Question

The number of ordered pairs (x, y) of real numbers satisfying $4x^2 - 4x + 2 = sin^2y$ and $x^2 + y^2 \le 3$, is equal to :

• A. 0
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (B)

2

Answer 2

The first equation is $4x^2 - 4x + 2 = \sin^2y$. Completing the square for $x$: $4(x - \frac{1}{2})^2 + 1 = \sin^2y$. Since $4(x - \frac{1}{2})^2 \ge 0$, the left side is $\ge 1$. The right side, $\sin^2y$, has a maximum value of 1. Thus, for equality, $4(x - \frac{1}{2})^2 = 0 \implies x = \frac{1}{2}$, and $\sin^2y = 1 \implies \sin y = \pm 1$. This implies $y = \frac{\pi}{2} + n\pi$ for any integer $n$. The second condition is $x^2 + y^2 \le 3$. Substituting $x=\frac{1}{2}$: $(\frac{1}{2})^2 + y^2 \le 3 \implies \frac{1}{4} + y^2 \le 3 \implies y^2 \le \frac{11}{4}$. So, $-\frac{\sqrt{11}}{2} \le y \le \frac{\sqrt{11}}{2}$. We need to find values of $y = \frac{\pi}{2} + n\pi$ in $[-\frac{\sqrt{11}}{2}, \frac{\sqrt{11}}{2}]$. Approximate values: $\frac{\pi}{2} \approx 1.57$, $\frac{\sqrt{11}}{2} \approx 1.66$. For $n=0$, $y = \frac{\pi}{2} \approx 1.57$, which is in the interval. For $n=-1$, $y = -\frac{\pi}{2} \approx -1.57$, which is in the interval. For other integer values of $n$, $y$ falls outside the interval. Thus, there are two possible values for $y$, and for each, $x$ is fixed at $\frac{1}{2}$. The ordered pairs are $(\frac{1}{2}, \frac{\pi}{2})$ and $(\frac{1}{2}, -\frac{\pi}{2})$.