Question

The real values of $x$ which satisfy $x^2-4x+3\ge 0$ and $x^2-3x-4\le 0$ is

• A. $(-1,1)\cup (3,4)$
• B. $[-1,1]\cap [3,4]$
• C. $[-1,1]\cup [3,4]$
• D. $(-1,1)\cup [3,4]$

Answer 1

Answer: (C)

$[-1,1]\cup [3,4]$

Given: $x^2-4x+3\ge 0$ and $x^2-3x-4\le 0$

Consider, $x^2-4x+3\ge 0$

$\Rightarrow (x-1)(x-3)\ge 0$

$\Rightarrow x\le 1$ or $x\ge 3...\left(1\right)$

and , $x^2-3x-4\le 0$

$\Rightarrow (x+1)(x-4)\le 0$

$\Rightarrow -1\le x\le 4...\left(2\right)$

Combining $\left(1\right)$ and $\left(2\right)$, we get

$x\in [-1,1]\cup [3,4]$

So, $\text{C}$ is the correct option.