Let A = {x \in (0, \pi) - { \pi } : log (2/\pi) | sin x|+ log (2/\pi) | cos x| = 2} and 2 B = {x \ge 0 : \sqrtx(\sqrtx - 4) - 3|\sqrtx - 2| + 6 = 0} . Then n(A \cup B) is equal to :
Step-by-Step Solution
Key Concept: Apply the core result for equivalence relations and set operations and simplify using the given constraints.
A : log 2\pi | sin x| + log 2\pi | cos x| = 2 (2) \Rightarrow log 2\pi (| sin x ⋅ cos x|) = 2 8 \Rightarrow | sin 2x| = 2 \pi Number of solution 4 B : let \sqrtx = t < 2 Then \sqrtx(\sqrtx - 4) + 3(\sqrtx - 2) + 6 = 0 2 \Rightarrow t - 4t + 3t - 6 + 6 = 0 2 \Rightarrow t - t = 0, t = 0, t = 1 x = 0, x = 1 again let \sqrtx = t > 2 then t - 4t - 3t + 6 + 6 = 0 2 2 \Rightarrow t - 7t + 12 = 0 \Rightarrow t = 3, 4 x = 9, 16 Total number of solutions n( A \cup B) = 4 + 4 = 8
Correct Answer: 2