The least value of a for which the equation `4/(sinx) + 1/(1 - sinx)= a` has at least one solution on the interval `(0, pi/2)` is

Options

A. 9
B. 4
C. 8
D. 1

Answer

To find the least value of \( a \) for which the equation

\[ \frac{4}{\sin x} + \frac{1}{1 - \sin x} = a \]

has at least one solution on the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Define the function
Let

\[ f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x} \]

We need to find the minimum value of \( f(x) \) on the interval \( (0, \frac{\pi}{2}) \).

### Step 2: Differentiate the function
To find the critical points, we differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1 - \sin x)^2} \]

### Step 3: Set the derivative to zero
Setting the derivative equal to zero to find critical points:

\[ -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1 - \sin x)^2} = 0 \]

Factoring out \( \cos x \) (noting that \( \cos x \neq 0 \) in the interval \( (0, \frac{\pi}{2}) \)):

\[ \frac{1}{(1 - \sin x)^2} = \frac{4}{\sin^2 x} \]

### Step 4: Cross multiply
Cross multiplying gives:

\[ \sin^2 x = 4(1 - \sin x)^2 \]

### Step 5: Expand and rearrange
Expanding the right-hand side:

\[ \sin^2 x = 4(1 - 2\sin x + \sin^2 x) \]

This simplifies to:

\[ \sin^2 x = 4 - 8\sin x + 4\sin^2 x \]

Rearranging gives:

\[ 0 = 3\sin^2 x - 8\sin x + 4 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3, b = -8, c = 4 \):

\[ \sin x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]

Calculating the discriminant:

\[ \sqrt{64 - 48} = \sqrt{16} = 4 \]

Thus,

\[ \sin x = \frac{8 \pm 4}{6} \]

This gives:

\[ \sin x = \frac{12}{6} = 2 \quad \text{(not valid in } (0, 1)\text{)} \quad \text{or} \quad \sin x = \frac{4}{6} = \frac{2}{3} \]

### Step 7: Find the minimum value of \( f(x) \)
Now, we evaluate \( f(x) \) at \( \sin x = \frac{2}{3} \):

\[ f\left(\sin^{-1}\left(\frac{2}{3}\right)\right) = \frac{4}{\frac{2}{3}} + \frac{1}{1 - \frac{2}{3}} = 6 + 3 = 9 \]

### Conclusion
The least value of \( a \) for which the equation has at least one solution on the interval \( (0, \frac{\pi}{2}) \) is

\[ \boxed{9} \]