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Solution for The Function F(X) = Xx Decreases on the Interval (A) (0, E) (B) (0, 1) (C) (0, 1/E) (D) None of These - CBSE (Science) Class 12 - Mathematics

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Question

The function f(x) = xx decreases on the interval
(a) (0, e)
(b) (0, 1)
(c) (0, 1/e)
(d) none of these

Solution

(c) (0, 1/e)

\[\text { Given }: \hspace{0.167em} f\left( x \right) = x^x \]

\[\text { Applying log with baseeon both sides, we get }\]

\[\log \left( f\left( x \right) \right) = x \log_e x\]

\[\frac{f'\left( x \right)}{f\left( x \right)} = 1 + \log_e x\]

\[f'\left( x \right) = f\left( x \right)\left( 1 + \log_e x \right) = x^x \left( 1 + \log_e x \right)\]

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

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

\[ \Rightarrow x^x \left( 1 + \log_e x \right) < 0\]

\[\text { Here, logaritmic function is defined for positive values of x } . \]

\[ \Rightarrow x^x > 0\]

\[ \Rightarrow 1 + \log_e x < 0 \left[ \text { Since } x^x > 0, x^x \left( 1 + \log_e x \right) < 0 \Rightarrow 1 + \log_e x < 0 \right] \]

\[ \Rightarrow \log_e x < - 1\]

\[ \Rightarrow x < e^{- 1} \left[ \because l {og}_a x < N \Rightarrow x < a^N \text { for }a > 1 \right]\]

\[\text { Here }, \]

\[e > 1\]

\[ \Rightarrow \log_e x < - 1 \Rightarrow x < e^{- 1} \]

\[ \Rightarrow x \in \left( 0, e^{- 1} \right)\]

\[\text { So,f(x) is decreasing on }\left( 0, \frac{1}{e} \right).\]

  Is there an error in this question or solution?
Solution The Function F(X) = Xx Decreases on the Interval (A) (0, E) (B) (0, 1) (C) (0, 1/E) (D) None of These Concept: Increasing and Decreasing Functions.
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