The function f(x)=2 (x-2)-x^2+4x+1 increases on the interval: 1. (1, 2) 2. (2, 3) 3. (1, 3) 4. (2, 4)

2. (1, 3) Solution: Given, f(x)=2 (x-2)-x^2+4x+1 Domain of f(x) is (2, ). f'(x)=2 x-2 -2x+4 = 2-2x^2+4x+4x-8 x-2 = -2x^2+8x-6 x-2 = -2(x^2-4x+3) x-2 For f(x) to be increasing, we must have f'(x)>0 -2(x^2-4x+3) x-2 >0 ^2-4x+3 0 &-2<0] (x-1)(x-3)<0 1<x<3 (1,3) Also, the domain of f(x) is (2, ). (1,3) (2, ) (1,3)

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

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Question

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

(1, 2)
(2, 3)
(1, 3)
(2, 4)

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Answer

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

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