The function f(x)=2 (x-2)-x^2+4x+1 increases on the interval: - Vidyadip

MCQ

The function $\text{f}(\text{x})=2\log(\text{x}-2)-\text{x}^2+4\text{x}+1$ increases on the interval:

A
$(1, 2)$
B
$(2, 3)$
✓
$(1, 3)$
D
$(2, 4)$

✓

Answer

Correct option: C.

$(1, 3)$

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

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