Question

Domain of the function $f\left(x\right)=\frac{1}{\sqrt{4x-|x^2-10x+9|}}$, is

• A. $(7-\sqrt{40},7+\sqrt{40})$
• B. $(0,7+\sqrt{40})$
• C. $(7-\sqrt{40},\infty )$
• D. none of these

Answer 1

Answer: (D)

none of these

Here, $f\left(x\right)=\frac{1}{\sqrt{4x-|x^2-10x+9|}}$ would exist, if
$4x-|x^2-10x+9|>0$
ie, $|x^2-10x+9|<4x$,
where $|x^2-10x+9|=\{\begin{array}{cc}{x}^2-10x+9& x\le 1\text{or}x\ge 9\\ -(x^2-10x+9)& 1<x<9\end{array}$
Case I: When $x\le 1$ or $x\ge 9$
$\therefore x^2-10x+9<4x$
$\Rightarrow x^2-14x+9<0\Rightarrow (x-7{)}^2<40$
$\Rightarrow x\in (7-\sqrt{40},7+\sqrt{40})$ (But $x\le 1$ or $x\ge 9$)
$\Rightarrow x\in (7-\sqrt{40},1]\cup [9,7+\sqrt{40})$ .....(i)
Case II: When $1<x<9$
$-x^2+10x-9<4x\Rightarrow x^2-6x+9>0$
$\Rightarrow (x-3{)}^2>0$ which is always true except $x=\left\{3\right\}$
$\therefore x\in (1,9)=\left\{3\right\}$.....(ii)
From Eqs. (i) and (ii), domain of $f\left(x\right)\in (7-\sqrt{40},7+\sqrt{40})-\left\{3\right\}$
Hence, (d) is the correct answer.