Question

If $f:R\to R$ is a twice differentiable function such that $f^{{}^{\prime \prime }}\left(x\right)>0$ for all $x\in R$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}$, $f\left(1\right)=1$, then

• A. $f^{{}^{\prime }}\left(1\right)>1$
• B. $f^{{}^{\prime }}\left(1\right)\le 0$
• C. $\frac{1}{2}<f^{{}^{\prime }}\left(1\right)\le 1$
• D. $0<f^{{}^{\prime }}\left(1\right)\le \frac{1}{2}$

Answer 1

Answer: (A)

$f^{{}^{\prime }}\left(1\right)>1$

Since $f^{\prime \prime }\left(x\right)>0\to f^{\prime }\left(x\right)$ is increasing.

Using LMVT, There exists $c$ such that $\frac{1}{2}<c<1$ and

$f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(\frac{1}{2}\right)}{1-\frac{1}{2}}=\frac{1-\frac{1}{2}}{1-\frac{1}{2}}=1$

It is given that $f^{\prime \prime }\left(x\right)>0$, hence $f^{\prime }\left(x\right)$ is an increasing function.

Since $f^{\prime }\left(x\right)$ is increasing and $c<1$

then $f^{\prime }\left(1\right)>f^{\prime }\left(c\right)$

Therefore $f^{\prime }\left(1\right)>1$

Hence, option $A$ is correct.