Question
Let $$f$$ be differentiable for all $$x.$$ If $$f\left( 1 \right) = - 2\,\,\& \,f'\left( x \right) \geqslant 2$$ for $$x \in \left[ {1,\,6} \right],$$ then-
A.
$$f\left( 6 \right) \geqslant 8$$
B.
$$f\left( 6 \right) < 8$$
C.
$$f\left( 6 \right) < 5$$
D.
$$f\left( 6 \right) = 5$$
Answer :
$$f\left( 6 \right) \geqslant 8$$
Solution :
$${\text{As }}f\left( 1 \right) = - 2\,\,\&\,\, f'\left( x \right) \geqslant 2\,\forall \,x \in \left[ {1,\,6} \right]$$
Applying Lagrange’s mean value theorem
$$\eqalign{
& \frac{{f\left( 6 \right) - f\left( 1 \right)}}{5} = f'\left( c \right) \geqslant 2 \cr
& \Rightarrow f\left( 6 \right) \geqslant 10 + f\left( 1 \right) \cr
& \Rightarrow f\left( 6 \right) \geqslant 10 - 2 \cr
& \Rightarrow f\left( 6 \right) \geqslant 8 \cr} $$