If f: R → R is a twice differentiable function such that f ^ ′′ (x) > 0 for all x ∈ R and f((1)/(2)) = (1)/(2), f(1) = 1 then

Correct answer: C. f ^ ′ (1) > 1 Using LMVT on f(x) for x ∈ [(1)/(2), 1] (f(1) - f((1)/(2)))/(1 - (1)/(2)) = f ^ ' (c), where c ∈ ((1)/(2), 1) (1 - (1)/(2))/((1)/(2)) = f ^ ' (c) ⇒ f ^ ' (c) = 1, where c ∈ ((1)/(2), 1) ∵ f ^ ″ (x) > 0 x ∈ R ∴ f ^ ' (x) is an increasing function x ∈ R ∴ f ^ ′ (1) > 1

JEE Advanced 2017Paper 2

Exam: JEE Advanced
Year: 2017
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2017 · Paper 2 · Mathematics · Application of Derivatives

If f: R → R is a twice differentiable function such that f ^ ′′ (x)...

f: R \to R

is a twice differentiable function such that

{f}^{′′}\left(x\right) > 0

for all

x \in R

and

f\left(\dfrac{1}{2}\right) = \dfrac{1}{2}, f\left(1\right) = 1

then

Options

A $0 < {f}^{′}\left(1\right) \le \dfrac{1}{2}$
B ${f}^{′}\left(1\right) \le 0$
C ${f}^{′}\left(1\right) > 1$
D $\dfrac{1}{2} < {f}^{′}\left(1\right) \le 1$

Correct answer

C. ${f}^{′}\left(1\right) > 1$

Step-by-step solution

Using LMVT on $f\left(x\right)$ for $x \in \left[\dfrac{1}{2}, 1\right]$

$\dfrac{f\left(1\right) - f\left(\dfrac{1}{2}\right)}{1 - \dfrac{1}{2}} = {f}^{'}\left(c\right),$ where $c \in \left(\dfrac{1}{2}, 1\right)$

$\dfrac{1 - \dfrac{1}{2}}{\dfrac{1}{2}} = {f}^{'}\left(c\right) ⇒ {f}^{'}\left(c\right) = 1,$ where $c \in \left(\dfrac{1}{2}, 1\right)$

∵ {\text{f}}^{″} \left(x\right) > 0 \forall x \in \text{R}

$∴ {f}^{'}\left(x\right)$ is an increasing function $\forall x \in R$

∴ {\text{f}}^{′} \left(1\right) > 1

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