The function f(x)=( sin x) sin x attains its minimum value when x equal

Question

The function f(x)=( sin x) sin x attains its minimum value when x equal (A)  cos -1(0) (B)  sin -1((1/2)) (C)  sin -1((2/ e )) (D)  sin -1((1/ e ))

Tardigrade Admin · Accepted Answer

y=( sin x) sin x=e sin x ln ( sin x) (d y/d x)=( sin x) sin x[ sin x ⋅ cot x+ ln ( sin x) ⋅ cos x]=0 hence cos x[1+ ln ( sin x)]=0 cos x=0 or ln ( sin x)=-1 sin x =(1/ e ) ; x = sin -1 (1/ e ) or x =2 n π+(π/2)

Q. The function $f (x) = (sin x)^{s i n x}$ attains its minimum value when $x$ equal

$cos^{- 1} (0)$

$sin^{- 1} (\frac{1}{2})$

$sin^{- 1} (\frac{2}{e})$

$sin^{- 1} (\frac{1}{e})$

$y = (sin x)^{s i n x} = e^{s i n x l n (s i n x)}$
$\frac{d y}{d x} = (sin x)^{s i n x} [sin x \cdot cot x + ln (sin x) \cdot cos x] = 0$
hence $cos x [1 + ln (sin x)] = 0$
$cos x = 0$ or $ln (sin x) = - 1$
$sin x = \frac{1}{e}; x = sin^{- 1} \frac{1}{e}$ or $x = 2 nπ + \frac{π}{2}$

Q. The function f(x)=(sinx)sinx attains its minimum value when x equal

cos−1(0)

sin−1(21​)

sin−1(e2​)

sin−1(e1​)