Principle deep dive

Piecewise definitions — continuity and differentiability at the join

A function given by cases, where everything happens at the join. Match one-sided limits to make it continuous; match one-sided derivatives to make it differentiable — that is how NDA hides a two-equation solve for a and b inside a continuity question. Concentrated in Limits & Continuity, with Differentiation, Functions and App of Derivatives.

questions in the bank
23
tagged HARD
9%
chapter spread
4
worked examples below
4

When to reach for it

f is given by cases, and the question asks about continuity, a limit, or differentiability at a join.

Why this principle matters

A function defined by cases is easy everywhere except at the joins, and the joins are the only place NDA ever asks. Away from a breakpoint the function is just whichever ordinary expression applies, so all the work is at the boundary between two pieces.

Two tests, in strict order. For CONTINUITY at c you need the left limit, the right limit and f(c) itself to be one and the same number — and f(c) is the piece the definition assigns AT c, which is frequently a third, separate line of the definition. For DIFFERENTIABILITY at c you need continuity first, and then the left and right derivatives to agree as well. Differentiability implies continuity; continuity never implies differentiability.

This is how NDA hides simultaneous equations inside a limits question. 'Find a and b such that f is continuous' gives you one equation per join; 'such that f is differentiable' gives you two per join — one matching values, one matching slopes. A three-case definition with two unknowns is a two-equation solve wearing a disguise, and the printed answer is usually a + b rather than either separately.

4 worked examples from the bank

Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.

Example 1Limits & ContinuityEASY
The value of kk which makes f(x)={sin⁡xxx≠0kx=0f(x)=\begin{cases}\dfrac{\sin x}{x} & x\neq 0 \\ k & x=0\end{cases} continuous at x=0x=0, is

[Q92 · Apr · 2019]

Example 2Limits & ContinuityEASY
For what value of kk is the function f(x)={2x+14,x<0k,x=0(x+12)2,x>0f(x) = \begin{cases} 2x + \dfrac{1}{4}, & x < 0 \\ k, & x = 0 \\ \left(x + \dfrac{1}{2}\right)^{2}, & x > 0 \end{cases} continuous ?

[Q98 · Sep · 2019]

Example 3Limits & ContinuityMODERATE
If the function f(x)={a+bx,x<15,x=1b−ax,x>1f(x)=\begin{cases}a+bx, & x<1\\5, & x=1\\b-ax, & x>1\end{cases} is continuous, then what is the value of (a+b)(a+b)?

[Q93 · Apr · 2021]

Example 4DifferentiationHARD
Let f(x)=axx+1+b, x<1f(x)=\frac{ax}{x+1}+b,\ x<1 and x−1, 1≤x≤2\sqrt{x-1},\ 1\leq x\leq2.
If f(x)f(x) is differentiable at x=1x=1, then what is the value of (a+b)(a+b)?

[Q63 · Sep · 2023]

Variants to recognise

Same principle, different surfaces. Pattern-match these on test day.

  • Continuity at the join

    Left limit = right limit = f(c). Gives one equation per breakpoint. Missing that f(c) is its own case is the usual error.

  • Differentiability at the join

    Continuity first, then left derivative = right derivative. Two conditions, so two equations — enough to pin two unknowns.

  • Solving for parameters

    'Find a and b so that f is continuous/differentiable' is a simultaneous-equation problem. The answer asked for is often a + b.

  • Removable vs jump discontinuity

    If the one-sided limits agree but differ from f(c), redefining f(c) repairs it. If they disagree, nothing can.

  • Definitions given in prose

    Not every piecewise function is typeset with a brace — 'f(x) = ax/(x+1) + b, x < 1 and √(x−1), 1 ≤ x ≤ 2' is the same object.

Drill every piecewise definitions — continuity and differentiability at the join question

23 questions from the bank — paginated, with cart and Word-export support.

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Related principles

Often combined with this one — drill these next if you found the examples above tractable.