STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

$f(x) = \frac{{{x^2}}}{{({x^3} + 200)}}$.....(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.