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 Question:
We are given a piecewise defined function $f(x)$ with two transition boundary points, $x = -1$ and $x = 1$.
For $f(x)$ to be continuous at all points on the real line $\mathbb{R}$, it must be continuous at these boundary points.
This requires that the left-hand limit, right-hand limit, and the function value be equal at both $x = -1$ and $x = 1$.

Step 2: Key Formula or Approach: We set up limit equations for continuity:
1. Continuity at $x = -1$:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) \] 2. Continuity at $x = 1$:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \] We then solve the resulting system of equations for the parameters $a$ and $b$.

Step 3: Detailed Explanation:

• Let us establish continuity at $x = -1$:
The left-hand limit is:
\[ \lim_{x \to -1^-} f(x) = -a(-1) - b = a - b \] The right-hand limit and function value is:
\[ \lim_{x \to -1^+} f(x) = f(-1) = 5(-1) + 1 = -4 \] Equating these two values:
\[ a - b = -4 \implies b = a + 4 \quad \text{--- (Equation 1)} \]
• Let us establish continuity at $x = 1$:
The left-hand limit and function value is:
\[ \lim_{x \to 1^-} f(x) = f(1) = 5(1) + 1 = 6 \] The right-hand limit is:
\[ \lim_{x \to 1^+} f(x) = a^2(1) + 3b = a^2 + 3b \] Equating these two values:
\[ a^2 + 3b = 6 \quad \text{--- (Equation 2)} \]
• Now, we substitute $b = a + 4$ from Equation 1 into Equation 2:
\[ a^2 + 3(a + 4) = 6 \] \[ a^2 + 3a + 12 = 6 \] \[ a^2 + 3a + 6 = 0 \]
• This is a quadratic equation in terms of $a$. Let us compute its discriminant $D$:
\[ D = 3^2 - 4(1)(6) = 9 - 24 = -15 \] Since the discriminant is negative ($D < 0$), there are no real values of $a$ that satisfy this quadratic equation.
Since $a$ must be a real number, no such real parameter exists, meaning there are no valid pairs of real numbers $(a, b)$.

Step 4: Final Answer:
There are 0 pairs of real numbers $(a, b)$ for which $f(x)$ is continuous everywhere.