Question

For the equation $|x^2-2x-3|=b$ which statement or statements are true

• A. For $b<0$ there are no solution
• B. For $b=0$ there are three solution
• C. For $b>0$ and $b<1$ there are four solution
• D. For $b=1$ there are two solution
• E. For $b>1$ there are no solution
• F. None of these

Answer 1

The correct option is A For $b<0$ there are no solution

$\because |x^2-2x-3|=b$
(A) For $b<0$, $|x^2-2x-3|<0$
Impossible, there are no solution.
(B) For $b=0$, $x^2-2x-3=0$
$\therefore x=3,-1$ there are two solutions.
(C) For $0<b<1$
$\therefore 0<|x^2-2x-3|<1$
$\Rightarrow 0<x^2-2x+3<1$ and $-1<x^2-2x+3<0$
$\Rightarrow x\in (1-\sqrt{5},-1)\cup (3,1+\sqrt{5})$ and $x\in (-1,1-\sqrt{3})\cup (1+\sqrt{3},3)$
(D) For $b=1$
$\therefore |x^2-2x-3|=1\Rightarrow x^2-2x-3=\pm 1\Rightarrow x=1\pm \sqrt{5}$ and $x=1\pm \sqrt{3}$
there are four solution
(E) For $b>1$ $|x^2-2x-3|>1$
$\therefore x^2-2x-3>1$ and $x^2-2x-3<-1$
$\Rightarrow {(x-1)}^2>5$ and ${(x-1)}^2<3$
$\therefore x-1>\sqrt{5}$ and $x-1<-\sqrt{5}$ and $1-\sqrt{3}<x1+\sqrt{3}$
$\Rightarrow$ $x\in (-\infty ,1-\sqrt{5})\cup (1+\sqrt{5},\infty )$ and $x\in (1-\sqrt{3},1+\sqrt{3})$