If f(x)=|x|^{| x|}, then f'≤ft(-π/6) is equal to

Correct Answer is : π6(−32log⁡π6−3π){π/6} ≤ft( -{{3}}{2} π/6 - 3/π )6π(−23log6π−π3)

{π/6} ≤ft( -{{3}}{2} π/6 + 3/π )

{π/6} ≤ft( -{{3}}{2} π/6 - 3/π )

{π/6} ≤ft( {{3}}{2} π/6 - 3/π )

Correct Answer is : π6(−32log⁡π6−3π){π/6} ≤ft( -{{3}}{2} π/6 - 3/π )6π(−23log6π−π3)

Test Index

CGPET 2019 Solved Paper

Section: Mathematics

Question : 107 of 150

Marks: +1, -0

f(x)=|x|^{|\sin x|}

f'\left(-\frac{\pi}{6}\right)

is equal to

$\sqrt{\frac{\pi}{6}} \left( -\frac{\sqrt{3}}{2} \log \frac{\pi}{6} + \frac{3}{\pi} \right)$
$\sqrt{\frac{\pi}{6}} \left( -\frac{\sqrt{3}}{2} \log \frac{\pi}{6} - \frac{3}{\pi} \right)$
$\sqrt{\frac{\pi}{6}} \left( \frac{\sqrt{3}}{2} \log \frac{\pi}{6} - \frac{3}{\pi} \right)$
None of the above

Solution:

Since,

f(x)=|x| \sin x \|

y=|x|^{|\sin x|} \text{[Let } f(x)=y]

Taking log both sides, we get

\log y = |\sin x| \log |x| \left[ \because \log A^{M} = M \log A \right]

Differentiate w.r. to

'x'

, we get

\frac{1}{y} \frac{dy}{dx} = \left[ \frac{|\sin x|}{\sin x} \cos x \cdot \log |x| + \frac{|\sin x|}{|x|} \cdot \frac{|x|}{x} \right]

\Rightarrow \frac{dy}{dx} = y \left[ \frac{|\sin x|}{\sin x} \cos x \cdot \log |x| + \frac{|\sin x|}{x} \right]

\Rightarrow f'(x)=|x|^{|\sin x|}\left[ \frac{|\sin x|}{\sin x} \cos x \right]

\left[ \cdot \log |x| + \frac{|\sin x|}{x} \right]

\Rightarrow f'\left(-\frac{\pi}{6}\right) = \left|\frac{-\pi}{6}\right|^{|\sin(-\pi/6)|}

\left[ \frac{ \left| \sin\left(-\frac{\pi}{6}\right) \right| }{ \sin\left(-\frac{\pi}{6}\right) } \cos\left(-\frac{\pi}{6}\right) \cdot \log \left| -\frac{\pi}{6} \right| + \frac{ \left| \sin\left(-\frac{\pi}{6}\right) \right| }{ -\frac{\pi}{6} } \right]

\Rightarrow f'\left(-\frac{\pi}{6}\right) = \left(\frac{\pi}{6}\right)^{1/2} \left[ \frac{-\sqrt{3}}{2} \log \frac{\pi}{6} \right]

\left[+ \frac{1/2}{-\pi/6} \right]

\Rightarrow f'\left(-\frac{\pi}{6}\right)

= \sqrt{\frac{\pi}{6}} \left[ \frac{-\sqrt{3}}{2} \log \frac{\pi}{6} - \frac{3}{\pi} \right]

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