Question

If $f,g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f\left(x\right)=x^2{e}^{x^2}+e^{-x^2}$. If $a$, $b$ and $c$ denote respectively, the absolute maximum of $f,g$ and $h$ on $[0,1]$, then

• A. $a=b$ and $c\ne b$
• B. $a=c$ and $a\ne b$
• C. $a\ne b$ and $c\ne b$
• D. $a=b=c$

Answer 1

The correct option is D

$a=b=c$

Explanation for the correct option:

Step 1. Find $f\text{'}\left(x\right),g\left(x\right)$ and $h\left(x\right)$:

$f\left(x\right)=x^2{e^x}^2+{e^{-x}}^2$

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 2x\left({e^x}^2\right)–2x\left({e^{-x}}^2\right)\\ & =& 2x({e^x}^2–e^{-x})>0\end{array}$

$g\left(x\right)=x{e^x}^2+x{e^{-x}}^2\times 2x+{e^{-x}}^2(-2x)>0$

$h\left(x\right)=x{e^x}^2+{e^{-x}}^2$

$h\left(2x\right)=2x{e^x}^2+x^2{e^x}^2\times 2x+{e^{-x}}^2(-2x)>0$

$\therefore$$f\left(x\right),g\left(x\right)$and $h\left(x\right)$ are increasing for $x\in [0,1]$

Step 2. Find the max $f\left(1\right),g\left(1\right)$ and $h\left(1\right)$:

$\begin{array}{rcl}a& =& f\left(1\right)\\ & =& e+\frac{1}{e}\end{array}$

$\begin{array}{rcl}b& =& g\left(1\right)\\ & =& e+\frac{1}{e}\end{array}$

$\begin{array}{rcl}c& =& h\left(1\right)\\ & =& e+\frac{1}{e}\end{array}$

$\mathbf{\therefore }\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}$

Hence, Option ‘D’ is Correct.