Question

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

• A. a = v 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

$f\left(x\right)=e^{x^2}+e^{-x^2}\Rightarrow f^{\prime }\left(x\right)=2x(e^{x^2}-e^{-x^2})\ge 0.\forall x\in [0,1]$
$\therefore$ f(x) is an increasing function on [0,1]
Hence $f_{\text{max}}=f\left(1\right)=e+\frac{1}{e}=a$
$g\left(x\right)=x{e}^{x^2}+e^{-x^2}\Rightarrow g^{\prime }\left(x\right)=(2x+1)e^{x^2}-2x{e}^{-x^2}\ge 0,\forall x\in [0,1]$
$\therefore g\left(x\right)$ is an increasing function on [0,1]
$\therefore g_{\text{max}}=g\left(1\right)=e+\frac{1}{e}=bh\left(x\right)=x^2{e}^{x^2}+e^{-x^2}{h}^{\prime }\left(x\right)=2x\left[e^{x^2}\right(1+x^2)-e^{-x^2}]\ge 0,\forall x\in [0,1]$
$\therefore$ h(x) is an increasing function on [0,1]
$\therefore h_{max}=h\left(1\right)=e+\frac{1}{e}=c$
Hence a =b = c.