$$
\text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$
$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }$
Then the number of elements in $\mathrm{S} \cap \mathrm{T}$ is :
Solution
<p>$|{x^2}| - 7|x| + 9 \le 0$</p>
<p>$$ \Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]$$</p>
<p>As $x \in Z$</p>
<p>So, x can be $\pm \,2, \pm \,3, \pm \,4, \pm \,5$</p>
<p>Out of these values of x,</p>
<p>$x = 3, - 4, - 5$</p>
<p>satisfy S as well</p>
<p>$n(S \cap T) = 3$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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