JEE Advanced 2013 · Paper 2 · Mathematics Q54

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MathematicsSingle correct+3 / -1Question 54 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 $f''(x)-2f'(x)+f(x)\ge e^x$ , $x\in[0,1]$ .

If the function $e^{-x}f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac14$ , which of the following is true?

f'(x)<f(x)

\frac14<x<\frac34

f'(x)>f(x)

0<x<\frac14

f'(x)<f(x)

0<x<\frac14

f'(x)<f(x)

\frac34<x<1

Answer and worked solution

Official answer: C

Minimum of g at ¼. $g$ is convex, so $g'<0$ on $\left(0,\frac14\right)$ and $g'>0$ on $\left(\frac14,1\right)$ .

Since $g'=e^{-x}(f'-f)$ : $f'<f$ on $\left(0,\frac14\right)$ and $f'>f$ on $\left(\frac14,1\right)$ .

(C)

Topics:Monotonicity

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JEE Advanced 2013 · Paper 2, Mathematics question 54