Question

If the domain of the function
$f\left(x\right)={\left[lo{g}_{10}\left(\frac{5x-x^2}{4}\right)\right]}^{-1/2}$ is $a\le x\le b$, then $a.b$ is

Answer 1

Answer: 4

We have, $f\left(x\right)={\left[lo{g}_{10}\left(\frac{5x-x^2}{4}\right)\right]}^{-1/2}....\left(i\right)$
From (i), clearly f(x) is defined for those values of $x$ for which
$lo{g}_{10}\left[\frac{5x-x^2}{4}\right]\ge 0$
$\Rightarrow \left(\frac{5x-x^2}{4}\right)\ge 10^0$
$\Rightarrow \left(\frac{5x-x^2}{4}\right)\ge 1$
$\Rightarrow x^2-5x+4\le 0$
$\Rightarrow \left(x-1\right)\left(x-4\right)\le 0$
Hence, domain of the function is $\left[1,4\right]ie.1\le x\le 4$