Question

If $a,b,c$ are the integral values of $x(a<b<c)$ satisfying $\sqrt{-x^2+10x-16}<x-2,$ then which of the following statements is (are) FALSE?

• A. The minimum value of $|x-a|+|x-b|+|x-c|$ is $2$
• B. The quadratic equation whose roots are $a$ and $b$ is $x^2-15x+56=0$
• C. $2a+3b-4c=0$
• D. $2b=a+c$

Answer 1

Answer: (B), (C)

$2a+3b-4c=0$

$\sqrt{-x^2+10x-16}<x-2$ is meaningful when
$-x^2+10x-16\ge 0$
$\Rightarrow x^2-10x+16\le 0\Rightarrow (x-2)(x-8)\le 0\Rightarrow x\in [2,8]\cdots \left(1\right)$

Now, $\sqrt{-x^2+10x-16}<x-2$
$\Rightarrow -x^2+10x-16<(x-2{)}^2\Rightarrow 2x^2-14x+20>0\Rightarrow x^2-7x+10>0$
$\Rightarrow (x-2)(x-5)>0$
$\Rightarrow x\in (-\infty ,2)\cup (5,\infty )\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$x\in (5,8]$
$\therefore a=6,b=7,c=8$

The minimum value of $|x-6|+|x-7|+|x-8|$ occurs at $x=\text{median}\{6,7,8\}$
i.e., the minimum value of $|x-6|+|x-7|+|x-8|$ occurs at $x=7$ and the minimum value is $2.$

The quadratic equation whose roots are $a$ and $b$ is,
$x^2-(a+b)+ab=0\Rightarrow x^2-13x+42=0$

Also, $2a+3b-4c=12+21-32=1$
and $2b=14=a+c$