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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.

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    $\begingroup$ It is not true that $f''(x)>0$ implies $f'(x)>c$ for all $x$. However, you can integrate from $x_0$ to $x_1$ and conclude that $f'(x_1)-f'(x_0)>0$ for $x_1>x_0$. $\endgroup$ – almagest Dec 17 '19 at 10:38
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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.

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  • $\begingroup$ sorry, but I am asking about the method I used, can you comment upon that? $\endgroup$ – user3290550 Dec 17 '19 at 10:27
  • $\begingroup$ The way you are integrating inequalities is not correct. I have edited my answer. $\endgroup$ – Kavi Rama Murthy Dec 17 '19 at 10:33

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