These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator and exact answers. Notation: lim (x → c) f(x) means “the limit as x approaches c of f(x)”; x → c⁻ means from the left and x → c⁺ means from the right. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
A function f satisfies lim (x → 4⁻) f(x) = 3, lim (x → 4⁺) f(x) = 3 and f(4) = 1. Which statement is true?
- (A) f has a removable discontinuity at x = 4.
- (B) f has a jump discontinuity at x = 4.
- (C) f has a vertical asymptote at x = 4.
- (D) f is continuous at x = 4, because f(4) is defined.
Answer and explanation
Answer: (A). Both one-sided limits equal 3, so lim (x → 4) f(x) = 3 exists. But f(4) = 1 ≠ 3. A limit that exists but disagrees with the function value is a removable discontinuity: the graph has a hole at (4, 3) and a separate point at (4, 1).
- (B) A jump needs the one-sided limits to be different. Here they are equal.
- (C) A vertical asymptote needs unbounded behaviour on at least one side. Both limits here are finite.
- (D) Being defined is only one condition. The limit must also equal f(4), and 3 ≠ 1.
Question 2 (multiple choice · core)
Let h(x) = (x² − 16)/(x² − 3x − 4). Which statement describes the discontinuities of h?
- (A) A removable discontinuity at x = 4 and a vertical asymptote at x = −1
- (B) Vertical asymptotes at both x = 4 and x = −1
- (C) A removable discontinuity at x = −1 and a vertical asymptote at x = 4
- (D) Removable discontinuities at both x = 4 and x = −1
Answer and explanation
Answer: (A). Factor: h(x) = (x − 4)(x + 4)/((x − 4)(x + 1)). For x ≠ 4 and x ≠ −1, h(x) = (x + 4)/(x + 1). At x = 4 the factor cancels and the limit is 8/5, but h(4) is undefined: a removable discontinuity (a hole at (4, 8/5)). At x = −1 the factor (x + 1) stays in the denominator, and substitution gives 3/0, so h is unbounded near x = −1: a vertical asymptote.
- (B) assumes every zero of the denominator gives an asymptote, without factoring the numerator.
- (C) swaps the two points. The factor that cancels, (x − 4), gives the hole.
- (D) assumes every zero of the denominator gives a hole. Only the factor shared with the numerator cancels.
Question 3 (multiple choice · core)
What kind of discontinuity does q(x) = (x² − 5x)/|x − 5| have at x = 5?
- (A) Removable
- (B) Jump
- (C) Vertical asymptote
- (D) None: q is continuous at x = 5.
Answer and explanation
Answer: (B). Write x² − 5x = x(x − 5). For x > 5, |x − 5| = x − 5, so q(x) = x and lim (x → 5⁺) q(x) = 5. For x < 5, |x − 5| = −(x − 5), so q(x) = −x and lim (x → 5⁻) q(x) = −5. Both one-sided limits are finite but different, so the discontinuity is a jump.
- (A) treats |x − 5| as if it were x − 5 and cancels, getting x → 5 from both sides. The absolute value changes sign on the left.
- (C) sees a zero denominator and stops. After rewriting, both sides are finite.
- (D) q(5) is undefined, so q cannot be continuous at 5.
Question 4 (multiple choice · core)
A function r has r(2) = 0. The table shows some values of r near x = 2.
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| r(x) | −10 | −100 | −1000 | 1000 | 100 | 10 |
Which conclusion is most consistent with the table?
- (A) r has a removable discontinuity at x = 2, because the values on the two sides cancel out to 0.
- (B) r has a jump discontinuity at x = 2.
- (C) r has a discontinuity due to a vertical asymptote at x = 2.
- (D) r is continuous at x = 2, because r(2) is defined.
Answer and explanation
Answer: (C). As x gets closer to 2, the values grow in size without settling: −10, −100, −1000 on the left and 1000, 100, 10 on the right (reading towards 2). This suggests r(x) → −∞ from the left and r(x) → +∞ from the right, which is the behaviour of a vertical asymptote. A table only suggests this; a formula or graph would be needed to justify it.
- (A) averages the two sides. A limit is not an average; each side must approach the same finite number.
- (B) A jump needs each side to approach a finite number. Neither side does here.
- (D) A defined value does not make a function continuous. The limit must exist and equal r(2).
Question 5 (graph · core)
The graph of a function f is shown for −4 ≤ x ≤ 5.
(a) For each of c = −1, c = 1 and c = 3, state lim (x → c⁻) f(x), lim (x → c⁺) f(x) and f(c). (b) Classify the discontinuity at each of these three points, and justify each answer using your results from (a). (c) Is f continuous at x = 0? Give a reason.
Worked solution and suggested Marlbridge rubric
(a)
| c | Left-hand limit | Right-hand limit | f(c) |
|---|---|---|---|
| −1 | 1 | 1 | 3 |
| 1 | 3 | 0.5 | 0.5 |
| 3 | +∞ (unbounded) | −∞ (unbounded) | undefined |
(b) At x = −1: removable. Both one-sided limits are 1, so lim (x → −1) f(x) = 1 exists, but f(−1) = 3 ≠ 1. At x = 1: jump. The one-sided limits are 3 and 0.5, both finite but different. (f(1) = 0.5 matches the right side, but that does not repair the break.) At x = 3: vertical asymptote. f(x) → +∞ as x → 3⁻ and f(x) → −∞ as x → 3⁺.
