MCQ
The value of $\int_0^2 {\frac{{{3^{\sqrt x }}}}{{\sqrt x }}} \,dx$ is
- ✓$\frac{2}{{\log 3}}.({3^{\sqrt 2 }} - 1)$
- B$0$
- C$2.\frac{{\sqrt 2 }}{{\log 3}}$
- D$\frac{{{3^{\sqrt 2 }}}}{{\sqrt 2 }}$
Also, as $x = 0 $ to $2$ so, $t = 0$ to $\sqrt 2 $
Therefore, $\int_0^2 {\frac{{{3^{\sqrt x }}}}{{\sqrt x }}\,} dx $
$= 2\int_0^{\sqrt 2 } {{3^t}} dt $
$= 2\left[ {\frac{{{3^t}}}{{\log 3}}} \right]_0^{\sqrt 2 }$
$= \frac{2}{{\log 3}}({3^{\sqrt 2 }} - 1)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $f$ is continuous, then which of the following hold$(s)$ for all $n$ ?
$(A)$ $a_{n-1}-b_{n-1}=0$ $(B)$ $a_n-b_n=1$ $(C)$ $a_n-b_{n+1}=1$ $(D)$ $a_{n-1}-b_n=-1$