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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
Let $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$ If $\int\limits_{\cos x}^{1} t^{2} f(t) d t=\sin ^{3} x+\cos x-1$ then $\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to
  • A
    $6-9 \sqrt{2}$
  • B
    $\frac{9}{\sqrt{2}}-6$
  • C
    $\frac{9}{2}-6 \sqrt{2}$
  • ✓
    $6-\frac{9}{\sqrt{2}}$
✓

Answer

Correct option: D.
$6-\frac{9}{\sqrt{2}}$
d
$\int\limits_{\cos x}^{1} t^{2} f(t) d t=\sin ^{3} x+\cos x-1$

Calculation for option

differentiating both sides

$-\cos ^{2} x f(\cos x) \cdot(-\sin x)=3 \sin ^{2} x \cdot \cos x-\sin x$

$\Rightarrow f (\cos x )=3 \tan x -\sec ^{2} x$

$\Rightarrow f^{\prime}(\cos x)(-\sin x)=3 \sec ^{2} x-2 \sec ^{2} x \tan x$

$\Rightarrow f^{\prime}(\cos x) \cos x=\frac{2}{\cos ^{2} x}-\frac{3}{\sin x \cdot \cos x}$

When $\cos x=\frac{1}{\sqrt{3}} ; \sin x=\frac{\sqrt{2}}{\sqrt{3}}$

$\therefore f^{\prime}\left(\frac{1}{\sqrt{3}}\right) \frac{1}{\sqrt{3}}=6-\frac{9}{\sqrt{2}}$

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