Q36 / 56 · IAT (IISER) 2026 · Mathematics
For real numbers $a$ and $b$, consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ given by
$f(x) = \begin{cases} -ax - b & \text{if } x < -1, \\ 5x + 1 & \text{if } -1 \leq x \leq 1, \\ a^2x + 3b & \text{if } x > 1. \end{cases}$
How many pairs $(a, b)$ are there for which $f$ is continuous at every point of $\mathbb{R}$?
$f(x) = \begin{cases} -ax - b & \text{if } x < -1, \\ 5x + 1 & \text{if } -1 \leq x \leq 1, \\ a^2x + 3b & \text{if } x > 1. \end{cases}$
How many pairs $(a, b)$ are there for which $f$ is continuous at every point of $\mathbb{R}$?
- A$2$
- B$1$
- Cinfinitely many
- Correct answerD$0$
Solution
For $f(x)$ to be continuous at $x = -1$, the left-hand limit must be equal to $f(-1)$.
$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-ax - b) = a - b$
$f(-1) = 5(-1) + 1 = -4$
Equating them, we get $a - b = -4 \Rightarrow b = a + 4$.
For $f(x)$ to be continuous at $x = 1$, the right-hand limit must be equal to $f(1)$.
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (a^2x + 3b) = a^2 + 3b$
$f(1) = 5(1) + 1 = 6$
Equating them, we get $a^2 + 3b = 6$.
Substituting $b = a + 4$ into the second equation:
$a^2 + 3(a + 4) = 6$
$a^2 + 3a + 12 = 6$
$a^2 + 3a + 6 = 0$
The discriminant of this quadratic equation is $\Delta = 3^2 - 4(1)(6) = 9 - 24 = -15$.
Since $\Delta < 0$, there are no real solutions for $a$.
Therefore, there are $0$ pairs of $(a, b)$ for which $f$ is continuous on $\mathbb{R}$.
Answer: $0$
$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-ax - b) = a - b$
$f(-1) = 5(-1) + 1 = -4$
Equating them, we get $a - b = -4 \Rightarrow b = a + 4$.
For $f(x)$ to be continuous at $x = 1$, the right-hand limit must be equal to $f(1)$.
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (a^2x + 3b) = a^2 + 3b$
$f(1) = 5(1) + 1 = 6$
Equating them, we get $a^2 + 3b = 6$.
Substituting $b = a + 4$ into the second equation:
$a^2 + 3(a + 4) = 6$
$a^2 + 3a + 12 = 6$
$a^2 + 3a + 6 = 0$
The discriminant of this quadratic equation is $\Delta = 3^2 - 4(1)(6) = 9 - 24 = -15$.
Since $\Delta < 0$, there are no real solutions for $a$.
Therefore, there are $0$ pairs of $(a, b)$ for which $f$ is continuous on $\mathbb{R}$.
Answer: $0$
Hence, the correct answer is option: 'D'.