JEE Mains 6-Apr-2023 Shift 2 Solved Paper

(1) Bonus
‌∵z+z=2‌Re(z)‌‌‌ If ‌z=x+iy
‌⇒z+z=2x
‌z2+(z)2=2(x2−y2)
‌(az2+bz)+(az2+bz)=2a . . . (1)
‌(bz2+az)+(bz2+az)=2b . . . (2)
‌‌ add (1) and (2) ‌
‌(a+b)z2+(a+b)z+(a+b)z2+(a+b)z=2(a+b)
‌(a+b)[z2+z+(z)2+z]=2(a+b)
‌‌ sub. (1) and (2) ‌
‌(a−b)[z2−z+z2−z]=2(a−b) . . . (3)
‌z2+z2−z−z=2 . . . (4)
‌‌ Case I: If ‌a+b≠0
‌‌ From (3) & (4) ‌
‌2x+2(x2−y2)=2‌⇒x2−y2+x=1 . . . (5)
2(x2−y2)−2x=2‌⇒x2−y2−x=1 . . . (6)
‌(5)−(6)
‌2x=0⇒x=0
‌‌ from (5) ‌y2=−1‌‌⇒‌ not possbible ‌
‌∴‌ Ans ‌=0
Case II: If a+b=0 then infinite number of solution.
So, the set X have infinite number of elements.