Medium MCQ +4 / -1 PYQ · JEE Mains 2022

$$ \text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$

$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }$

Then the number of elements in $\mathrm{S} \cap \mathrm{T}$ is :

  1. A 7
  2. B 5
  3. C 4
  4. D 3 Correct answer

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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