Question

If r > 0, -$\pi$ ≤ $\theta$ ≤ $\pi$ and (r, $\theta$) satisfy r sin$\theta$ = 3 and r = 4(1 + sin$\theta$) then the number of possible solutions of the pair ( r, $\theta$) is

• A. 2
• B. 4
• C. 0
• D. Infinite

Answer 1

The correct option is A

2

r = 4(1 + $\frac{3}{r}$) ⇒ (r - 6)(r + 2) = 0 ⇒ r = 6, r ≠ -2 as r > 0

$\therefore$ sin $\theta$ = $\frac{1}{2}$ ⇒ $\theta$ = $\frac{\pi }{6}$, $\frac{5\pi }{6}$

So ($r,\theta )$ = (6, $\frac{\pi }{6}$), (6, $\frac{5\pi }{6}$)