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Solution for The Function F(X) = 2 Log (X − 2) − X2 + 4x + 1 Increases on the Interval (A) (1, 2) (B) (2, 3) (C) (1, 3) (D) (2, 4) - CBSE (Science) Class 12 - Mathematics

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Question

The function f(x) = 2 log (x − 2) − x2 + 4x + 1 increases on the interval
(a) (1, 2)
(b) (2, 3)
(c) (1, 3)
(d) (2, 4)

Solution

(b) (2, 3)

\[\text { Given: } \hspace{0.167em} f\left( x \right) = 2 \log \left( x - 2 \right) - x^2 + 4x + 1\]

\[\text { Domain of f }\left( x \right) is\left( 2, \infty \right).\]

\[f'\left( x \right) = \frac{2}{x - 2} - 2x + 4\]

\[ = \frac{2 - 2 x^2 + 4x + 4x - 8}{x - 2}\]

\[ = \frac{- 2 x^2 + 8x - 6}{x - 2}\]

\[ = \frac{- 2 \left( x^2 - 4x + 3 \right)}{x - 2}\]

\[\text { For f(x) to be increasing, we must have }\]

\[f'\left( x \right) > 0\]

\[ \Rightarrow \frac{- 2 \left( x^2 - 4x + 3 \right)}{x - 2} > 0\]

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

\[ \Rightarrow \left( x - 1 \right)\left( x - 3 \right) < 0\]

\[ \Rightarrow 1 < x < 3\]

\[ \Rightarrow x \in \left( 1, 3 \right)\]

\[\text { Also, the domain of f }\left( x \right)is\left( 2, \infty \right).\]

\[ \Rightarrow x \in \left( 1, 3 \right) \cap \left( 2, \infty \right)\]

\[ \Rightarrow x \in \left( 2, 3 \right)\]

  Is there an error in this question or solution?
Solution for question: The Function F(X) = 2 Log (X − 2) − X2 + 4x + 1 Increases on the Interval (A) (1, 2) (B) (2, 3) (C) (1, 3) (D) (2, 4) concept: Increasing and Decreasing Functions. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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