Question

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

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

Answer 1

Answer: (D)

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

Given: $f$ is twice differentiable function

$f^{\prime \prime }\left(x\right)>0$

$\therefore f^{\prime \prime }\left(x\right)=k>0$ $\forall x\in R$

$\therefore f^{\prime }\left(x\right)=kx+a$ (by integrating)

$\int f^{\prime }\left(x\right)=\int (kx+a)dx$

$f\left(x\right)=k\frac{x^2}{2}+ax+b$

$f\left(\frac{1}{2}\right)=\frac{1}{2}$ , $f\left(1\right)=1$

$\frac{k}{8}+\frac{a}{2}+b=\frac{1}{2}$ .... $\left(1\right)$

$\frac{k}{2}+a+b=1$ ...... $\left(2\right)$

$\left(2\right)-\left(1\right),$ we get

$\Rightarrow \frac{3k}{8}+\frac{a}{2}=\frac{1}{2}$

$3k+4a=4$

$k+a=1+\frac{k}{4}$

$f^{\prime }\left(1\right)=k+a=1+\frac{k}{4}$

$\therefore f^{\prime }\left(1\right)>1$ $(\because k>0)$