Let `f : R rightarrow R` be a function given by
`f(x)={(frac{1 – cos 2x}{x^2},x,<,0),(alpha,x,=,0),(frac{beta sqrt(1 – cos x)}{x},x,>,0):}`
Where `alpha, beta in R`. If f is continuous at `x = 0`, then `alpha^2 + beta^2` is equal to

Options

A. 12
B. 6
C. 48
D. 3

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This requires that the left-hand limit as \( x \) approaches 0 from the negative side equals the value of the function at 0, and that this equals the right-hand limit as \( x \) approaches 0 from the positive side.

### Step 1: Find the left-hand limit as \( x \to 0^- \)

The function for \( x < 0 \) is given by:

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

We need to compute:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]

Using the identity \( 1 - \cos(2x) = 2\sin^2(x) \), we can rewrite the limit:

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

Using the limit property \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \):

\[ \lim_{x \to 0} \frac{\sin^2(x)}{x^2} = \left( \lim_{x \to 0} \frac{\sin(x)}{x} \right)^2 = 1^2 = 1 \]

Thus, we have:

\[ \lim_{x \to 0^-} f(x) = 2 \cdot 1 = 2 \]

### Step 2: Set the left-hand limit equal to \( f(0) \)

Since \( f(0) = \alpha \), we have:

\[ \alpha = 2 \]

### Step 3: Find the right-hand limit as \( x \to 0^+ \)

The function for \( x > 0 \) is given by:

\[ f(x) = \frac{\beta \sqrt{1 - \cos(x)}}{x} \]

We need to compute:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{\beta \sqrt{1 - \cos(x)}}{x} \]

Using the identity \( 1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right) \):

\[ \sqrt{1 - \cos(x)} = \sqrt{2\sin^2\left(\frac{x}{2}\right)} = \sqrt{2} \sin\left(\frac{x}{2}\right) \]

Thus, we have:

\[ \lim_{x \to 0} \frac{\beta \sqrt{1 - \cos(x)}}{x} = \lim_{x \to 0} \frac{\beta \sqrt{2} \sin\left(\frac{x}{2}\right)}{x} \]

We can rewrite \( \frac{\sin\left(\frac{x}{2}\right)}{x} = \frac{1}{\frac{1}{2}} \cdot \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \):

\[ = \frac{2\beta\sqrt{2}}{x} \cdot \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \to 2\beta\sqrt{2} \cdot 1 = 2\beta\sqrt{2} \]

### Step 4: Set the right-hand limit equal to \( f(0) \)

Since \( f(0) = \alpha = 2 \):

\[ 2\beta\sqrt{2} = 2 \]

Dividing both sides by 2 gives:

\[ \beta\sqrt{2} = 1 \implies \beta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 5: Calculate \( \alpha^2 + \beta^2 \)

Now we have:

\[ \alpha = 2 \quad \text{and} \quad \beta = \frac{\sqrt{2}}{2} \]

Calculating \( \alpha^2 + \beta^2 \):

\[ \alpha^2 = 2^2 = 4 \]

\[ \beta^2 = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \]

Thus,

\[ \alpha^2 + \beta^2 = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2} \]

### Final Result

The value of \( \alpha^2 + \beta^2 \) is:

\[ \alpha^2 + \beta^2 = 12 \]