Question

The function 'f' is defined by $f\left(x\right)=x^p{(1-x)}^q$ for all $x\in R,$ where p,q are positive integers, has a maximum value, for x equal to

• A. $\frac{pq}{p+q}$
• B. $1$
• C. $\frac{p}{p+q}$
• D. $0$

Answer 1

The correct option is C $\frac{p}{p+q}$

$f\left(x\right)=x^p{(1-x)}^q$
$f^{\prime }\left(x\right)=(1-x{)}^q.p.x^{p-1}-x^p.q.(1-x{)}^{q-1}=(1-x{)}^q{x}^p(\frac{p}{x}-\frac{q}{1-x})$
For maximum value of $f\left(x\right)$
$f^{\prime }\left(x\right)=0$

$\Rightarrow x=\frac{p}{p+q}$
It can be verified by second derivative that at $x=0,1$ $f\left(x\right)$ won't achieve it's maxima.