Q.

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

1 0 only when θ0  
2 0 for all real θ  
3 0 for all real θ  
4 0 only when θ0  

Ans.

(3)

f(θ)=sinθ(sinθ+sin3θ)

=sinθ(sinθ+3sinθ-4sin3θ)

=sinθ(4sinθ-4sin3θ)=sin2θ(4-4sin2θ)

=4sin2θ(1-sin2θ)

=4sin2θcos2θ=(2sinθcosθ)2=(sin2θ)20, which is true for all θ.