Suppose `3 sin^(-1) ( log _(2) x) + cos^(-1) ( log _(2) y) =pi //2` and `sin^(-1) ( log _(2) x ) + 2 cos^(-1) ( log_(2) y) = 11 pi //6`. then the value of `1/x^(-2) + 1/y^(-2)` equals .

Options

A. 6
B. 7
C. 5
D. `7/2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let:
- \( p = \sin^{-1}(\log_2 x) \)
- \( q = \cos^{-1}(\log_2 y) \)

### Step 2: Rewrite the Given Equations
From the problem statement, we have the following equations:
1. \( 3p + q = \frac{\pi}{2} \) (Equation 1)
2. \( p + 2q = \frac{11\pi}{6} \) (Equation 2)

### Step 3: Solve for \( p \) and \( q \)
We can solve these equations simultaneously.

First, we can express \( q \) from Equation 1:
\[ q = \frac{\pi}{2} - 3p \]

Now, substitute this expression for \( q \) into Equation 2:
\[ p + 2\left(\frac{\pi}{2} - 3p\right) = \frac{11\pi}{6} \]

### Step 4: Simplify the Equation
Expanding the equation gives:
\[ p + \pi - 6p = \frac{11\pi}{6} \]
\[ -5p + \pi = \frac{11\pi}{6} \]

### Step 5: Isolate \( p \)
Now, isolate \( p \):
\[ -5p = \frac{11\pi}{6} - \pi \]
Convert \( \pi \) to sixths:
\[ -5p = \frac{11\pi}{6} - \frac{6\pi}{6} = \frac{5\pi}{6} \]
\[ p = -\frac{\pi}{6} \]

### Step 6: Substitute Back to Find \( q \)
Now substitute \( p \) back into the expression for \( q \):
\[ q = \frac{\pi}{2} - 3\left(-\frac{\pi}{6}\right) \]
\[ q = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Step 7: Find \( \log_2 x \) and \( \log_2 y \)
Now that we have \( p \) and \( q \):
1. From \( p = \sin^{-1}(\log_2 x) = -\frac{\pi}{6} \):
\[ \log_2 x = \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \]
Therefore, \( x = 2^{-1/2} = \frac{1}{\sqrt{2}} \).

2. From \( q = \cos^{-1}(\log_2 y) = \pi \):
\[ \log_2 y = \cos(\pi) = -1 \]
Therefore, \( y = 2^{-1} = \frac{1}{2} \).

### Step 8: Calculate \( \frac{1}{x^{-2}} + \frac{1}{y^{-2}} \)
Now we need to find:
\[ \frac{1}{x^{-2}} + \frac{1}{y^{-2}} = x^2 + y^2 \]
Calculating \( x^2 \) and \( y^2 \):
\[ x^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus:
\[ x^2 + y^2 = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \]

### Final Answer
The value of \( \frac{1}{x^{-2}} + \frac{1}{y^{-2}} \) is:
\[ \frac{3}{4} \]