-
$\begin{array}{rcl}
y & = & (7x-2x^2)e^{-x} \\
\dfrac{dy}{dx} & = & (7x-2x^2) (-e^{-x}) + e^{-x}(7-4x) \\
\dfrac{dy}{dx} & = &e^{-x}(2x^2-7x-4x+7) \\
\dfrac{dy}{dx} & = & e^{-x}(2x^2-11x+7)
\end{array}$$\begin{array}{rcl}
\dfrac{dy}{dx} & = & e^{-x}(2x^2-11x+7) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(4x-11)+(2x^2-11x+7)(-e^{-x}) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(-2x^2+11x-7+4x-11) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(-2x^2+15x-18) \\
\dfrac{d^2y}{dx^2} & = & -e^{-x}(2x-3)(x-6)
\end{array}$ - For the point of inflexion of the graph,
$\begin{array}{rcl}
\dfrac{d^2y}{dx^2} & = & 0 \\
-e^{-x}(2x-3)(x-6) & = & 0
\end{array}$$\therefore x=\dfrac{3}{2}$ or $x=6$.
$\begin{array}{|l|c|c|c|c|c|} \hline
x & x< \dfrac{3}{2} & x=\dfrac{3}{2} & \dfrac{3}{2} < x < 6 & x=6 & x > 6 \\ \hline
\dfrac{d^2y}{dx^2} & -ve & 0 & +ve & 0 & -ve \\ \hline
y& \text{concave downwards} & \text{pt. of inflexion} & \text{concave upwards} & \text{pt. of inflexion} & \text{concave downwards} \\ \hline
\end{array}$Therefore, there are two points of inflexion of the graph.
Hence, the claim is agreed.
2022-M2-04
Ans: (a) $\dfrac{dy}{dx} = e^{-x}(2x^2-11x+7)$, $\dfrac{d^2y}{dx^2}=e^{-x}(-2x^2+15x-18)$ (b) Yes
