Question

The minimum value of ${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$ is
A $e$
B $\dfrac{1}{e}$
C 0
D 1

Answer 1

Hint: In this problem, first we need to find the minimum values of the functions $2{x^2} - 2x + 1$ and ${\sin ^2}x$. Next, substitute the obtained values in the given expression to find the minimum value.

Complete step-by-step answer:
The given expression is ${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$.
The minimum value of the function ${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$ is obtained by calculating the minimum values of function $2{x^2} - 2x + 1$ and ${\sin ^2}x$, and then substitute the obtained minimum values into expression ${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$.
Now, consider the function $2{x^2} - 2x + 1$ as ${y_1}$.
${y_1} = 2{x^2} - 2x + 1$
Calculate the first derivative of the above function as substitute it equal to 0 to obtain the critical values.

\,\,\,\,\,\,{{y'}_1} = 4x - 2 \\\ \Rightarrow 0 = 4x - 2 \\\ \Rightarrow 4x = 2 \\\ \Rightarrow x = \dfrac{1}{2} \\\ \end{gathered}$$ Further, calculate the second derivative of the function $${y_1} = 2{x^2} - 2x + 1$$. $${y''_1} = 4\left( { + ve} \right)$$ Since, the second derivative is positive, $$x = \dfrac{1}{2}$$ is a point of minima. The minimum value of the function $${y_1} = 2{x^2} - 2x + 1$$ is calculated as follows: $$\begin{gathered} \,\,\,\,\,\,{y_1} = 2{\left( {\dfrac{1}{2}} \right)^2} - 2\left( {\dfrac{1}{2}} \right) + 1 \\\ \Rightarrow {y_1} = \dfrac{1}{2} - 1 + 1 \\\ \Rightarrow {y_1} = \dfrac{1}{2} \\\ \end{gathered}$$ Now, consider the function $${\sin ^2}x$$ as $${y_2}$$. $${y_2} = {\sin ^2}x$$ Since, $${y_2} = {\sin ^2}x$$ is an even function, the minimum value of the function $${y_2} = {\sin ^2}x$$ is 0. Now, substitute $$\dfrac{1}{2}$$ for $$2{x^2} - 2x + 1$$ and 0 for $${\sin ^2}x$$ in function $${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$$to obtain the minimum value as shown below. $$\begin{gathered} \,\,\,\,\,{e^{\left( {\dfrac{1}{2}} \right)\left( 0 \right)}} \\\ \Rightarrow {e^0} \\\ \Rightarrow 1 \\\ \end{gathered}$$ Thus, the minimum value of the function $${e^{\left( {2{x^2} - 2x + 1} \right){{\sin }^2}x}}$$ is 1, hence, option (D) is correct answer. Note: The minimum or maximum values of any function are obtained using the derivative formula. Calculate the first derivative of the given function, and substitute it equal to 0 to obtain the critical point. Next find the second derivative and substitute the obtained critical values. If the second derivative is positive, its point of minima and vice-versa.