If `alpha` and `beta` are roots of `x^(2)-(sqrt(1-cos 2 theta))x+theta=0`, where `0 lt theta lt (pi)/2`. Then `lim_(theta to 0^(+))(1/(alpha)+1/(beta))` is

Options

A. (a)`1/(sqrt(2))`
B. (b)`-sqrt(2)`
C. (c)`sqrt(2)`
D. (d)None of these

Answer

To solve the problem, we need to find the limit:

\[ \lim_{\theta \to 0^+} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation:

\[ x^2 - \sqrt{1 - \cos 2\theta} x + \theta = 0 \]

### Step 1: Identify the coefficients
The quadratic equation can be written in the standard form \(ax^2 + bx + c = 0\), where:
- \(a = 1\)
- \(b = -\sqrt{1 - \cos 2\theta}\)
- \(c = \theta\)

### Step 2: Use the sum of roots formula
The sum of the roots \(\alpha + \beta\) is given by:

\[ \alpha + \beta = -\frac{b}{a} = \sqrt{1 - \cos 2\theta} \]

### Step 3: Use the product of roots formula
The product of the roots \(\alpha \beta\) is given by:

\[ \alpha \beta = \frac{c}{a} = \theta \]

### Step 4: Rewrite the limit
We can rewrite the expression for the limit:

\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{\sqrt{1 - \cos 2\theta}}{\theta} \]

### Step 5: Substitute into the limit
Now we substitute this into the limit:

\[ \lim_{\theta \to 0^+} \left( \frac{\sqrt{1 - \cos 2\theta}}{\theta} \right) \]

### Step 6: Simplify \(\sqrt{1 - \cos 2\theta}\)
Using the trigonometric identity, we know that:

\[ 1 - \cos 2\theta = 2\sin^2 \theta \]

Thus,

\[ \sqrt{1 - \cos 2\theta} = \sqrt{2\sin^2 \theta} = \sqrt{2} |\sin \theta| \]

Since \(\theta\) approaches \(0^+\), \(\sin \theta\) is positive, so we can write:

\[ \sqrt{1 - \cos 2\theta} = \sqrt{2} \sin \theta \]

### Step 7: Substitute back into the limit
Now we substitute back into the limit:

\[ \lim_{\theta \to 0^+} \frac{\sqrt{2} \sin \theta}{\theta} \]

### Step 8: Evaluate the limit
Using the standard limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):

\[ \lim_{\theta \to 0^+} \frac{\sqrt{2} \sin \theta}{\theta} = \sqrt{2} \cdot \lim_{\theta \to 0^+} \frac{\sin \theta}{\theta} = \sqrt{2} \cdot 1 = \sqrt{2} \]

### Final Answer
Thus, the limit is:

\[ \lim_{\theta \to 0^+} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = \sqrt{2} \]