MCQ
The product $(32)(32)^{1/6}(32)^{1/36} ...... to\,\, \infty $ is
- A$16$
- B$32$
- ✓$64$
- D$0$
= ${(32)^{1 + \frac{1}{6} + \frac{1}{{36}} + .....\infty }}$
$ = {(32)^{\frac{1}{{1 - (1/6)}}}}$
$ = {(32)^{\frac{1}{{5/6}}}} = {(32)^{6/5}}$
$ = {2^6} = 64$.
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$\lim _{x \rightarrow \infty} \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e(1+x)^\beta\right.}=0$
Then which of the following is (are) correct?
$(A)$ $(-1,3) \in S$ $(B)$ $(-1,1) \in S$ $(C)$ $(1,-1) \in S$ $(D)$ $(1,-2) \in S$