JEE Advanced 2013 · Paper 2 · Mathematics Q53

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MathematicsSingle correct+3 / -1Question 53 of 60

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Paragraph for Questions 53 and 54

Let $f:[0,1]\to\mathbb R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies $f''(x)-2f'(x)+f(x)\ge e^x$ , $x\in[0,1]$ .

Which of the following is true for $0<x<1$ ?

0<f(x)<\infty

-\dfrac12<f(x)<\dfrac12

-\dfrac14<f(x)<1

-\infty<f(x)<0

Answer and worked solution

Official answer: D

Let $g(x)=e^{-x}f(x)$ . Then $g''(x)=e^{-x}\left(f''-2f'+f\right)\ge1>0$ , so $g$ is convex with $g(0)=g(1)=0$ .

So $g<0$ on $(0,1)$ , hence $f(x)<0$ there.

(D)

Topics:Monotonicity Higher-order derivatives

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JEE Advanced 2013 · Paper 2, Mathematics question 53