The range of the function `f(x)=1+sinx+sin^(3)x+sin^(5)x+……` when `x in (-pi//2,pi//2)`, is

Options

A. (0,1)
B. R
C. (-2,2)
D. none of these

Answer

To find the range of the function \( f(x) = 1 + \sin x + \sin^3 x + \sin^5 x + \ldots \) for \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we can follow these steps:

### Step 1: Identify the series
The function can be expressed as an infinite series:
\[ f(x) = 1 + \sin x + \sin^3 x + \sin^5 x + \ldots \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \sin x \).

### Step 2: Find the sum of the series
The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
provided that \( |r| < 1 \). In our case:
\[ f(x) = \frac{1}{1 - \sin x} \quad \text{for } |\sin x| < 1 \]
Since \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), \( \sin x \) ranges from -1 to 1, but we need to ensure \( |\sin x| < 1 \).

### Step 3: Analyze the limits of \( f(x) \)
- As \( x \) approaches \( -\frac{\pi}{2} \), \( \sin x \) approaches -1:
\[ f\left(-\frac{\pi}{2}\right) = \frac{1}{1 - (-1)} = \frac{1}{2} \]
- As \( x \) approaches \( \frac{\pi}{2} \), \( \sin x \) approaches 1:
\[ f\left(\frac{\pi}{2}\right) = \frac{1}{1 - 1} \quad \text{(undefined, approaches infinity)} \]

### Step 4: Determine the range of \( f(x) \)
Since \( \sin x \) varies continuously from -1 to 1 as \( x \) varies from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( f(x) \) will take all values from \( \frac{1}{2} \) to \( +\infty \).

### Conclusion
Thus, the range of the function \( f(x) \) is:
\[ \left(\frac{1}{2}, +\infty\right) \]