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 the absolute maximum of f, g and h on $[0, 1]$ then

a = b

and

b \neq c

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c = a

and

a \neq b

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a \neq b

and

b \neq c

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a = b = c

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Solution

The correct option is A $a = b = c$
We have,
$f (x) = e^{x^{2}} + e^{{- x}^{2}} \Rightarrow f^{'} (x) f (x)$
$= 2 x (e^{x^{2}} - e^{{- x}^{2}}) \geq 0 \forall x \in [0, 1]$
Clearly for $0 \leq x \leq 1, f (x) \geq g (x) \geq h (x)$
$∵ f (1) = g (1) = h (1) = e + \frac{1}{e}$
and f(1) is the greatest
$∴ a = b = c = e + \frac{1}{e} \Rightarrow a = b = c$

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Similar questions

Q. Let

f, g

and

h

be real-valued functions defined on the interval

[0, 1]

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}}

a, b

and

c

denote, respectively, the absolute maximum of

f, g

and

h

[0, 1]

respectively then

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