The minimum value of e^ (2 x^2 - 2x + 1) ^2 x is - Vidyadip

MCQ

The minimum value of ${e^{(2{x^2} - 2x + 1){{\sin }^2}x}}$ is

A
$e$
B
$1/e$
✓
$1$
D
$0$

✓

Answer

Correct option: C.

$1$

c
(c) Given $y = {e^{(2{x^2} - 2x + 1){{\sin }^2}x}}$

For minima or maxima, $\frac{{dy}}{{dx}} = 0$

$\therefore {e^{(2{x^2} - 2x + 1){{\sin }^2}x}}[(4x - 2){\sin ^2}x + 2(2{x^2} - 2x + 1)\sin x\cos x] = 0$

==> $[(4x - 2){\sin ^2}x + 2(2{x^2} - 2x + 1)\sin x\cos x] = 0$

==> $2\sin x[(2x - 1)\sin x + (2{x^2} - 2x + 1)\cos x] = 0$

==> $\sin x = 0$

$\therefore y$ is minimum for $\sin x = 0$

Thus minimum value of $ y = {e^{(2{x^2} - 2x + 1)(0)}} = {e^0} = 1$.

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