Question

If$f\left(x\right)={\int }_0^e{e}^{\frac{-x^2}{2}}(1-t^2)dt$, minimum at $x=?$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is B

$-1$

Explanation for the correct option :-

Step 1: Find the value of $x$:

Given,

$$\begin{gathered} f\left(x\right)={\int }_0^e{e}^{\frac{-x^2}{2}}(1-t^2)dt \\ f\text{'}\left(x\right)=e^{\frac{-x^2}{2}}(1–x^2) \end{gathered}$$

For minima,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \Rightarrow (1–x^2)=0 \\ \Rightarrow x=\pm 1 \end{gathered}$$

Step 2: Find the double derivative:

Differentiate $f\text{'}\left(x\right)$ with respect to $x$

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=e^{\frac{-x^2}{2}}(-2x)+(1–x^2)e^{\frac{-x^2}{2}}\left(\frac{-2x}{2}\right) \\ =-e^{\frac{-x^2}{2}}(2x+x{(1–x}^2) \end{gathered}$$

Step 2: Find the point of minima:

Now, check minima

At $x=1$

$f\text{'}\text{'}\left(1\right)<0$, so $x=1$ is the point of maxima.

At $x=-1$

$f\text{'}\text{'}\left(1\right)>0,$ so $x=-1$ is the point of minima.

Hence, the correct option is B.