)+g(0)=0$
  • B
    $g(1)=\sqrt{2} g(0)$
  • C
    $g(1)=g(0)$
  • ✓
    $\sqrt{2} g(1)=g(0)$
✓

Answer

Correct option: D.
$\sqrt{2} g(1)=g(0)$
d
$g(t)=\int_{-\pi / 2}^{\pi / 2} \cos \left(\frac{\pi}{4} t+f(x)\right) \,d x$

$g(t)=\pi \cos \frac{\pi}{4} t+\int_{-\pi / 2}^{\pi / 2} f(x) \,d x$

$g(t)=\pi \cos \frac{\pi}{4} t$

$g(1)=\frac{\pi}{\sqrt{2}}, g(0)=\pi$

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