STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

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$.