Question

The function f(x) = 2 log (x − 2) − x² + 4x + 1 increases on the interval
(a) (1, 2)
(b) (2, 3)
(c) (1, 3)
(d) (2, 4)

Answer 1

(b) (2, 3)

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}f\left(x\right)=2\log \left(x-2\right)-x^2+4x+1 \\ \text{Domain of}f\left(x\right)\text{is}\left(2,\infty \right)\text{.} \\ f\text{'}\left(x\right)=\frac{2}{x-2}-2x+4 \\ =\frac{2-2x^2+4x+4x-8}{x-2} \\ =\frac{-2x^2+8x-6}{x-2} \\ =\frac{-2\left(x^2-4x+3\right)}{x-2} \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be increasing, we must have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \frac{-2\left(x^2-4x+3\right)}{x-2}>0 \\ \Rightarrow x^2-4x+3+<0\left[\because \left(x-2\right)>0\&-2<0\right] \\ \Rightarrow \left(x-1\right)\left(x-3\right)<0 \\ \Rightarrow 1<x<3 \\ \Rightarrow x\in \left(1,3\right) \\ \mathrm{Also}\text{, the domain of}\mathrm{f}\left(\mathrm{x}\right)\text{is}\left(2,\infty \right)\text{.} \\ \Rightarrow x\in \left(1,3\right)\cap \left(2,\infty \right) \\ \Rightarrow x\in \left(2,3\right) \end{gathered}$$