JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 1)
$$
\text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$
$$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }
$$
Then the number of elements in $$\mathrm{S} \cap \mathrm{T}$$ is :
7
5
4
3
Explanation
$$|{x^2}| - 7|x| + 9 \le 0$$
$$ \Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]$$
As $$x \in Z$$
So, x can be $$ \pm \,2, \pm \,3, \pm \,4, \pm \,5$$
Out of these values of x,
$$x = 3, - 4, - 5$$
satisfy S as well
$$n(S \cap T) = 3$$
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