if for all real value of `x,(4x^2+1)/(64x^2-96x.sina+5)<1/32`,then a lies in the interval

Options

A. `(0, (pi)/(3))`
B. `((pi)/(3), (2pi)/(3))`
C. `((4pi)/(3), (5pi)/(3))`
D. none of these

Answer

To solve the inequality \(\frac{4x^2 + 1}{64x^2 - 96x \sin a + 5} < \frac{1}{32}\) for all real values of \(x\), we will follow these steps:

### Step 1: Cross-multiply the inequality
We start with the given inequality:
\[ \frac{4x^2 + 1}{64x^2 - 96x \sin a + 5} < \frac{1}{32} \]
Cross-multiplying gives:
\[ 32(4x^2 + 1) < 64x^2 - 96x \sin a + 5 \]

### Step 2: Simplify the inequality
Expanding the left-hand side:
\[ 128x^2 + 32 < 64x^2 - 96x \sin a + 5 \]
Now, rearranging all terms to one side:
\[ 128x^2 - 64x^2 + 96x \sin a + 32 - 5 < 0 \]
This simplifies to:
\[ 64x^2 + 96 \sin a x + 27 < 0 \]

### Step 3: Identify the quadratic form
The inequality is now in the standard quadratic form \(Ax^2 + Bx + C < 0\) where:
- \(A = 64\)
- \(B = 96 \sin a\)
- \(C = 27\)

### Step 4: Determine the condition for the quadratic to be less than zero
For the quadratic \(64x^2 + 96 \sin a x + 27\) to be less than zero for all \(x\), the discriminant must be less than zero:
\[ B^2 - 4AC < 0 \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ (96 \sin a)^2 - 4 \cdot 64 \cdot 27 < 0 \]

### Step 5: Calculate the discriminant
Calculating the discriminant:
\[ 9216 \sin^2 a - 6912 < 0 \]
This simplifies to:
\[ 9216 \sin^2 a < 6912 \]
Dividing both sides by 9216:
\[ \sin^2 a < \frac{6912}{9216} \]
This simplifies to:
\[ \sin^2 a < \frac{3}{4} \]

### Step 6: Determine the range for \(a\)
Taking the square root gives:
\[ -\frac{\sqrt{3}}{2} < \sin a < \frac{\sqrt{3}}{2} \]
The values of \(a\) that satisfy this condition can be found using the inverse sine function:
\[ \sin^{-1}(-\frac{\sqrt{3}}{2}) < a < \sin^{-1}(\frac{\sqrt{3}}{2}) \]
This corresponds to:
\[ -\frac{\pi}{3} < a < \frac{\pi}{3} \]
However, since we are looking for \(a\) in the range of \(0\) to \(2\pi\), we also consider the intervals where \(\sin a\) is positive and negative.

### Step 7: Final intervals for \(a\)
The intervals where \(\sin a\) is between \(-\frac{\sqrt{3}}{2}\) and \(\frac{\sqrt{3}}{2}\) are:
1. From \(0\) to \(\frac{\pi}{3}\)
2. From \(\frac{2\pi}{3}\) to \(\frac{4\pi}{3}\)
3. From \(\frac{5\pi}{3}\) to \(2\pi\)

Thus, the intervals for \(a\) are:
\[ a \in \left(0, \frac{\pi}{3}\right) \cup \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) \cup \left(\frac{5\pi}{3}, 2\pi\right) \]