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ExamsJEE AdvancedMaths

Let A be the set of all real solutions of x(x² + 3|x| + 5|x - 1| + |6x - 2|) = 0, and B be the set of all real solutions of x² - |x| - 12 = 0. How many subsets does the set A x B have?

  1. 2
  2. 4
  3. 8
  4. 16

Correct answer: 8

Solution

For set A: x(x² + 3|x| + 5|x-1| + |6x-2|) = 0. Either x = 0, or the bracket = 0. For x > 0: bracket = x² + 3x + 5(x-1) + (6x-2) = x² + 14x - 7 for x > 1/3. Discriminant = 196 + 28 = 224 > 0 but both roots are irrational and checking signs shows no positive real root makes bracket zero... Actually let me recheck. For x > 1: bracket = x² + 3x + 5(x-1) + (6x-2) = x² + 14x - 7. Setting = 0: x = (-14 + sqrt(224))/2 which is negative, so no positive root > 1. For 0 < x < 1/3: bracket = x² + 3x + 5(1-x) + (2-6x) = x² - 8x + 7. At x approaching 0+: bracket approaches 7 > 0. For 1/3 < x < 1: bracket = x² + 3x + 5(1-x) + (6x-2) = x² + 4x + 3 > 0. So bracket > 0 for all x > 0. By symmetry or direct check, bracket > 0 for x < 0 too. Hence A = {0}. For B: |x|² - |x| - 12 = 0, so t² - t - 12 = 0, (t-4)(t+3) = 0, t = 4 (since t >= 0), so |x| = 4, B = {-4, 4}. A x B = {(0, -4), (0, 4)}, which has 2 elements. Number of subsets = 2² = 4.

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