Question

For real numbers $a$ and $b$, consider the function $f : \mathbb{R} \to \mathbb{R}$ given by \[ f(x) = \begin{cases} -ax - b ;& \text{if } x \\ 5x + 1 ;& \text{if } -1 \le x \le 1, \\ a^2x + 3b ;& \text{if } x > 1 . \end{cases} \] How many pairs $(a, b)$ are there for which $f$ is continuous at every point of $\mathbb{R}$?

• A. 0
• B. 1
• C. 2
• D. infinitely many

Hint:

For systems of equations containing quadratic terms, always calculate the discriminant $D = b^2 - 4ac$ first to determine if real solutions exist before attempting to find the roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a piecewise function to be continuous everywhere, the left-hand limit (LHL) must equal the right-hand limit (RHL) at every transition point.

Step 2: Detailed Explanation:
1. Continuity at $x = -1$:
LHL: $-a(-1) - b = a - b$.
RHL: $5(-1) + 1 = -4$.
Equation 1: $a - b = -4 \implies b = a + 4$.
2. Continuity at $x = 1$:
LHL: $5(1) + 1 = 6$.
RHL: $a^{2}(1) + 3b = a^{2} + 3b$.
Equation 2: $a^{2} + 3b = 6$.
3. Solve for a:
Substitute $b = a + 4$ into Equation 2:
$a^{2} + 3(a + 4) = 6 \implies a^{2} + 3a + 12 = 6$.
$a^{2} + 3a + 6 = 0$.
4. Check for real solutions:
Discriminant $D = b^{2} - 4ac = 3^{2} - 4(1)(6) = 9 - 24 = -15$.
Since $D < 0$, there are no real values of $a$ that satisfy this. Since $a$ and $b$ must be real, no pairs exist.

Step 3: Final Answer:
The number of pairs is 0.
This matches option (A).