MCQ
વિધેય ${x^x}$ એ . . .. અંતરાલમાં વધતું છે.
- ✓$x > {1 \over e}$
- B$x < {1 \over e}$
- C$x < 0$
- Dદરેક $x$ માટે
For $\frac{{dy}}{{dx}} > 0$;
${x^x}(1 + \log x) > 0$
==> $1 + \log x > 0 \Rightarrow {\log _e}x > {\log _e}\frac{1}{e}$
For this to be positive $ x $ should be greater than $\frac{1}{e}$.
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