Single Choice

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 $$[0,1]$$ respectively then

A$$a=b$$ and $$c\neq b$$
B$$a=c$$ and $$a\neq b$$
C$$a\neq b$$ and $$c\neq b$$
D$$a=b=c$$
Correct Answer

Solution

$$1\geqslant x \geqslant x^2 \forall x\in [0,1]$$

$$e^{x^2} \geqslant xe^{x^2} \geqslant x^2e^{x^2}$$ i.e.

$$e^{-x^2}+e^{x^2} \geqslant e^{-x^2}+xe^{x^2} \geqslant e^{-x^2}+x^2e^{x^2}$$
Equality holds when $$x=1$$

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

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

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

Hence, $$a=b=c$$.

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