Question
The function $$f\left( x \right) = 2\,\log \left( {x - 2} \right) - {x^2} + 4x + 1$$ increases on the interval :
A.
$$\left( {1,\,2} \right)$$
B.
$$\left( {2,\,3} \right)$$
C.
$$\left( {\frac{1}{2},\,3} \right)$$
D.
$$\left( {2,\,4} \right)$$
Answer :
$$\left( {2,\,3} \right)$$
Solution :
$$\eqalign{
& f\left( x \right) = 2\,\log \left( {x - 2} \right) - {x^2} + 4x + 1 \cr
& \Rightarrow f'\left( x \right) = \frac{2}{{x - 2}} - 2x + 4 \cr
& \Rightarrow f'\left( x \right) = 2\left[ {\frac{{1 - {{\left( {x - 2} \right)}^2}}}{{x - 2}}} \right] \cr
& \Rightarrow f'\left( x \right) = - 2\frac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{x - 2}} \cr
& \Rightarrow f'\left( x \right) = - \frac{{2\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} \cr
& \therefore \,f'\left( x \right) > 0 \cr
& \Rightarrow - 2\left( {x - 1} \right)\left( {x - 3} \right)\left( {x - 2} \right) > 0 \cr
& \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) < 0 \cr
& \Rightarrow x\, \in \left( { - \infty ,\,1} \right) \cup \left( {2,\,3} \right) \cr} $$
Thus, $$f\left( x \right)$$ is increasing on $$\left( { - \infty ,\,1} \right) \cup \left( {2,\,3} \right).$$