IIT-JEE 2010 · Paper 1 · Mathematics Q36

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MathematicsSingle correct+3 / -1Question 36 of 84

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 $h(x)=x^2e^{x^2}+e^{-x^2}$ . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then

a=b

and

c\ne b

a=c

and

a\ne b

a\ne b

and

c\ne b

a=b=c

Answer and worked solution

Official answer: D

Compare on $[0,1]$ . Since $0\le x\le1$ ,
$h(x)=x^2e^{x^2}+e^{-x^2}\le g(x)=xe^{x^2}+e^{-x^2}\le f(x)=e^{x^2}+e^{-x^2}.$

Maximum of $f$ . $f=2\cosh(x^2)$ increases on $[0,1]$ , so $a=f(1)=e+\frac1e$ .

All three reach it. $g(1)=h(1)=e+\frac1e$ as well, and none can exceed $f$ 's maximum. So $a=b=c=e+\frac1e$ .

(D)

Topics:Maxima and minima

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IIT-JEE 2010 · Paper 1, Mathematics question 36