Given x×y=x2+y3
Now (x×1)×1=x×(1×1)
(x2+13)×1=x×(12+13)
(x2+1)2+13=x2+23
x4+1+2x2+1=x2+8
x4+x2−6=0
Let x2=t
t2+t−6=0
(t+3)(t−2)=0
t=−3 so t=2 or x2=2
Now 2sin−1(x4+x2+2x4+x2−2)=2sin−1(4+2+24+2−2)
=2sin−1(21)=6π×2=3π
JEE Main 2022 — Mathematics Trigonometry
Let x×y=x2+y3 and (x×1)×1=x×(1×1). Then a value of 2sin−1(x4+x2+2x4+x2−2) is
Held on 24 Jun 2022 · Verified 6 Jul 2026.
4π
3π
6π
π
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