Question

The set of values of ${}^{\prime }{x}^{\prime }$ which satisfies the inequation $\sqrt{x^2-18x+72}<(x-1)$ is

• A. $\varphi$
• B. $[1,2)$
• C. $[12,\infty )$
• D. $(1,2]$

Answer 1

Answer: (C)

$[12,\infty )$

Given inequation is

$\sqrt{x^2-18x+72}<x-1$

For $\sqrt{x^2-18x+72}$ to be real

$x^2-18x+72\ge 0$

$x^2-6x-12x+72\ge 0$

$x(x-6)-12(x-6)\ge 0$

$(x-12)(x-6)\ge 0$
$x\in (-\infty ,6)\cup (12,\infty )$ ..........(1)

Now squaring both sides of the inequation,

$x^2-18x+72<x^2+1-2x$

$16x-71>0$

$x>\frac{71}{16}>6$ ..............(2)

but $\frac{71}{16}<12$

taking common regions

$\therefore xϵ[12,\infty )$