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 $[0, 1]$ respectively then

a = b

and

c \neq b

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a = c

and

a \neq b

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a \neq b

and

c \neq b

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a = b = c

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Solution

The correct option is D $a = b = c$
$1 ⩾ x ⩾ x^{2} \forall x \in [0, 1]$

$e^{x^{2}} ⩾ x e^{x^{2}} ⩾ x^{2} e^{x^{2}}$ i.e.

$e^{- x^{2}} + e^{x^{2}} ⩾ e^{- x^{2}} + x e^{x^{2}} ⩾ e^{- x^{2}} + x^{2} e^{x^{2}}$
Equality holds when $x = 1$

i.e. $f (x) ⩾ g (x) ⩾ h (x) \forall x ϵ [0, 1]$

Thus, $f (1)$ is the greatest.

Thus, $a = b = c = e + \frac{1}{e}$

Hence, $a = b = c$ .

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Let f, g and h be real-valued functions defined on the interval [0,1] by f(x)=ex2+e−x2, g(x)=xex2+e−x2 and h(x)=x2ex2+e−x2 If a, b and c denote, respectively, the absolute maximum of f, g and h on [0,1] respectively then

The correct option is D a=b=c1⩾x⩾x2∀x∈[0,1]ex2⩾xex2⩾x2ex2 i.e.e−x2+ex2⩾e−x2+xex2⩾e−x2+x2ex2Equality holds when x=1i.e. f(x)⩾g(x)⩾h(x)∀xϵ[0,1]Thus, f(1) is the greatest.Thus, a=b=c=e+1eHence, a=b=c.