MCQ
જો$y = {\log _{10}}{x^2}$, તો ${{dy} \over {dx}} = . . . .$
- A${2 \over x}$
- ✓${2 \over {x{{\log }_e}10}}$
- C${1 \over {x{{\log }_e}10}}$
- D${1 \over {10x}}$
$y = \frac{{{{\log }_e}{x^2}}}{{{{\log }_e}10}}$, $\left( \because {{\log }_{a}}b=\frac{{{\log }_{e}}b}{{{\log }_{e}}a} \right)$
$y = \frac{{2{{\log }_e}x}}{{{{\log }_e}10}}$,
$\therefore \frac{{dy}}{{dx}} = \frac{2}{{x{{\log }_e}10}}$.
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($1$) $P(X>Y)$ is
($A$) $\frac{1}{4}$ ($B$) $\frac{5}{12}$ ($C$) $\frac{1}{2}$ ($D$) $\frac{7}{12}$
($2$) $P(X=Y)$ is
($A$). $\frac{11}{36}$ ($B$) $\frac{1}{3}$ ($C$) $\frac{13}{36}$ ($D$) $\frac{1}{2}$
Given the answer quetion ($1$) and ($2$)