If `f(x) = {(ax^(2)-b, ",", "when" 0 le x le 1),(2, ",","when"x = 1),(x+1, ",","when"1 lt x le 2):}` is continuous at x = 1, then the most suitable values of a, b are

Options

A. `a = 2, b = 0`
B. `a = 1, b = -1`
C. `a = 4, b = 2`
D. All of the above

Answer

To determine the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous at \( x = 1 \), we need to ensure that the left-hand limit and the right-hand limit at \( x = 1 \) are equal to the function value at \( x = 1 \).

### Step 1: Identify the function pieces around \( x = 1 \)

The function \( f(x) \) is defined as follows:
- \( f(x) = ax^2 - b \) for \( 0 \leq x < 1 \)
- \( f(1) = 2 \)
- \( f(x) = x + 1 \) for \( 1 < x \leq 2 \)

### Step 2: Calculate the left-hand limit as \( x \) approaches 1

The left-hand limit as \( x \) approaches 1 from the left is given by:
\[ \lim_{x \to 1^-} f(x) = a(1)^2 - b = a - b \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1

The right-hand limit as \( x \) approaches 1 from the right is given by:
\[ \lim_{x \to 1^+} f(x) = 1 + 1 = 2 \]

### Step 4: Set the limits equal to ensure continuity

For the function to be continuous at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]
This gives us the equation:
\[ a - b = 2 \]

### Step 5: Solve for one variable in terms of the other

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

### Step 6: Determine suitable values for \( a \) and \( b \)

Since there are no additional conditions given in the problem, we can choose any value for \( a \) and then calculate \( b \). For example, if we let \( a = 2 \):
\[ b = 2 - 2 = 0 \]
Thus, one suitable pair of values is \( (a, b) = (2, 0) \).

### Final Answer

The most suitable values of \( a \) and \( b \) are \( a = 2 \) and \( b = 0 \).

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