The function $\frac{\sin (x+\alpha)}{\sin (x+\beta)}$ has no extrema if (n ∈ Z)
Answer & explanation
Correct answer: option 1
$f(x)=\frac{\sin (x+\alpha)}{\sin (x+\beta)}$
$\Rightarrow f'(x)=\frac{\sin (x+\beta) \cos (x+\alpha)-\sin (x+\alpha) \cos (x+\beta)}{\sin ^2(x+\beta)}$
$=\frac{\sin (\beta-\alpha)}{\sin ^2(x+\beta)}$
∴ f has no extrema if f'(x) ≠ 0
i.e. if sin (β – α) ≠ 0
i.e. if sin (α – β) ≠ 0, i.e. if α – β ≠ nπ, n ∈ Z.