Question
If f: $$R \to R$$ is a twice differentiable function such that $$f''\left( x \right) > 0$$ for all $$x \in R,$$ and $$f\,f\left( {\frac{1}{2}} \right) = \frac{1}{2},f\left( 1 \right) = 1,$$ then
A.
$$f'\left( 1 \right) \leqslant 0$$
B.
$$0 < f'\left( 1 \right) \leqslant \frac{1}{2}$$
C.
$$\frac{1}{2} < f'\left( 1 \right) \leqslant 1$$
D.
$$f'\left( 1 \right) > 1$$
Answer :
$$f'\left( 1 \right) > 1$$
Solution :
$$\eqalign{
& f''\left( x \right) > 0,\forall x \in R \cr
& f\left( {\frac{1}{2}} \right) = \frac{1}{2},f\left( 1 \right) = 1 \cr} $$
∴ $$f'$$ is an increasing function on $$R.$$
By Lagrange's Mean Value theorem.
$$\eqalign{
& f'\left( \alpha \right) = \frac{{f\left( 1 \right) - f\left( {\frac{1}{2}} \right)}}{{1 - \frac{1}{2}}},\alpha \in \left( {\frac{1}{2},1} \right) \cr
& \Rightarrow f'\left( \alpha \right) = 1\,{\text{for}}\,{\text{some}}\,\alpha \in \left( {\frac{1}{2},1} \right) \cr
& \Rightarrow f'\left( 1 \right) > 1 \cr} $$