y=4cos3θ−3cosθ+8cos2θ−4+5cosθ+22cosθ+2cos2θ−1y=(2cos2θ+2cosθ−1)(2cosθ+2)(2cos2θ+2cosθ−1)y=21(1+cosθ1)⇒θ=2πy=21y′=21((1+cosθ)2−1×(−sinθ))⇒θ=2πy=21 y′′=21[(1+cosθ)4cosθ(1+cosθ)2−sinθ(2)(1+cosθ)(−sinθ)]⇒θ=2πy=1
If y(θ)= 2 cos θ+cos 2 θ cos 3 θ+4 cos 2 θ+5 cos θ+2, then at θ= π 2, y^prime prime+y^prime+y is equal to :
JEE Main 2024 — Mathematics Calculus
If y(θ)=cos3θ+4cos2θ+5cosθ+22cosθ+cos2θ, then at θ=2π,y′′+y′+y is equal to :
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
21
- B
1
- C
2
- D
23
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