Continuity of piecewise functions: Class 12 practice
A function is continuous at a if the left-hand limit, right-hand limit and f(a) are all equal. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Limits
The idea behind continuity of piecewise functions
A function is continuous at a if the left-hand limit, right-hand limit and f(a) are all equal. For a piecewise function, set the two pieces equal at the joining point to find an unknown constant.
Step-by-step method
- Find the left-hand limit from the left piece.
- Find the right-hand limit / f(a) from the right piece.
- Set them equal and solve.
Worked example 1 (Basic)
Find k so that f(x) = 3x + 3 for x ≤ -3, and 2x² + k for x > -3, is continuous at x = -3.
- LHL = 3(-3) + 3 = -6
- RHL = 2(-3)² + k = 18 + k
- k = -24
- Answer: k = -24
Worked example 2 (Standard)
Find k so that f(x) = 3x + 1 for x ≤ 1, and 2x² + k for x > 1, is continuous at x = 1.
- LHL = 3(1) + 1 = 4
- RHL = 2(1)² + k = 2 + k
- k = 2
- Answer: k = 2
Worked example 3 (Challenge)
Find k so that f(x) = 1x − 3 for x ≤ 1, and 2x² + k for x > 1, is continuous at x = 1.
- LHL = 1(1) − 3 = -2
- RHL = 2(1)² + k = 2 + k
- k = -4
- Answer: k = -4
Common mistakes and misconceptions
- Checking only one side
- Using the wrong piece at the boundary
Practice questions
- (Basic) Find k so that f(x) = 1x + 4 for x ≤ 1, and 3x² + k for x > 1, is continuous at x = 1.
- (Basic) Find k so that f(x) = 5x − 4 for x ≤ -3, and 1x² + k for x > -3, is continuous at x = -3.
- (Standard) Find k so that f(x) = 4x − 5 for x ≤ -1, and 3x² + k for x > -1, is continuous at x = -1.
- (Standard) Find k so that f(x) = 1x + 3 for x ≤ 3, and 3x² + k for x > 3, is continuous at x = 3.
- (Challenge) Find k so that f(x) = 5x + 2 for x ≤ 0, and 4x² + k for x > 0, is continuous at x = 0.
- (Challenge) Find k so that f(x) = 2x + 1 for x ≤ 3, and 1x² + k for x > 3, is continuous at x = 3.
Answer key with full working
- Q1: LHL = 1(1) + 4 = 5 → RHL = 3(1)² + k = 3 + k → k = 2 → Answer: k = 2
- Q2: LHL = 5(-3) − 4 = -19 → RHL = 1(-3)² + k = 9 + k → k = -28 → Answer: k = -28
- Q3: LHL = 4(-1) − 5 = -9 → RHL = 3(-1)² + k = 3 + k → k = -12 → Answer: k = -12
- Q4: LHL = 1(3) + 3 = 6 → RHL = 3(3)² + k = 27 + k → k = -21 → Answer: k = -21
- Q5: LHL = 5(0) + 2 = 2 → RHL = 4(0)² + k = 0 + k → k = 2 → Answer: k = 2
- Q6: LHL = 2(3) + 1 = 7 → RHL = 1(3)² + k = 9 + k → k = -2 → Answer: k = -2
Syllabus note
This skill is listed under "Continuity and Differentiability" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is continuity of piecewise functions in?
"Continuity and Differentiability", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.