Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1567 Upvotes
Verified Answer
The correct answer is:
$[1,3]$
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, $A \cap B=R \cap[1,3]=[1,3]$
$\left(\begin{array}{l}
\because x^{2}-x+2=x^{2}-x+\frac{1}{4}+\frac{7}{4} \\
=\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4} \geq \frac{7}{4}
\end{array}\right)$
and $B=\left\{x: x^{2}-4 x+3 \leq 0\right\}$
$=\{x:(x-1)(x-3) \leq 0)=[1,3]$
Hence, $A \cap B=R \cap[1,3]=[1,3]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.