Let `f(x)={(frac{1-cos2x}{x^2},x, <,0),(alpha,x,=,0),(beta (frac{sqrt(1-cosx)}{x}),x, >,0):}`
If f(x) is continuous at x = 0 then value of `4|alpha^2 + beta^2|` is

Options

A. 48
B. 36
C. 28
D. 16

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the left-hand limit as \( x \) approaches 0 from the negative side must equal the right-hand limit as \( x \) approaches 0 from the positive side, and both must equal \( f(0) \).

The function is defined as follows:

\[ f(x) = \begin{cases} \frac{1 - \cos(2x)}{x^2} & \text{if } x < 0 \\ \alpha & \text{if } x = 0 \\ \beta \cdot \frac{\sqrt{1 - \cos(x)}}{x} & \text{if } x > 0 \end{cases} \]

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

We need to find:

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

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

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

Thus, we have:

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

### Step 2: Calculate the right-hand limit as \( x \to 0^+ \)

Now we need to find:

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

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

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

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \beta \cdot \frac{\sqrt{2} \cdot \sin\left(\frac{x}{2}\right)}{x} \]

To simplify this, we can rewrite \( \frac{\sin\left(\frac{x}{2}\right)}{x} \) as follows:

\[ \frac{\sin\left(\frac{x}{2}\right)}{x} = \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \frac{1}{2} \to 1 \cdot \frac{1}{2} = \frac{1}{2} \quad \text{as } x \to 0 \]

So, we have:

\[ \lim_{x \to 0^+} f(x) = \beta \cdot \sqrt{2} \cdot \frac{1}{2} = \frac{\beta \sqrt{2}}{2} \]

### Step 3: Set the limits equal to ensure continuity

For \( f(x) \) to be continuous at \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This gives us:

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

From the second equation, we can solve for \( \beta \):

\[ \beta \sqrt{2} = 4 \implies \beta = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 4: Calculate \( 4 |\alpha^2 + \beta^2| \)

Now we have \( \alpha = 2 \) and \( \beta = 2\sqrt{2} \). We calculate:

\[ \alpha^2 = 2^2 = 4 \]
\[ \beta^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]

Thus:

\[ \alpha^2 + \beta^2 = 4 + 8 = 12 \]

Finally, we compute:

\[ 4 |\alpha^2 + \beta^2| = 4 \cdot 12 = 48 \]

### Final Answer

The value of \( 4 |\alpha^2 + \beta^2| \) is \( \boxed{48} \).