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

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)
✓
(2, 3)
C
(1, 3)
D
(2, 4)

✓

Answer

Correct option: B.

(2, 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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