KVPY Sample Paper KVPY Stream-SX Model Paper-5

The positive, integer k for which \[\frac{{{k}^{2}}}{3{{k}^{3}}+500}\] is a maximum is

A) 5

B) 6

C) 7

D) 8

Correct Answer: C

Solution :

[c]

Let \[f(x)=\frac{{{x}^{2}}}{3{{x}^{3}}+500}\] and \[g(x)=3x+\frac{500}{{{x}^{2}}}\]

If \[g(x)\]is minimum, then \[f(x)\]is maximum \[g'(x)=3-\frac{1000}{{{x}^{3}}}\]\[\Rightarrow \]\[g''(x)\frac{3000}{{{x}^{4}}}>0\]

Minimum value of \[g(x)\]at \[x={{\left( \frac{1000}{3} \right)}^{1/3}}\]\[={{(333.33)}^{1/3}}\]

Where \[6<x<7\]

So greatest term of the sequence \[\left( \frac{{{k}^{2}}}{3{{k}^{3}}+500} \right)\] is either \[k=6\] or \[k=7\]

\[\therefore \]\[f(6)=\frac{36}{1148}\]

\[f(7)=\frac{49}{1529}\]

\[\therefore \]\[f(7)>f(6)\]

\[\therefore \] A \[k=7\]is maximum.