JEE Advanced 2017 · Paper 2 · Mathematics Q4

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MathematicsSingle correct+3 / -1Question 40 of 54

If $f: \mathbb{R} \to \mathbb{R}$ is a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbb{R}$ , and $f\left(\frac{1}{2}\right) = \frac{1}{2}$ , $f(1) = 1$ , then

f'(1) \le 0

0 < f'(1) \le \frac{1}{2}

\frac{1}{2} < f'(1) \le 1

f'(1) > 1

Answer and worked solution

Official answer: D

Mean value theorem. On $[\tfrac12,1]$ there is $c\in(\tfrac12,1)$ with
$f'(c)=\frac{1-\frac12}{1-\frac12}=1.$

Convexity. $f''>0$ , so $f'$ is strictly increasing. Since $1>c$ , $f'(1)>f'(c)=1$ .

(D)

Topics:Rolle's and mean value theorems Monotonicity

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JEE Advanced 2017 · Paper 2, Mathematics question 4