Verified Solution • 2025–26 Session moderate

QJEEM22Ma3823: JEE Main 2022 Mathematics - Class 11 JEEM22Ma3823 Solution: Then the number of elements in is :

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

T={x∈Z:x2−7∣x∣+9≤0}. T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }

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

Multiple Choice Options

  • A 7
  • B 5
  • C 4
  • D 3

Step-by-Step Verified Solution

∣x2∣−7∣x∣+9≤0|{x^2}| - 7|x| + 9 \le 0

⇒∣x∣∈[7−132,7+132]\Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]

As x∈Zx \in Z

So, x can be ± 2,± 3,± 4,± 5\pm \,2, \pm \,3, \pm \,4, \pm \,5

Out of these values of x,

x=3,−4,−5x = 3, - 4, - 5

satisfy S as well

n(S∩T)=3n(S \cap T) = 3

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