Statement-1: The function $x^2\left(e^x+e^{-x}\right)$ is increasing for all $x0$. Statement-2: The functions $x^2 e^x$ and $x^2 e^{-x}$ are increasing for all $x0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b)$.

Question

Statement-1: The function $x^2\left(e^x+e^{-x}\right)$ is increasing for all $x>0$. Statement-2: The functions $x^2 e^x$ and $x^2 e^{-x}$ are increasing for all $x>0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b)$.

Aicharya · Accepted Answer

Let $y=x^2 \cdot e^{-x}$ For increasing function, $ \begin{aligned} & \frac{d y}{d x}>0 \Rightarrow x\left[(2-x) e^{-x} ight]>0 \ & \because x>0, herefore(2-x) e^{-x}>0 \ & \Rightarrow(2-x) \frac{1}{e^x}>0 \ & ext { For } 0 < x < 2,(2-x) < 0 \ & herefore \frac{1}{e^x} < 0, ext { but it is not possible } \end{aligned} $ Hence the statement- 2 is false

Statement-1: The function $x^2\left(e^x+e^{-x}\right)$ is increasing for all $x>0$. Statement-2: The…

Solution