Question

The domain of $\log [1-{\log }_{10}(x^2-5x+16)]$ is

• A. $2<x<3$
• B. $1<x<3$
• C. $-2<x<3$
• D. $-2<x<2$

Answer 1

The correct option is A $2<x<3$

Let $f\left(x\right)=\log [1-{\log }_{10}(x^2-5x+16)]$

For $f$ to be defined, we must have $x^2-5x+16>0$ which is true as
$\Delta =$ -ive and the sign of first term is +ive. Again by definition of $\log x$ we must have
$1-{\log }_{10}(x^2-5x+16)>0$

$\Rightarrow {\log }_{10}(x^2-5x+16)<1={\log }_{10}10$

$\Rightarrow x^2-5x+16<10$

or $x^2-5x+6<0$ or $\Rightarrow (x-2)(x-3)<0$

or $2<x<3$

$\therefore x\in (2,3)$