Q.

Let f(θ)=sinθ(sinθ+sin3θ). Then f(θ) is             [2000]

1 ≥0 only when θ≥0  
2 ≤0 for all real θ  
3 ≥0 for all real θ  
4 ≤0 only when θ≤0  

Ans.

(3)

f(θ)=sinθ(sinθ+sin3θ)

=sinθ(sinθ+3sinθ-4sin3θ)

=sinθ(4sinθ-4sin3θ)=sin2θ(4-4sin2θ)

=4sin2θ(1-sin2θ)

=4sin2θcos2θ=(2sinθcosθ)2=(sin2θ)2≥0, which is true for all θ.