MCQ
The range of the function $f(x)=(\sin x)^{\sin x}$ defined on $(0, \pi)$ is
- A$(0,1)$
- B$\left(e^{-1 / e}, 1\right)$
- C$\left[e^{-1 / e}, 1\right)$
- ✓$\left[e^{-1 / e}, 1\right]$
We have,
$f(x) =(\sin x)^{\sin x}, x \in(0, \pi)$
$x =\frac{\pi}{2}$
$\Rightarrow \quad f\left(\frac{\pi}{2}\right) =1$
Hence, maximum value of $f(x)$ is $1.$
Taking log both sides
$\log f(x)=\sin x \log \sin x$
On differentiating w.r.t. $x$, we get $f^{\prime}(x)=f(x)[\cos x+\cos x \log \sin x]$ put $f^{\prime}(x)=0$
$\cos x(1+\log \sin x)=0$
$\cos x=0, \log \sin x=-1$
$x=\frac{\pi}{2}, \sin x=e^{-1}$
$\therefore$ Range of $f(x)$ is $\left[\left(\frac{1}{e}\right)^{1 / e}, 1\right]=\left[e^{-1 / e}, 1\right]$
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${l_1} = \left( {3 + t} \right)\hat i + \left( { - 1 + 2t} \right)\hat j + \left( {4 + 2t} \right)\hat k,\, - \infty < t < \infty $
${l_2} = \left( {3 + 2s} \right)\hat i + \left( {3 + 2s} \right)\hat j + \left( {2 + s} \right)\hat k,\, - \infty < s < \infty $
Statement $1$ : Line $l$ and $l_2$ are coplaner lines
Statement $2$ : Line $l$ and $l_2$ are intersecting lines