Question

The number of integral values of $x$ satisfying $\sqrt{-x^2+10x-16}<x-2$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

$\sqrt{-x^2+10x-16}<x-2$

$\therefore -x^2+10x-16\ge 0$

$x^2-10x+16\le 0(x-2)(x-8)\le 0$

$x\in [2,8]\dots \left(1\right)$

$\sqrt{-x^2+10x-16}<x-2$

$squaringonbothsides$

$-x^2+10x-16<x^2-4x+4$

$2x^2-14x+20>0$

$x^2-7x+10>0$

$(x-5)(x-2)>0$

$x\in (-\infty ,2)U(5,\infty )\dots \left(2\right)$

Points which lie in both the interval $\left(1\right)$ and $\left(2\right)$ are $6,7and8$

$\therefore x=6,7,8$

$so3valuesofx$