Question

Find the domain of $f\left(x\right)=\sqrt{{\log }_{1/2}\left(\frac{5x-x^2}{4}\right)}$

• A. $(0,1]\cup [4,5)$
• B. $(0,2]\cup [4,5)$
• C. $(0,1]\cup [3,5)$
• D. $(0,3]\cup [4,5)$

Answer 1

The correct option is A $(0,1]\cup [4,5)$

$lo{g}_{1/2}\left(\frac{5x-x^2}{4}\right)$
$=\frac{ln\left(\frac{5x-x^2}{4}\right)}{-ln2}$
$=-\frac{ln\left(\frac{5x-x^2}{4}\right)}{ln2}$
Therefore for real value of f(x),
$0<\left(\frac{5x-x^2}{4}\right)<1$
Or
$0<5x-x^2<4$
Now
$5x-x^2>0$
or
$x(5-x)>0$
$x>0$ and $x<5$
$xϵ(0,5)$ ...(i)
And
$5x-x^2<4$ implies
$x^2-5x+4>0$
$(x-4)(x-1)>0$
$x>4$ and $x<1$
Hence
$xϵ(-\infty ,-1)\cup (4,\infty )$ ...(ii)
From i and ii
$xϵ(0,1)\cup (4,5)$.