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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
On the interval $\left[ {\frac{{5\pi }}{3},\,\,\frac{{7\pi }}{4}} \right],$ the greatest value of the function $f(x) = \int_{5\pi /3}^x {(6\cos t - 2\sin t)\,dt = } $
  • A
    $3\sqrt 3 + 2\sqrt 2 + 1$
  • ✓
    $3\sqrt 3 - 2\sqrt 2 - 1$
  • C
    Does not exist
  • D
    None of these
✓

Answer

Correct option: B.
$3\sqrt 3 - 2\sqrt 2 - 1$
b
(b) $f'(x) = (6\cos x - 2\sin x)1 - 0$

$ = 2[3\cos x - \sin x] > 0$ in $\left[ {\frac{{5\pi }}{3},\frac{{7\pi }}{4}} \right]$

$\therefore$ $f(x)$ is an increasing function, hence $f(x)$ has greatest value at $x = \frac{{7\pi }}{4}$

$\therefore$ Greatest $f(x) = f\left( {\frac{{7\pi }}{4}} \right) = [6\sin t + 2\cos t]_{5\pi /3}^{7\pi /4}$

$ = - 6\frac{1}{{\sqrt 2 }} + \frac{2}{{\sqrt 2 }} + 6\frac{{\sqrt 3 }}{2} - 1 $

$= 3\sqrt 3 - 2\sqrt 2 - 1$.

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