Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
7.2 Continuity: The 3-Step Verification, Discontinuity Types, and Economic Thresholds
In Applied Mathematics and econometric modeling, a continuous function models smooth, unbroken processes—such as steady asset price trajectories, smooth cost curves, and uninterrupted compounding. Understanding continuity is the prerequisite for differentiation, because only continuous curves can possess defined tangent slopes.
The 3 Fundamental Conditions of Continuity
A real-valued function \(f(x)\) is continuous at a point \(x = c\) in its domain if and only if all three conditions are satisfied:
Existence of Function Value: \(f(c)\) is defined (yields a finite real number).
Existence of Two-Sided Limit: The limit \(\lim_{x \to c} f(x)\) exists:
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L \]
Equivalence of Limit and Value:
\[ \lim_{x \to c} f(x) = f(c) \]
In concise operator form: \(\text{LHL} = \text{RHL} = f(c)\).
Taxonomy of Discontinuities
Discontinuity Type
Mathematical Criterion
Visual / Graphical Behavior
Real-World / Business Example
Removable Discontinuity
\(\lim_{x \to c} f(x)\) exists, but \(\lim_{x \to c} f(x) \neq f(c)\) (or \(f(c)\) is undefined)
A single missing "hole" on an otherwise smooth curve
An pricing formula undefined at exactly zero quantity due to \(0/0\) division
Jump (Step) Discontinuity
\(\text{LHL} \neq \text{RHL}\) (both finite but unequal)
The curve abruptly steps up or down to a different height
Income tax slabs under non-marginal regimes, postal weight brackets, or parking fee hour blocks
Infinite Discontinuity
One or both one-sided limits approach \(\pm \infty\)
Vertical asymptote
Average fixed cost \(AFC(x) = \frac{FC}{x}\) as output \(x \to 0^+\)
Comprehensive Exam Example: Determining Unknown Parameters for Continuity
Problem: Determine the values of \(a\) and \(b\) such that the function \(f(x)\) defined by:
\[ f(x) = \begin{cases} 5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10 \end{cases} \]
is continuous everywhere.
Solution:
For \(f(x)\) to be continuous everywhere, it must be continuous at the transition points \(x = 2\) and \(x = 10\).
For continuity at \(x = 10\):
\[ 10a + b = 21 \quad \text{--- (Equation 2)} \]
3. Solving the Linear System:
Subtract Equation (1) from Equation (2):
\[ (10a + b) - (2a + b) = 21 - 5 \implies 8a = 16 \implies a = 2 \]
Substitute \(a = 2\) into Equation (1):
\[ 2(2) + b = 5 \implies 4 + b = 5 \implies b = 1 \]
Answer: \(a = 2\) and \(b = 1\).
Progress Tracker & Self-Assessment Checklist
Track your conceptual mastery and diagnostic skills across Continuity and Parameter Determinations.
Conceptual Milestones
Problem Solving Milestones
Practice Worksheet: Graded Exercises & Case Problems
Work through these graded problems from foundational mechanics to applied multi-stage scenarios.
Problem 1 (Finding Constant \(k\) for Continuity)
Find the value of \(k\) so that the function \(f(x)\) is continuous at \(x = 0\):
\[ f(x) = \begin{cases} \displaystyle \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & \text{if } -1 \le x < 0 \\[10pt] \displaystyle \frac{2x + 1}{x - 1}, & \text{if } 0 \le x \le 1 \end{cases} \]
For \(f(x)\) to be continuous at \(x = 0\), we must have \(\text{LHL} = f(0)\):
\[ k = -1. \]
Answer: \(k = -1\).
Problem 2 (Greatest Integer Function Discontinuity)
Prove that the greatest integer function \(f(x) = [x]\) is discontinuous at all integral points.
View Step-by-Step Solution
Solution:
Let \(c\) be any integer (\(c \in \mathbb{Z}\)).
\(f(c) = [c] = c\).
For \(x \to c^-\), \(x\) lies strictly between \(c - 1\) and \(c\) (e.g. \(c - 0.001\)).
Hence \([x] = c - 1\).
\[ \text{LHL} = \lim_{x \to c^-} [x] = c - 1. \]
For \(x \to c^+\), \(x\) lies strictly between \(c\) and \(c + 1\) (e.g. \(c + 0.001\)).
Hence \([x] = c\).
\[ \text{RHL} = \lim_{x \to c^+} [x] = c. \]
Since \(\text{LHL} = c - 1 \neq c = \text{RHL}\), the limit \(\lim_{x \to c} [x]\) does not exist at any integer \(c\).
Therefore, \(f(x) = [x]\) has a jump discontinuity of size 1 at every integer point.
Problem 3 (Modulus Function Continuity)
Examine the continuity of \(f(x) = |x - 3|\) at \(x = 3\).
View Step-by-Step Solution
Solution:
By definition of absolute value:
\[ f(x) = \begin{cases} -(x - 3) = 3 - x, & x < 3 \\ x - 3, & x \ge 3 \end{cases} \]