(c) Yes. f(0) = 2, and the graph is an unbroken line through (0, 2), so lim (x → 0) f(x) = 2 = f(0).
| Point | What earns it |
|---|---|
| 1 | All limits and values in (a) correct at x = −1 and x = 1 |
| 1 | Removable at x = −1, citing that the limit exists (1) and differs from f(−1) = 3 |
| 1 | Jump at x = 1, citing two different finite one-sided limits |
| 1 | Vertical asymptote at x = 3, citing unbounded behaviour on at least one side, and a correct reason for (c) |
Writing “∞” in (a) for x = 3 is acceptable if the student also says the limit does not exist as a number.
Question 6 (constructed response · core)
Let R(x) = (2x² − 2x − 12)/(x² − 9).
(a) Find every value of x at which R is discontinuous. (b) Classify each discontinuity. Justify each answer using limits. (c) Give the coordinates of any hole in the graph of R. (d) A student says: “R(3) and R(−3) are both undefined, so the two discontinuities must be the same type.” Explain the student’s error.
Worked solution and suggested Marlbridge rubric
(a) x² − 9 = (x − 3)(x + 3) is zero at x = 3 and x = −3. R is a rational function, so these are its only discontinuities.
(b) Factor the numerator: 2x² − 2x − 12 = 2(x² − x − 6) = 2(x − 3)(x + 2). For x ≠ ±3,
R(x) = 2(x − 3)(x + 2)/((x − 3)(x + 3)) = 2(x + 2)/(x + 3).
At x = 3: lim (x → 3) R(x) = 2(5)/6 = 5/3. The limit exists but R(3) is undefined, so the discontinuity is removable.
At x = −3: the simplified form gives 2(−1)/0 = −2/0, a nonzero number over 0. For x slightly greater than −3, x + 3 is small and positive while 2(x + 2) is close to −2, so R(x) → −∞ as x → −3⁺. From the left, x + 3 is small and negative, so R(x) → +∞ as x → −3⁻. This is a vertical asymptote.
(c) The hole is at (3, 5/3).
(d) Being undefined only tells you where to look. The type depends on the limits. At x = 3 the factor (x − 3) cancels and the limit is the finite number 5/3. At x = −3 the factor (x + 3) remains in the denominator and R is unbounded.
| Point | What earns it |
|---|---|
| 1 | Both x = 3 and x = −3, from factoring the denominator |
| 1 | Simplifies to 2(x + 2)/(x + 3) and states removable at x = 3 with limit 5/3 |
| 1 | Vertical asymptote at x = −3, with the sign of R on at least one side correctly explained |
| 1 | Hole at (3, 5/3) and an explanation in (d) that the type depends on the limit, not on R(c) being undefined |
A table of values such as R(−2.99) = −198 and R(−3.01) = 202 supports the asymptote but does not replace the algebra.
Question 7 (constructed response · stretch)
A function w is defined by
- w(x) = (x² − 4)/(x − 2) for x < 2
- w(2) = 3
- w(x) = x² for 2 < x ≤ 3
- w(x) = 1/(x − 5) for x > 3
(a) w has exactly three discontinuities. Find them, classify each one and justify each classification with limits. (b) A student says: “w has a vertical asymptote at x = 2, because the first formula has x − 2 in the denominator.” Explain why the student is wrong. (c) Explain why no change to the single value w(3) could make w continuous at x = 3.
Worked solution and suggested Marlbridge rubric
(a) Candidates: x = 2 and x = 3, where the rule changes, and x = 5, where the last denominator is zero.
At x = 2: for x < 2, (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2, so lim (x → 2⁻) w(x) = 4. On the right, lim (x → 2⁺) x² = 4. The limit is 4, but w(2) = 3. Removable.
At x = 3: w(3) = 3² = 9 and lim (x → 3⁻) w(x) = 9. On the right, lim (x → 3⁺) 1/(x − 5) = 1/(−2) = −1/2. Both finite, 9 ≠ −1/2. Jump.
At x = 5: as x → 5⁺, x − 5 is small and positive, so w(x) → +∞. As x → 5⁻, w(x) → −∞. Vertical asymptote.
(b) The factor (x − 2) cancels with the numerator, so near x = 2 on the left w behaves like x + 2, which approaches 4, a finite number. Nothing is unbounded, so there is no asymptote. The break at x = 2 is a hole at (2, 4) with a separate point at (2, 3).
(c) Continuity at x = 3 needs lim (x → 3) w(x) to exist. The one-sided limits are 9 and −1/2, so the two-sided limit does not exist. Changing w(3) changes only the single point, not the one-sided limits, so the break remains whatever value w(3) takes.
| Point | What earns it |
|---|---|
| 1 | Removable at x = 2: simplifies the first piece, finds both one-sided limits equal to 4 and compares with w(2) = 3 |
| 1 | Jump at x = 3: one-sided limits 9 and −1/2, both finite and different |
| 1 | Vertical asymptote at x = 5, with unbounded behaviour on at least one side |
| 1 | (b): the factor cancels, so the left-hand limit at 2 is finite (4) |
| 1 | (c): the two-sided limit does not exist, and the value w(3) does not affect the one-sided limits |
How did you do?
- Q1 or Q4 wrong: reread “The three types you need” and Figure 1 in the study guide.
- Q2 or Q6 wrong: redo Worked example 1 and the “Rational functions: factor first” section.
- Q3 wrong: see “Piecewise functions and absolute values” and rewrite the absolute value on each side.
- Q5 or Q7 wrong: redo Worked example 2 and practise the “Writing the justification” sentences.
Then tick off the topic checklist.