Question

If x and y are +ve and x + y = 1 then the minimum value of x log x + y log y is

• A. log 2
• B. -log 2
• C. 2 log 2
• D. 0

Answer 1

The correct option is B -log 2

$f\left(x\right)=x\log x+(1-x)log(1-x)$
$f^{\prime }\left(x\right)=\log x+1+\frac{1}{1-x}-log(1-x)-\frac{x}{1-x}$
$=2+\log x-\log (1-x)$
$e^{-2}=\frac{x}{1-x}$
$e^{-2}=(1+e^{-2})x$
$x=\frac{e^{-2}}{1+e^{-2}}$
$y=1-\frac{e^{-2}}{1+e^{-2}}$
$y=\frac{1}{e^{-2}+1}$
$=\frac{e^{-2}}{1+e^{-2}}(\log e^{-2}-\log (1+e^{-2}\left)\right)+\frac{1}{e^{-2}+1}(-\log (1+e^{-2}\left)\right)$
$=-2\frac{e^{-2}}{1+e^{-2}}-\log (1+e^{-2})$