Increasing and Decreasing Functions — Maths STD 12 Science — Question

2. (1, 3)

Solution:

Given, $\text{f}(\text{x})=2\log(\text{x}-2)-\text{x}^2+4\text{x}+1$

Domain of f(x) is $(2,\infty).$

$\text{f}'(\text{x})=\frac{2}{\text{x}-2}-2\text{x}+4$

$=\frac{2-2\text{x}^2+4\text{x}+4\text{x}-8}{\text{x}-2}$

$=\frac{-2\text{x}^2+8\text{x}-6}{\text{x}-2}$

$=\frac{-2(\text{x}^2-4\text{x}+3)}{\text{x}-2}$

For f(x) to be increasing, we must have

$\text{f}'(\text{x})>0$

$\Rightarrow\frac{-2(\text{x}^2-4\text{x}+3)}{\text{x}-2}>0$

$\Rightarrow\text{x}^2-4\text{x}+3<0$ $[\because\ \text{x}(\text{x}-2)>0\ \&-2<0]$

$\Rightarrow(\text{x}-1)(\text{x}-3)<0$

$\Rightarrow1<\text{x}<3$

$\Rightarrow\text{x}\in(1,3)$

Also, the domain of f(x) is $(2,\infty).$

$\Rightarrow\text{x}\in(1,3)\cap(2,\infty)$

$\Rightarrow\text{x}\in(1,3)$