A) Increasing for \[x>\,\frac{1}{4}\]and decreasing for \[x<\frac{1}{4}\]
B) Increasing for every value of x
C) Decreasing for every value of x
D) None of these
Correct Answer: A
Solution :
\[f(x)={{x}^{4}}-\frac{{{x}^{3}}}{3}\Rightarrow f'(x)=4{{x}^{3}}-{{x}^{2}}\] For increasing \[4{{x}^{3}}-{{x}^{2}}>0={{x}^{2}}(4x-1)>0\] Therefore, the function is increasing for \[x>\frac{1}{4}\] Similarly decreasing for \[x<\frac{1}{4}\].
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