Where the standard functions stay unbroken, and where they jump

What to assume they know

• Definition 1 and Definition 2 from the previous topic, and the three demands hidden in the first
• One-sided limits, and evaluating a branch as the input approaches from inside that branch
• The greatest integer of a real number, from Class XI
• The modulus function, and rewriting it as two branches
• Reading a piecewise definition and identifying every input where the rule changes
• Sets written in builder notation, and the union of two sets
• The idea that a function's domain may exclude an isolated input

What they should be able to do

• Split the real line at every input where a piecewise rule changes, and decide the easy regions in one line each
• Compute both one-sided limits at a join and state the verdict there
• Distinguish a genuine break from a case split whose two branches already agree at the boundary
• Explain why an input missing from the domain is not a point of discontinuity
• Give an example of a continuous function whose graph cannot be drawn in one stroke, and say why that is not a contradiction
• Read a step graph correctly, saying which end of each step is included
• Argue that the greatest integer function fails at every integer, using the two one-sided values rather than the picture
• Locate all points of discontinuity of a stated piecewise function and count them
• Recognise a function that is continuous everywhere despite an alarming definition

Where it usually goes wrong

• "Two cases means a break." Four exercises in this set — Q10, Q11, Q16 and Q34 — are written in cases and break nowhere. The cases mark where the rule changes, which need not be where the function changes.
• "One case means no break." The greatest integer function is a single rule and fails at every integer. So is the fractional part of Q19.
• "A gap in the domain is a break." Example 12 is the chapter's own answer: the function is continuous, and the graph still needs the pen lifted, because the lifting happens where nothing is defined.
• "Continuous means drawable in one stroke." Example 12 kills this on Part I p. 111 in the chapter's own words. Keep the pen picture only for the case where the domain is an interval.
• "The value at the join decides everything." In Example 11 the value at the join is set to a third number entirely and the verdict does not move, because the two sides already disagreed. Set the value last, not first.
• "A step graph includes both ends of each step." Each step of Fig 5.8 owns its left end and not its right. A student who reads both ends as included will claim the function takes two values at each integer, which no function does.
• "Failing at infinitely many points is worse than failing at one." It is not a different kind of failure. The verdict at each integer is reached by the same two-line comparison used at a single join.
• "If the two sides agree, I am finished." Only if the input is in the domain. Example 12 has both sides agreeing at zero and no verdict to give.

Questions to check understanding

• Find every point of discontinuity of a stated piecewise function — the form of Exercise 5.1 Q6 to Q16
• Decide continuity of a function whose branches are each simple, by comparing the two one-sided values at each join
• Explain why a stated function with an excluded input is continuous
• Show that the greatest integer function, or the fractional part, fails at every integer — the form of Exercise 5.1 Q19
• Given a step graph, read off what the rule returns at an integer, and what it returns just below that integer
• Decide continuity for a difference of two moduli, and justify the answer — the form of Exercise 5.1 Q34
• Produce a function that is continuous and whose graph cannot be drawn without lifting the pen

