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AP and TS EAMCET Engineering Model Paper 20

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Question : 68 of 160

Marks: +1, -0

The minimum value of

(\sin x)^{\sin x}

0 < x < \pi

is

$e^{\frac{1}{e}}$
$e^{-e}$
1
$e^{-\frac{1}{e}}$

Solution:

Let

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

and

s=\log f(x)=\sin x \cdot \log \sin x

\frac{ds}{dx}=\sin x \cdot \frac{\cos x}{\sin x}+(\log \sin x) \cos x

=\cos x(1+\log \sin x)

\frac{d^2 s}{dx^2}=\cos x \left( \frac{1}{\sin x} \cdot \cos x \right)

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

=\frac{\cos^2 x}{\sin x}-\sin x(1+\log \sin x)

Now

rac{ds}{dx}=0 \Rightarrow \cos x=0

or

1+\log \sin x=0

\Rightarrow x=\frac{\pi}{2}

or

\sin x=e^{-1}

\left( \frac{d^2 s}{dx^2} \right)_{x=\frac{\pi}{2}} =0-1<0 \Rightarrow f(x)

has maximum value

\left( \frac{d^2 s}{dx^2} \right)_{\sin x = e^{-1}} = \frac{1-\frac{1}{e^2}}{\frac{1}{e}}

-\frac{1}{e}(1+\log e^{-1})

=e \left(1-\frac{1}{e^2}\right) >0

∴ f(x) has minimum when

\sin x = \frac{1}{e}

Minimum value

[f(x)]_{\sin x = \frac{1}{e}} = \left( \frac{1}{e} \right)^{\frac{1}{e}}

=e^{-\frac{1}{e}}

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