JEE Advance - Mathematics (2010 - Paper 1 Offline - No. 24)
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 :
$f(x)=e^{x^2}+e^{-x^2}$,
$g(x)=x e^{x^2}+e^{-x^2}$
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 :
$a=b$ and $c \neq b$
$a=c$ and $a \neq b$
$a \neq b$ and $c \neq b$
$a=b=c$
Explanation
$$ \begin{aligned} & f(x)=e^{x^2}+e^{-x^2} \\\\ & \begin{aligned} f^{\prime}(x) & =e^{x^2} \frac{d}{d x}\left(x^2\right)+e^{-x^2} \frac{d}{d x}\left(-x^2\right) \\\\ & =e^{x^2}(2 x)+e^{-x^2}(-2 x) \\\\ & =2 x\left(e^{x^2}-e^{-x^2}\right) \geq 0 \forall x \in[0,1] \\\\ g(x) & =x e^{x^2}+e^{-x^2} \\\\ h(x) & =x^2 e^{x^2}+e^{-x^2} \end{aligned} \end{aligned} $$
Clearly for $0 \leq x \leq 1 \quad f(x) \geq g(x) \geq h(x)$
$\because f(1)=g(1)=h(1)=e+\frac{1}{e}$ and $f(1)$ is greatest
$$ \because a=b=c=e+\frac{1}{e} $$
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