JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023y=f(x)=sin3(3πcos(32π(−4x3+5x2+1)23)). Then, at x=1:
Choose the correct answer:
- A.
2y′−3π2y=0
2y′+3π2y=0
Explanation
Solution: x=1 par, (−4x3+5x2+1)=2.
Andar wali term 32π(2)3/2=32π(22)=32π ho jayegi.
Then y=sin3(3πcos32π)=sin3(3π⋅−21)=sin3(−6π)=−81.
Chain rule se differentiate karke x=1 rakhne par:
y′=3sin2(−6π)cos(−6π)⋅3π(−sin32π)⋅32π⋅23(2)1/2(−2)
Solve karne par equation 2y′+3π2y=0 satisfy hoti hai.
Sahi Option: (4)
Explanation
Solution: x=1 par, (−4x3+5x2+1)=2.
Andar wali term 32π(2)3/2=32π(22)=32π ho jayegi.
Then y=sin3(3πcos32π)=sin3(3π⋅−21)=sin3(−6π)=−81.
Chain rule se differentiate karke x=1 rakhne par:
y′=3sin2(−6π)cos(−6π)⋅3π(−sin32π)⋅32π⋅23(2)1/2(−2)
Solve karne par equation 2y′+3π2y=0 satisfy hoti hai.
Sahi Option: (4)

