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