If y = | x|^{|x|}, then the value of dy/dx at x = -π/6 is

Correct Answer is : (26)−π6[6log⁡2−3π]≤ft(2/6)^{-π/6} ≤ft[6 2 - {3} π](62)−6π[6log2−3π]

≤ft(2/6)^{-π/6} ≤ft[6 2 - {3} π]

≤ft(2/6)^{π/6} ≤ft[6 2 + {3} π]

≤ft(2/6)^{-π/6} ≤ft[6 2 + {3} π]

Correct Answer is : (26)−π6[6log⁡2−3π]≤ft(2/6)^{-π/6} ≤ft[6 2 - {3} π](62)−6π[6log2−3π]

Test Index

Manipal Entrance Test 2014 Math Paper

Question : 35 of 80

Marks: +1, -0

y = |\sin x|^{|x|},

\frac{dy}{dx}

x = -\frac{\pi}{6}

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

Solution:

Given

y = |\sin x|^{|x|}

In the neighbourhood of

-\frac{\pi}{6}, |x|

and

|\sin x|

both are negative

i.e.,

y = (-\sin x)^{-x}

Taking log on both sides, we get

\log y = -x \log(-\sin x)

⇒

\frac{1}{y} \frac{dy}{dx} = (-x) \frac{1}{-\sin x}(-\cos x)

+ \log(-\sin x)(-1)

= -x \cot x - \log(-\sin x)

= -\left[x \cot x + \log(-\sin x)\right]

⇒

\frac{dy}{dx} = -y\left[x \cot x + \log(-\sin x)\right]

∴

\left( \frac{dy}{dx} \right)_{x=\frac{\pi}{6}} = (2)^{-\frac{\pi}{6}} \frac{6 \log 2 - \sqrt{3} \pi}{6}

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