Question

Let $f$, $g$ and $h$ be real-valued functions defined on the interval $\left[0,1\right]$ by $f\left(x\right)=e^{x^2}+e^{-x^2}$ and $g\left(x\right)=x{e}^{x^2}+e^{-x^2}$ 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$\left[0,1\right]$ then

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

Answer 1

The correct option is D

$a=b=c$

Determine the relation between $a$, $b$ and $c$.

Step 1: From the equation $f\left(x\right)=e^{x^2}+e^{-x^2}$

Consider the given equation as:

$$\begin{gathered} f\left(x\right)=e^{x^2}+e^{-x^2} \\ f\text{'}\left(x\right)=e^{x^2}.2x+e^{-x^2}.\left(-2x\right) \\ f\text{'}\left(x\right)=2x\left(e^{x^2}-e^{-x^2}\right) \end{gathered}$$

From the above Equation, we get only positive values, so that $f\text{'}\left(x\right)\ge 0$ then the function $f\left(x\right)$ is increasing function than the $0\le x\le 1$

Considering the maximum value then the function $f\left(x\right)$ becomes: $f\left(1\right)=e+\frac{1}{e}$

Step 2: From the equation $g\left(x\right)=x{e}^{x^2}+e^{-x^2}$

Consider the given equation as:

$$\begin{gathered} g\left(x\right)=x{e}^{x^2}+e^{-x^2} \\ g\text{'}\left(x\right)=e^{x^2}+x{e}^{x^2}.2x+e^{-x^2}.\left(-2x\right) \\ g\text{'}\left(x\right)=e^{x^2}+2x\left(x{e}^{x^2}-e^{-x^2}\right) \end{gathered}$$

From the above Equation, we get only positive values, so that $g\text{'}\left(x\right)\ge 0$ then the function $g\left(x\right)$ is increasing function than the $0\le x\le 1$

Considering the maximum value then the function $g\left(x\right)$ becomes: $g\left(1\right)=e+\frac{1}{e}$

Step 3: From the equation $h\left(x\right)=x^2{e}^{x^2}+e^{-x^2}$

Consider the given equation as

$$\begin{gathered} h\left(x\right)=x^2{e}^{x^2}+e^{-x^2} \\ h\text{'}\left(x\right)=x^2{e}^{x^2}.2x+2x.e^{x^2}+e^{-x^2}.\left(-2x\right) \\ h\text{'}\left(x\right)=2x\left(x^2{e}^{x^2}+e^{x^2}-e^{-x^2}\right) \end{gathered}$$

From the above Equation, we get only positive values, so that $h\text{'}\left(x\right)\ge 0$ then the function $h\left(x\right)$ is increasing function than the $0\le x\le 1$

Considering the maximum value then the function $h\left(x\right)$ becomes: $h\left(1\right)=e+\frac{1}{e}$

Since, $a$, $b$ and $c$ denote respectively the absolute maximum of $f$, $g$ and $h$ on $\left[0,1\right]$ then $a=b=c$

Hence, the correct answer is Option D.