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. |
You need to login to perform this action.
You will be redirected in
3 sec