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.
\[\begin{gathered}
  \,\,\,\,\,\,{{y'}_1} = 4x - 2 \\
   \Rightarrow 0 = 4x - 2 \\
   \Rightarrow 4x = 2 \\
   \Rightarrow x = \dfrac{1}{2} \\
\end{gathered}\]
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.