KCET · Maths · Application of Derivatives
The function \(x^x ; x\gt0\) is strictly increasing at
- A \(\forall x \in R\)
- B \(x \lt \frac{1}{e}\)
- C \(x\gt\frac{1}{e}\)
- D \(x \lt 0\)
Answer & Solution
Correct Answer
(C) \(x\gt\frac{1}{e}\)
Step-by-step Solution
Detailed explanation
Let \(y=x^x\)
\(\frac{d y}{d x}=x^x(1+\log x)\)
For increasing function, \(\frac{d y}{d x}\gt0\).
\(\Rightarrow \quad x^x(1+\log x)\gt0\)
\(\Rightarrow \quad 1+\log x\gt0\)
\(\Rightarrow \quad \log _e x\gt\log _e \frac{1}{e}\)
Hence, \(x\gt\frac{1}{e}\)
\(\frac{d y}{d x}=x^x(1+\log x)\)
For increasing function, \(\frac{d y}{d x}\gt0\).
\(\Rightarrow \quad x^x(1+\log x)\gt0\)
\(\Rightarrow \quad 1+\log x\gt0\)
\(\Rightarrow \quad \log _e x\gt\log _e \frac{1}{e}\)
Hence, \(x\gt\frac{1}{e}\)
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