Question

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

• A. e
• B. 1/e
• C. 1
• D. 0

Answer 1

The correct option is C 1

$\text{Let}f\left(x\right)=e^{(2x^2-2x+1){\sin }^{2}x}$

$\text{Since,}{e}^x\text{is an increasing function.}$

Let $t=(2x^2-2x+1){\sin }^{2}x$

$\frac{dt}{dx}=(2x^2-2x+1)\left(2\sin x\cos x\right)+{\sin }^{2}x(4x-2)$

$\Rightarrow \frac{dt}{dx}=0=4x^2\sin x\cos x-4x\sin x\cos x$

$+2\sin x\cos x+4x{\sin }^{2}x-2{\sin }^{2}x$

$\Rightarrow \sin x[4x^2\cos x-4x\cos x+2\cos x+$

$4x\sin x-2\sin x]=0$

$\Rightarrow \mathrm{sinn}[2\cos n(2x^2-2x+1)+2\sin x(2x-1)=0$

Now,

$2\cos x(2x^2-2x+1)+2\sin x(2x-1)\ne 0$

$\sin x=0$

$x=n\pi$

$\text{50; minimum value of}t=0$

Hence,

$\begin{array}{cc}\text{minimum ralue of}f\left(x\right)& =e^0\\ & =1\end{array}$