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• Example 10 with Fig 5.4 (Part I p. 110). Add two below and at the join, subtract two above it, with the join at input one. The chapter runs three cases: below, above, and at. Verified: the two easy cases are one substitution each; at the join the left side gives three and the right gives minus one, so the join is the sole bad input. Read off a three-hundred-dot the printed page, the figure is two parallel segments with a filled dot at the upper endpoint and a hollow circle at the lower one, and the axis carries ticks at minus three, minus two, minus one, one, two and three.
• Example 11 with Fig 5.5 (Part I p. 110). The same two branches, but the join is given its own third case with value zero. Verified: both one-sided values are unchanged, so the verdict is unchanged and the join is still the only bad input. A printed omission a reviewer should see: on a six-hundred-dot the printed page the figure carries a hollow circle at both segment endpoints and draws no marker at all at the join's own value, which is the one thing that distinguishes this example from the previous one. Show the point; the textbook figure does not.
• Example 12 with Fig 5.6 (Part I p. 111). Add two below zero, subtract from two above zero, and nothing at zero. The chapter writes the domain as a union of two open pieces and shows continuity on each. Verified: the two branches agree in the limit at zero — both approach two — yet the function still has no value there, so there is no verdict to give at zero. The figure is an inverted V with a single hollow circle at the apex, confirmed on a three-hundred-dot the printed page. The chapter's own closing remark is the payoff: the pen must be lifted, but only where the function is absent.
• Example 13 with Fig 5.7 (Part I p. 111). Squaring below zero, the identity at and above zero. The chapter partitions the line into three sets — the negatives, the single point zero, and the positives — and handles each. Verified: both branches give zero at zero and both one-sided limits are zero, so the join is fine and the function is continuous everywhere. The figure shows the parabola arm meeting the ray at the origin with no marker at all, because none is needed. Labelled points on the figure include the pairs at input minus two, minus one, one and two.
• Example 14 (Part I p. 112). Every polynomial is continuous, in four lines, because the limit of a polynomial at an input is the polynomial evaluated there. This is the result that makes most of Exercise 5.1 one line long.
• Example 15 with Fig 5.8 (Part I pp. 112–113). The greatest integer function. The argument splits in two: at a non-integer input the value is locally constant, so the verdict is immediate; at an integer, the chapter produces a small positive quantity and evaluates the rule just below and just above, getting two values one apart. Verified: the two one-sided values at an integer differ by exactly one, so they can never agree, and the function therefore fails at every integer. Read off a three-hundred-dot the printed page, each step carries a filled dot at its left end and a hollow circle at its right end, and the axis ticks are labelled as coordinate pairs rather than as bare numbers — the horizontal ones as pairs with second entry zero, the vertical ones as pairs with first entry zero.
• Exercise 5.1 Q6 to Q9 (Part I p. 116). Q6 is two linear branches joining at two. Q7 is three branches with joins at minus three and three. Q8 is a modulus over the identity away from zero, and zero at zero. Q9 is the identity over a modulus below zero, and minus one at and above zero. Verified: Q6 fails only at two, where the sides give seven and one. Q7 is fine at minus three — all three quantities there are six — and fails at three, where the sides give minus six and twenty. Q8 fails only at zero, where the sides give minus one and one. Q9 is the trap: below zero the quotient is already minus one, so the function is the constant minus one on the whole line and has no bad input at all.
• Exercise 5.1 Q10 to Q13 (Part I p. 116). Q10 joins a linear branch to a quadratic at one; Q11 joins a cubic-minus-three to a quadratic-plus-one at two; Q12 joins a tenth power minus one to a square at one; Q13 joins a plus-five branch to a minus-five branch at one. Verified: Q10 and Q11 have no bad input — in Q10 both sides give two, in Q11 both give five. Q12 fails at one, where the sides give zero and one. Q13 fails at one, where the sides give six and minus four. Q10 and Q11 against Q12 and Q13 is the whole of section 4: four items of identical shape, two of which are continuous.
• Exercise 5.1 Q14 to Q16 (Part I p. 117). Three-branch functions with joins at named inputs. Verified: Q14 is defined on a closed interval and fails at both interior joins, where the sides give three against four and four against five. Q15 fails only at the upper join, where the value is zero and the branch above gives four. Q16 fails nowhere: at the lower join all three quantities are minus two, and at the upper join all three are two.
• Exercise 5.1 Q19 (Part I p. 117). The input minus its greatest integer, shown to fail at every integer. Verified: the difference is the fractional part, so just below an integer it is close to one and at the integer it is zero. The same two-case argument as Example 15, and the exercise.
• Exercise 5.1 Q23 (Part I p. 117). A sine over the identity below zero, and a linear branch at and above zero. Verified: the left side needs the standard limit of a sine over its own argument, which is one, and the right side and the value are both one, so there is no bad input. That standard limit is Class XI material and is not stated anywhere in this chapter — record the dependency rather than sourcing it here.
• Exercise 5.1 Q34 (Part I p. 118). A modulus minus a shifted modulus. Verified: both terms are continuous everywhere and a difference of continuous functions is continuous, so the answer is that there are no bad inputs. The machinery for that sentence is the next topic; here it is a good closing puzzle, because the definition looks like it must break somewhere and does not.

Figures to have open

• Redraws of Fig 5.4 and Fig 5.5 (Part I p. 110) as a matched pair, with the endpoint fills exactly as listed above and with the join's own value drawn in on the second, flagged as the explanation's addition.
• A redraw of Fig 5.6 (Part I p. 111): two rays rising and falling to a common apex carrying a single hollow circle, with the excluded input also marked on the horizontal axis.
• A redraw of Fig 5.7 (Part I p. 111): a parabola arm on the left meeting a ray on the right at the origin, with the four labelled points and their dotted guides.
• A redraw of Fig 5.8 (Part I p. 112) covering at least the steps from minus three to three, each with a filled left end and a hollow right end. Section 10 needs one step of this drawing enlarged.
• A four-panel comparison for section 4 built from Exercise 5.1 Q10 to Q13. The items are the chapter's; the side-by-side layout is added here.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 5, §5.2 Continuity, Examples 10 and 11 with Fig 5.4 and Fig 5.5, Part I p. 110
• Examples 12 and 13 with Fig 5.6 and Fig 5.7, Part I p. 111
• Examples 14 and 15 with Fig 5.8, Part I p. 112, and the two-case argument concluding Example 15, Part I p. 113
• Exercise 5.1, questions 6 to 16, Part I pp. 116–117
• Exercise 5.1, questions 19, 23 and 34, Part I pp. 117–118