Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
38 views
in Mathematics by (39.4k points)
closed ago by

The interval in which the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}, \mathrm{x}>0\) is strictly increasing is

(1) \(\left(0, \frac{1}{\mathrm{e}}\right]\)

(2) \(\left[\frac{1}{\mathrm{e}^{2}}, 1\right)\)

(3) \((0, \infty)\)

(4) \(\left[\frac{1}{\mathrm{e}}, \infty\right)\)

1 Answer

+2 votes
by (42.7k points)
selected ago by
 
Best answer

Correct option is (4) \(\left[\frac{1}{\mathrm{e}}, \infty\right)\)

\(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{x}} ; \mathrm{x}>0\)

lny = xlnx

\(\frac{1}{y} \frac{d y}{d x}=\frac{x}{x}+\ell n x\)

\(\frac{d y}{d x}=x^{x}(1+\ell n x)\)

for strictly increasing

\(\frac{\mathrm{dy}}{\mathrm{dx}} \geq 0 \Rightarrow \mathrm{x}^{\mathrm{x}}(1+\ell \mathrm{n} x) \geq 0\)

\(\Rightarrow \ell \mathrm{nx} \geq-1\)

\(\mathrm{x} \geq \mathrm{e}^{-1}\)

\(\mathrm{x} \geq \frac{1}{\mathrm{e}}\)

\(\mathrm{x} \in\left[\frac{1}{\mathrm{e}}, \infty\right)\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...