2023-M2-09

Furthermore,

$\begin{array}{rcl}
f'(x) & = & (1-2x^2)e^{-x^2} \\
f”(x) & = & (1-2x^2)e^{-x^2}\times(-2x) +e^{-x^2}\times(-4x) \\
f”(x) & = & (-2x+4x^3-4x)e^{-x^2} \\
f”(x) & = & 2x(2x^2-3)e^{-x^2}
\end{array}$

$\begin{array}{rcl}
f'(x) & = & 0 \\
(1-2x^2)e^{-x^2} & = & 0 \\
2x^2 & = & 1 \\
x^2 & = & \dfrac{1}{2} \\
x & = & \pm\dfrac{\sqrt{2}}{2}
\end{array}$

$\begin{array}{|l|c|c|c|c|c|} \hline
x & x< \dfrac{-\sqrt{2}}{2} & x=\dfrac{-\sqrt{2}}{2} & \dfrac{-\sqrt{2}}{2} < x < \dfrac{\sqrt{2}}{2} & x =\dfrac{\sqrt{2}}{2} & x >\dfrac{\sqrt{2}}{2} \\ \hline
f'(x) & -ve & 0 & +ve & 0 & -ve \\ \hline
f(x) & \text{decreasing} & \text{min. pt.} & \text{increasing} & \text{max. pt.} & \text{decreasing} \\ \hline
\end{array}$

Therefore, the maximum point of $G$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}e^\frac{-1}{2}\right)$.

The minimum point of $G$ is $\left(\dfrac{-\sqrt{2}}{2}, \dfrac{-\sqrt{2}}{2}e^\frac{-1}{2}\right)$.

1. Note that $f'(1)=-e^{-1}$. By the point-slope form, the equation of $L$ is

   $\begin{array}{rcl}
   \dfrac{y-e^{-1}}{x-1} & = & -e^{-1} \\
   y-e^{-1} & = & e^{-1}x +e^{-1} \\
   y & = & \dfrac{1}{e} x +\dfrac{2}{e}
   \end{array}$

2. Note that $L$ is the tangent to $G$ at $\left(1,\dfrac{1}{e}\right)$.

   For $x \in (0,1)$, $f”(x)=2x(2x^2-3)e^{-x^2}<0$. Hence, the graph of $y=f(x)$ is concave downwards on $(0,1)$.

   Therefore, $G$ lies below $L$ in the interval $(0,1)$.

3. The required area

   $\begin{array}{cl}
   = & \dint_0^1 \left( \dfrac{-1}{e}x+\dfrac{2}{e}-xe^{-x^2} \right) dx \\
   = & \dint_0^1 \left(\dfrac{-1}{e}x+\dfrac{2}{e}\right)dx -\dint_0^1xe^{-x^2}dx \\
   = & \left[\dfrac{-1}{2e}x^2+\dfrac{2}{e}x\right]_0^1 +\dfrac{1}{2}\dint_0^1 e^{-x^2}d(-x^2) \\
   = & \dfrac{-1}{2e}+\dfrac{2}{e} +\dfrac{1}{2}\left[e^{-x^2}\right]_0^1 \\
   = & \dfrac{3}{2e}+\dfrac{1}{2}(e^{-1}-1) \\
   = & \dfrac{2}{e}-\dfrac{1}{2}
   \end{array}$