Try this problem of TIFR GS-2010 from Real analysis, Differentiantiation and Maxima and Minima.
The maximum value of $f(x)=x^n(1-x)^n$ for natural number $n\geq 1$ and $0\leq x\leq1$
REAL ANALYSIS
MAXIMA AND MINIMA
DIFFERENTIATION
Answer:$\frac{1}{4^n}$
TIFR 2010|PART A |PROBLEM 1
AN INTRODUCTION TO ANALYSIS DIFFERENTIAL CALCULUS PART-I RK GHOSH, KC MAITY
Here first differentiate $f(x)$
Then equate the terms of $f'(x)$ containing $x$ to $0$ and find all possible values of $x$,since your answer is in terms of $n$ no need to perform any kind of operations on $n$
Now equating $x$ we get $x=0,\frac{1}{2},1$
Now put each of these values of $x$ in $f(x)$ and see for which value of $x$ you get the maximum value of $f(x)$
you will get the maximum value of $f(x)$ for $x=\frac{1}{2}$ that is $\frac{1}{4^n}$
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