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Question
The function f(x) = 2 log (x − 2) − x2 + 4x + 1 increases on the interval
Options
(1, 2)
(2, 3)
(1, 3)
(2, 4)
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Solution
(2, 3)
\[\text { Given: } \hspace{0.167em} f\left( x \right) = 2 \log \left( x - 2 \right) - x^2 + 4x + 1\]
\[\text { Domain of f }\left( x \right) is\left( 2, \infty \right).\]
\[f'\left( x \right) = \frac{2}{x - 2} - 2x + 4\]
\[ = \frac{2 - 2 x^2 + 4x + 4x - 8}{x - 2}\]
\[ = \frac{- 2 x^2 + 8x - 6}{x - 2}\]
\[ = \frac{- 2 \left( x^2 - 4x + 3 \right)}{x - 2}\]
\[\text { For f(x) to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow \frac{- 2 \left( x^2 - 4x + 3 \right)}{x - 2} > 0\]
\[ \Rightarrow x^2 - 4x + 3 + < 0 \left[ \because \left( x - 2 \right) > 0 \text { & }- 2 < 0 \right]\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) < 0\]
\[ \Rightarrow 1 < x < 3\]
\[ \Rightarrow x \in \left( 1, 3 \right)\]
\[\text { Also, the domain of f }\left( x \right)is\left( 2, \infty \right).\]
\[ \Rightarrow x \in \left( 1, 3 \right) \cap \left( 2, \infty \right)\]
\[ \Rightarrow x \in \left( 2, 3 \right)\]
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