Let `f(x)={{:(ax-b,","x le 1),(3x,"," 1 lt x lt 2),(bx^2-a,"," x ge 2):}` If f is continuous function then (a,b) is equal to

Options

A. (1,2)
B. (1,3)
C. (3,6)
D. (6,3)

Answer

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 \( x = 2 \).

### Step 1: Check continuity at \( x = 1 \)

The function is defined as:
- \( f(x) = ax - b \) for \( x \leq 1 \)
- \( f(x) = 3x \) for \( 1 < x < 2 \)

To ensure continuity at \( x = 1 \), we must have:
\[ f(1^-) = f(1^+) \]

Calculating \( f(1^-) \):
\[ f(1^-) = a(1) - b = a - b \]

Calculating \( f(1^+) \):
\[ f(1^+) = 3(1) = 3 \]

Setting these equal gives us our first equation:
\[ a - b = 3 \quad \text{(Equation 1)} \]

### Step 2: Check continuity at \( x = 2 \)

The function is defined as:
- \( f(x) = 3x \) for \( 1 < x < 2 \)
- \( f(x) = bx^2 - a \) for \( x \geq 2 \)

To ensure continuity at \( x = 2 \), we must have:
\[ f(2^-) = f(2^+) \]

Calculating \( f(2^-) \):
\[ f(2^-) = 3(2) = 6 \]

Calculating \( f(2^+) \):
\[ f(2^+) = b(2^2) - a = 4b - a \]

Setting these equal gives us our second equation:
\[ 4b - a = 6 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

We now have the following system of equations:
1. \( a - b = 3 \)
2. \( 4b - a = 6 \)

From Equation 1, we can express \( a \) in terms of \( b \):
\[ a = b + 3 \]

Substituting this into Equation 2:
\[ 4b - (b + 3) = 6 \]
\[ 4b - b - 3 = 6 \]
\[ 3b - 3 = 6 \]
\[ 3b = 9 \]
\[ b = 3 \]

Now substituting \( b = 3 \) back into Equation 1 to find \( a \):
\[ a - 3 = 3 \]
\[ a = 6 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ (a, b) = (6, 3) \]