Question

The equation (cos p – 1)x² + (cos p)x + sin p = 0 in variable x has real roots. Then p can take any value in the interval

• A. (0, 2p)
• B. (–p, 0)
• C. $\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$
• D. (0, p)

Answer 1

Answer: (D)

(0, p)

Answer 2

cos²p – 4 sin p(cos p – 1) ³ 0

(cos p – 2 sin p)² + 4 sin p(1 – sin p) ³ 0

\ 4 sin p (1 – sin p) > 0 for 0 < p < p

& (cos p – 2 sin p)² ³ 0 Ž \ 0 < p < p