Explanation

(b)
Given, $y = e^{\left( 2x^{2} - 2a + 1 \right)\sin^{2}x}$
For minimum or maximum, put $\frac{dy}{dx} = 0$
$\therefore\ \ e^{\left( 2x^{2} - 2x + 1 \right)\sin^{2}x}\left\lbrack (4x - 2)\sin^{2}x + 2\left( 2x^{2} - 2x + 1 \right)\sin x\cos x \right\rbrack = 0$
$\Rightarrow \ \ (4x - 2)\sin^{2}x + 2\left( 2x^{2} - 2x + 1 \right)\sin x\cos x = 0$
$\Rightarrow \ \ 2\sin x\left\lbrack (2x - 1)\sin x + \left( 2x^{2} - 2x + 1 \right)\cos x \right\rbrack = 0$
$\Rightarrow \ \ \ \sin x = 0$
$\therefore\ \ y$ is minimum for $\sin x = 0$
Thus, minimum value of
$y = e^{\left( 2x^{2} - 2x + 1 \right)(0)} = e^{0} = 1$
Alternate
Let $f(x) = e^{\left( 2x^{2} - 2x + 1 \right)\sin^{2}x}$
Since $\sin^{2}x \geq 0$
And $2x^{2} - 2x + 1 \geq \frac{4(2)(1) - 2^{2}}{4(2)}$ $\left( \because\ \ minima = \frac{4ac - b^{2}}{4a} \right)$
$\Rightarrow \ \ \ \ 2x^{2} - 2x + 1 \geq \frac{1}{2}$
$\therefore\ \ \ \left( 2x^{2} - 2x + 1 \right)\sin^{2}x \geq 0$
$\therefore\ \ \ f(x)_{\min} = e^{0} = 1$