Question

If it is given that $f:R\to R$ 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 which of the following is the correct option.
(a) $f'\left( 1 \right)>1$
(b) $f'\left( 1 \right)\le 0$
(c) $\dfrac{1}{2} < f'\left( 1 \right)\le 1$
(d) $0 < f'\left( 1 \right)\le \dfrac{1}{2}$

Answer 1

To solve this problem we will use the concept that our function is continuous as well as differentiable for all$x\in R$ so it is also differentiable and continuous in the interval $\left[ \dfrac{1}{2},1 \right]$, then according to the Lagrange’s mean value theorem there exists a point c such that $f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$, where b = 1 and a = $\dfrac{1}{2}$.

Complete step-by-step answer :
It is given that function f is twice differentiable for all $x\in R$, so obviously it is also continuous as well as differentiable in the interval $\left[ \dfrac{1}{2},1 \right]$.
Now we know that Lagrange’s mean value theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point,
x = c in this interval, such that
$f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$,
Now we can apply this theorem in the interval $\left[ \dfrac{1}{2},1 \right]$ because function f satisfy all the required conditions, so applying the theorem we get,
$f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$
Where b = 1 and a = $\dfrac{1}{2}$,
$f'\left( c \right)=\dfrac{f\left( 1 \right)-f\left( \dfrac{1}{2} \right)}{1-\dfrac{1}{2}}$
We are given that $f\left( \dfrac{1}{2} \right)=\dfrac{1}{2},f\left( 1 \right)=1$, so putting this in above equation we get
$f'\left( c \right)=\dfrac{1-\dfrac{1}{2}}{1-\dfrac{1}{2}}=1\,\,\,....\left( 1 \right)$
It Is also given to us that $f''\left( x \right)>0$, hence $f'\left( x \right)$ is an increasing function,
And we know that $\dfrac{1}{2} < c <1$,
Since $f\prime (x)~$ is increasing and c < 1 .
Now we if apply function f’ on both sides of the equation c < 1, we get
$f'\left( c \right) < f'\left( 1 \right)$
Therefore from equation 1, we get
$f'\left( 1 \right) > 1$
Our answer matches with the option (a) hence it is the correct answer.

Note : In the question only differentiability of the function is given not continuity so remember that if a function is continuous at some points then it may or may not differentiable at those points but if a function is differentiable at some points then we can say with certainty that it has to be continuous at those points.