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

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Application of Derivatives

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}}

. If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then

Answer: (D)

For

x \in [0, 1]

f (x) = e^{x^{2}} + e^{- x^{2}}

is increasing (since

f^{'} (x) = 2 x e^{x^{2}} - 2 x e^{- x^{2}} = 2 x (e^{x^{2}} - e^{- x^{2}}) \geq 0

for

x \geq 0

).
Maximum at

x = 1

a = e + e^{- 1}

g (x) = x e^{x^{2}} + e^{- x^{2}}

. We have

g^{'} (x) = e^{x^{2}} + 2 x^{2} e^{x^{2}} - 2 x e^{- x^{2}} = e^{x^{2}} (1 + 2 x^{2}) - 2 x e^{- x^{2}} > 0

for

x \in [0, 1]

.
Maximum at

x = 1

b = e + e^{- 1}

h (x) = x^{2} e^{x^{2}} + e^{- x^{2}}

. We have

h^{'} (x) = 2 x e^{x^{2}} + 2 x^{3} e^{x^{2}} - 2 x e^{- x^{2}} = 2 x e^{x^{2}} (1 + x^{2}) - 2 x e^{- x^{2}} > 0

for

x \in (0, 1]

.
Maximum at

x = 1

c = e + e^{- 1}

.

Therefore,

a = b = c = e + e^{- 1}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option