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Limits Continuity and Differentiability question

2017 · Shift 2 · Q19

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JEE Advanced
Mathematics
Limits Continuity and Differentiability
2017 · Shift 2 · Q19

Limits Continuity and Differentiability question

2017 · Shift 2 · Q19

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMCQ+3 / −1

If f : R

\to

R is a twice differentiable function such that f"(x) > 0 for all x

\in

R, and

f\left( {{1 \over 2}} \right) = {1 \over 2}

, f(1) = 1, then

A
f'(1) $\le$ 0
B
f'(1) > 1
C
0 < f'(1) $\le {1 \over 2}$
D
${1 \over 2}$ < f'(1) $\le$ 1

View written solutionFree

Correct answer: B

Use the condition $f''(x)>0$

Since $f''(x)>0$ for all $x\in \mathbb{R}$ , the function $f$ is strictly convex on $\mathbb{R}$ .

Equivalently, its derivative $f'(x)$ is strictly increasing.

Apply the Mean Value Theorem on $\left[\tfrac12,1\right]$

Given:

f\left(\frac12\right)=\frac12,\qquad f(1)=1

So the slope of the secant line from $x=\frac12$ to $x=1$ is

\frac{f(1)-f\left(\frac12\right)}{1-\frac12} =\frac{1-\frac12}{\frac12} =1.

By the Mean Value Theorem, there exists some $c\in\left(\frac12,1\right)$ such that

f'(c)=1.

Use the fact that $f'$ is strictly increasing

Since $c<1$ and $f'$ is strictly increasing,

f'(1)>f'(c)=1.

Therefore,

f'(1)>1.

Check the options

A: $f'(1)\le 0$
False, since $f'(1)>1$ .
B: $f'(1)>1$
True.
C: $0<f'(1)\le \frac12$
False.
D: $\frac12<f'(1)\le 1$
False, because actually $f'(1)>1$ .

Final answer

The correct option is

\boxed{\text{B}}

This matches the stored correct answer.

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