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(D) We are given the functions $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$,and $h(x)=x^2 e^{x^2}+e^{-x^2}$ on the interval $[0,1]$.For $x \in [0

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$a=b=c$

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(D) We are given the functions $f(x)=e^{x^2}+e^{-x^2}$,$g(x)=x e^{x^2}+e^{-x^2}$,and $h(x)=x^2 e^{x^2}+e^{-x^2}$ on the interval $[0,1]$.For $x \in [0,1]$,we have $0 \leq x^2 \leq x \leq 1$.Since $e^{x^2} > 0$ and $e^{-x^2} > 0$,we compare the functions:$f(x) - g(x) = e^{x^2} + e^{-x^2} - x e^{x^2} - e^{-x^2} = e^{x^2}(1-x) \geq 0$ for $x \in [0,1]$. Thus $f(x) \geq g(x)$.$g(x) - h(x) = x e^{x^2} + e^{-x^2} - x^2 e^{x^2} - e^{-x^2} = e^{x^2}(x-x^2) = x e^{x^2}(1-x) \geq 0$ for $x \in [0,1]$. Thus $g(x) \geq h(x)$.Therefore,$f(x) \geq g(x) \geq h(x)$ for all $x \in [0,1]$.At $x=1$,$f(1) = e^1 + e^{-1} = e + \frac{1}{e}$,$g(1) = 1 \cdot e^1 + e^{-1} = e + \frac{1}{e}$,and $h(1) = 1^2 \cdot e^1 + e^{-1} = e + \frac{1}{e}$.Since $f(x)$,$g(x)$,and $h(x)$ are increasing functions on $[0,1]$ (as their derivatives are non-negative),their maximum values occur at $x=1$.Thus,$a = f(1) = e + \frac{1}{e}$,$b = g(1) = e + \frac{1}{e}$,and $c = h(1) = e + \frac{1}{e}$.Hence,$a=b=c$.

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

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