Q62 / 180 · COMEDK 2017 · Mathematics

If $A=\left\{x: x^{2}-x+2>0\right\}$ and $B=\left\{x: x^{2}-4 x+3 \leq 0\right\}$, then $A \cap B$ is

  • Correct answer
    A
    $[1,3]$
  • B
    $(-\infty, \infty)$
  • C
    $(1,3)$
  • D
    $(-\infty, 1) \cup(3, \infty)$

Solution

Here, $A=\left\{x: x^{2}-x+2>0\right\}=R$ $\left(\begin{array}{l} \because x^{2}-x+2=x^{2}-x+\frac{1}{4}+\frac{7}{4} \\ =\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4} \geq \frac{7}{4} \end{array}\right)$ and $B=\left\{x: x^{2}-4 x+3 \leq 0\right\}$ $=\{x:(x-1)(x-3) \leq 0)=[1,3]$ Hence, $A \cap B=R \cap[1,3]=[1,3]$

Hence, the correct answer is option: 'A'.

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