If $f : R\rightarrow R$ is a twice differentiable function such that $f''(x) > 0$ $\forall$ $x\in R$ then which of the following is correct?

Question

If $f : R\rightarrow R$ is a twice differentiable function such that $f''(x) > 0$ $\forall$ $x\in R$ and $f\left(\dfrac{1}{2}\right)=\dfrac{1}{2},f(1)=1$, then which of the following is correct:-

i) $f'(1)\le0$

ii) $\dfrac{1}{2}< f'(1)\le1$

iii) $f'(1)>1$

iv) $0<f'(1)\le\dfrac{1}{2}$

My attempt is as follows:-

$$f''(x)>0$$

Integrating both sides

$$f'(x)>c_1$$

Again integrating on both sides

$$f(x)>c_1x+c_2$$

$$f\left(\dfrac{1}{2}\right)>\dfrac{c_1}{2}+c_2$$ $$1>c_1+2c_2\tag{1}$$

$$f(1)>c_1+c_2$$ $$1>c_1+c_2\tag{2}$$

Subtracting $(2)$ from $(1)$

$$c_2<0\tag{3}$$

Multiplying $(2)$ with $2$ and then subtracting from $(1)$

$$-1>-c_1$$ $$c_1>1\tag{4}$$

As we know $f'(x)>c_1$, so we can say $f'(x)>1$.

So answer should be $f'(1)>1$. But is this correct way of solving this question? Is this the correct method or am I violating something here? I doubt it because this question has been solved using Lagrange's theorem. Please help me in this.

Answer 1

Your argument is not correct. For example the second derivative of $x^{2}$ is greater than $0$ but you cannot say $2x>c_1$ for all real numbers $x$.

By MVT we have $1=\frac {f(1)-f(\frac 1 2)} {1-\frac 1 2}=f'(t)$ for some $t \in (\frac 1 2, 1)$. Since $f'' >0$ implies that $f'$ is strictly increasing we get $1 =f'(t) <f'(1)$. So iii) is true and all other options are false.