Question

Consider the system of equations $\cos^{-1}x + (\sin^{-1}y)^2 = \frac{p\pi^2}{4}$ and $(\cos^{-1}x) \cdot (\sin^{-1}y)^2 = \frac{\pi^4}{16}, p \in Z.$

The value of $p$ for which system has a solution is:

• A. 1
• B. 2
• C. 0
• D. -1

Answer 1

Answer: (B)

2

Answer 2

Let $A = \cos^{-1}x$ and $B = (\sin^{-1}y)^2$. The given system of equations is:

1. $A + B = \frac{p\pi^2}{4}$
2. $A \cdot B = \frac{\pi^4}{16}$

$A$ and $B$ are the roots of the quadratic equation $t^2 - (A+B)t + AB = 0$. Substituting the expressions from the given equations: $t^2 - \frac{p\pi^2}{4}t + \frac{\pi^4}{16} = 0$

For real solutions for $t$ (i.e., for $A$ and $B$), the discriminant must be non-negative: $D = \left(\frac{p\pi^2}{4}\right)^2 - 4 \cdot 1 \cdot \frac{\pi^4}{16} \ge 0$ $D = \frac{p^2\pi^4}{16} - \frac{\pi^4}{4} \ge 0$ $\frac{\pi^4}{16}(p^2 - 4) \ge 0$ Since $\frac{\pi^4}{16} > 0$, we must have $p^2 - 4 \ge 0$, which implies $p^2 \ge 4$. So, $p \ge 2$ or $p \le -2$.

Since $A = \cos^{-1}x \ge 0$ and $B = (\sin^{-1}y)^2 \ge 0$, both roots must be non-negative. The sum $A+B = \frac{p\pi^2}{4}$ must be non-negative, which implies $p \ge 0$. Combining $p \ge 0$ with ($p \ge 2$ or $p \le -2$), we conclude that $p \ge 2$. Since $p \in Z$, possible values for $p$ are $2, 3, 4, \dots$.

After checking the ranges of A and B, we find that the only possible value is $p=2$.