MCQ
The largest term in the sequence ${a_n} = {{{n^2}} \over {{n^3} + 200}}$ is given by
- A${{529} \over {49}}$
- B${8 \over {89}}$
- ✓${{49} \over {543}}$
- DNone of these
$f(x) = \frac{{{x^2}}}{{({x^3} + 200)}}$.....(i)
$f'(x) = x\frac{{(400 - {x^3})}}{{{{({x^3} + 200)}^2}}} = 0$
When $x = {(400)^{1/3}}$ $( \because x \ne 0)$
$x = {(400)^{1/3}} - h \Rightarrow f'(x) > 0$
$x = {(400)^{1/3}} + h \Rightarrow f'(x) < 0$
$\therefore $$f(x)$ has maxima at $x = {(400)^{1/3}}$
Since $7 < {(400)^{1/3}} < 8,$ either ${a_7}$ or ${a_8}$ is the greatest term of the sequence.
$ \because {a_7} = \frac{{49}}{{543}}$ and ${a_8} = \frac{8}{{89}}$ and $\frac{{49}}{{543}} > \frac{8}{{89}}$
$\therefore $ ${a_7} = \frac{{49}}{{543}}$ is the greatest term.
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$\frac{{dy}}{{dx}} + \frac{1}{x}\sin 2y = {x^3}\,{\cos ^2}\,y$ represented by family of curves which is is givey by
| $X$ | $0$ | $2$ | $4$ | $6$ | $8$ |
| $P(X)$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
is $ \frac{46}{9}$ , then the variance of the distribution is