Let f(x) be a continuous function fiven by `f(x)={(2x",", |x|le1),(x^(2)+ax+b",",|x|gt1):},"then "`
If f(x) is continuous for all real x then the value of `a^(2)+b^(2)` is

Options

A. 3
B. 4
C. 5
D. 11

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points where the definition of the function changes, specifically at \( x = -1 \) and \( x = 1 \).

### Step 1: Define the function

The function is defined as follows:
\[ f(x) = \begin{cases} 2x & \text{if } |x| \leq 1 \\ x^2 + ax + b & \text{if } |x| > 1 \end{cases} \]

### Step 2: Continuity at \( x = -1 \)

To ensure continuity at \( x = -1 \), we set the left-hand limit equal to the right-hand limit:
\[ f(-1) = 2(-1) = -2 \]
For \( x < -1 \):
\[ f(-1) = (-1)^2 + a(-1) + b = 1 - a + b \]
Setting these equal gives us:
\[ 1 - a + b = -2 \]
Rearranging this, we get:
\[ -a + b = -3 \quad \text{(Equation 1)} \]

### Step 3: Continuity at \( x = 1 \)

To ensure continuity at \( x = 1 \), we again set the left-hand limit equal to the right-hand limit:
\[ f(1) = 2(1) = 2 \]
For \( x > 1 \):
\[ f(1) = 1^2 + a(1) + b = 1 + a + b \]
Setting these equal gives us:
\[ 1 + a + b = 2 \]
Rearranging this, we get:
\[ a + b = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations

Now we have a system of two equations:
1. \( -a + b = -3 \)
2. \( a + b = 1 \)

We can solve these equations simultaneously.

From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = 1 - a \]

Substituting this into Equation 1:
\[ -a + (1 - a) = -3 \]
Simplifying:
\[ -2a + 1 = -3 \]
\[ -2a = -4 \]
\[ a = 2 \]

Now substituting \( a = 2 \) back into Equation 2 to find \( b \):
\[ 2 + b = 1 \]
\[ b = 1 - 2 = -1 \]

### Step 5: Calculate \( a^2 + b^2 \)

Now that we have \( a = 2 \) and \( b = -1 \), we can calculate \( a^2 + b^2 \):
\[ a^2 + b^2 = 2^2 + (-1)^2 = 4 + 1 = 5 \]

### Final Answer

Thus, the value of \( a^2 + b^2 \) is:
\[ \boxed{5} \]