Question
Let $$f,g$$ and $$h$$ be real-valued functions defined on the interval $$\left[ {0,1} \right]$$ by $$f\left( x \right) = {e^{{x^2}}} + {e^{ - x^2}},g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}}$$ and $$h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}}.$$ If $$a,b$$ and $$c$$ denote, respectively, the absolute maximum of $$f,g$$ and $$h$$ on $$\left[ {0,1} \right],$$ then
A.
$$a = b\,{\text{and}}\,c \ne b$$
B.
$$a = c\,{\text{and}}\,a \ne b$$
C.
$$a \ne b\,{\text{and}}\,c \ne b$$
D.
$$a = b\,{\text{ = }}\,c$$
Answer :
$$a = b\,{\text{ = }}\,c$$
Solution :
$$f\left( x \right) = {e^{{x^2}}} + {e^{ - {x^2}}} \Rightarrow f'\left( x \right) = 2x\left( {{e^{{x^2}}} - {e^{ - {x^2}}}} \right) \geqslant 0,\forall {\text{x}} \in \left[ {0,1} \right]$$
$$\therefore f\left( x \right)$$ is an increasing function on $$\left[ {0,1} \right]$$
$$\eqalign{
& {\text{Hence}}\,{f_{\max }} = f\left( 1 \right) = e + \frac{l}{e} = a \cr
& g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}} \cr
& \Rightarrow g'\left( x \right) = \left( {2{x^2} + 1} \right){e^{{x^2}}} - 2x{e^{ - {x^2}}} \geqslant 0,\forall x \in \left[ {0,1} \right] \cr} $$
$$\therefore g\left( x \right)$$ is an increasing function on $$\left[ {0,1} \right]$$
$$\eqalign{
& \therefore {g_{\max }} = g\left( 1 \right) = e + \frac{1}{e} = b \cr
& h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}} \cr
& h'\left( x \right) = 2x\left[ {{e^{{x^2}}}\left( {1 + {x^2}} \right) - {e^{ - {x^2}}}} \right] \geqslant 0,\forall x \in \left[ {0,1} \right] \cr} $$
$$\therefore h\left( x \right)$$ is an increasing function on $$\left[ {0,1} \right]$$
$$\therefore {h_{\max }} = h\left( 1 \right) = e + \frac{1}{e} = c$$
Hence $$a = b = c$$