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, angles in radians, and exact answers unless stated. Notation: lim (x → c) f(x) means “the limit as x approaches c of f(x)”; c⁻ and c⁺ mean from the left and from the right. “Continuous at c” uses the three-condition definition: f(c) exists, lim (x → c) f(x) exists, and the two are equal. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let f(x) = (x² − 4)/(x − 2) for x ≠ 2, and f(2) = 3. Which statement is true?
- (A) f is not continuous at x = 2, because f(2) is undefined.
- (B) f is not continuous at x = 2, because lim (x → 2) f(x) does not exist.
- (C) f is not continuous at x = 2, because lim (x → 2) f(x) exists but is not equal to f(2).
- (D) f is continuous at x = 2.
Answer and explanation
Answer: (C). Condition 1: f(2) = 3 is given, so f(2) exists. Condition 2: for x ≠ 2, f(x) = (x − 2)(x + 2)/(x − 2) = x + 2, so lim (x → 2) f(x) = 4. Condition 3: 4 ≠ 3, so this condition fails.
- (A) substitutes 2 into the fraction and gets 0/0. But the fraction is only used for x ≠ 2; f(2) = 3 is given separately.
- (B) treats 0/0 as “no limit”. After factoring, the limit is 4.
- (D) checks only that f(2) exists and that the limit exists, and forgets to compare them.
Question 2 (multiple choice · core)
Let
- f(x) = x³ + 1 for x < 1
- f(x) = 3x − 1 for x ≥ 1
Which statement is true?
- (A) f is continuous at x = 1.
- (B) f is not continuous at x = 1, because f(1) is undefined.
- (C) f is not continuous at x = 1, because the one-sided limits at x = 1 are different.
- (D) f is not continuous at x = 1, because a function given by two formulas cannot be continuous where the formula changes.
Answer and explanation
Answer: (A). f(1) = 3(1) − 1 = 2, from the second piece (x ≥ 1). From the left, lim (x → 1⁻) (x³ + 1) = 2. From the right, lim (x → 1⁺) (3x − 1) = 2. So lim (x → 1) f(x) = 2 = f(1), and f is continuous at x = 1.
- (B) misses that “x ≥ 1” includes x = 1, so the second piece defines f(1).
- (C) assumes two different formulas must give two different limits. Work them out: both are 2.
- (D) is a common belief, but pieces can meet. The definition, not the number of formulas, decides continuity.
Question 3 (multiple choice · core)
The table gives information about a function f at four values of c.
| c | f(c) | lim (x → c⁻) f(x) | lim (x → c⁺) f(x) |
|---|---|---|---|
| 1 | 2 | 2 | 3 |
| 2 | undefined | 4 | 4 |
| 3 | 5 | 5 | 5 |
| 4 | 1 | 6 | 6 |
At which value of c is f continuous?
- (A) c = 1
- (B) c = 2
- (C) c = 3
- (D) c = 4
Answer and explanation
Answer: (C). At c = 3, f(3) = 5, both one-sided limits are 5, so lim (x → 3) f(x) = 5 = f(3). All three conditions hold.
- (A) f(1) exists, but the one-sided limits are 2 and 3, so the limit does not exist (condition 2 fails). Matching the left-hand limit is not enough.
- (B) the limit is 4, but f(2) is undefined (condition 1 fails).
- (D) f(4) = 1 and the limit is 6. Both exist, but they differ (condition 3 fails).
Question 4 (multiple choice · core)
Which of the following functions is continuous at x = 0?
- (A) f(x) = (sin x)/x
- (B) f(x) = (sin x)/x for x ≠ 0, and f(0) = 1
- (C) f(x) = |x|/x for x ≠ 0, and f(0) = 0
- (D) f(x) = 1/x² for x ≠ 0, and f(0) = 0
Answer and explanation
Answer: (B). From Topic 1.8, lim (x → 0) (sin x)/x = 1. In (B), f(0) = 1 is defined, the limit is 1, and they are equal.
- (A) has the right limit, but (sin x)/x is undefined at x = 0, so condition 1 fails.
- (C) f(0) = 0 exists, but |x|/x is −1 for x < 0 and 1 for x > 0. The one-sided limits are −1 and 1, so condition 2 fails.
- (D) f(0) = 0 exists, but 1/x² grows without bound from both sides. An infinite limit does not exist as a number, so condition 2 fails.
Question 5 (constructed response · core)
Let f be defined by
- f(x) = x² − 2 for x < −1
- f(x) = x for −1 ≤ x < 2
- f(x) = 5 − x² for x ≥ 2
(a) Use the definition of continuity to decide whether f is continuous at x = −1. (b) Use the definition of continuity to decide whether f is continuous at x = 2. (c) For any point where f is not continuous, name the condition that fails and the type of discontinuity.
Worked solution and suggested Marlbridge rubric
(a) f(−1) = −1, from the middle piece (−1 ≤ x). From the left, lim (x → −1⁻) (x² − 2) = 1 − 2 = −1. From the right, lim (x → −1⁺) x = −1. So lim (x → −1) f(x) = −1 = f(−1). f is continuous at x = −1.
(b) f(2) = 5 − 4 = 1, from the last piece (x ≥ 2). From the left, lim (x → 2⁻) x = 2. From the right, lim (x → 2⁺) (5 − x²) = 1. The one-sided limits are 2 and 1, which are not equal, so lim (x → 2) f(x) does not exist. f is not continuous at x = 2.
(c) At x = 2, condition 2 fails (the limit does not exist). Condition 1 holds, since f(2) = 1. The one-sided limits are finite but different, so this is a jump discontinuity.
| Point | What earns it |
|---|---|
| 1 | At x = −1: f(−1) = −1 and both one-sided limits equal −1, each found from the correct piece |
| 1 | Concludes continuous at x = −1 because the limit equals f(−1) (comparison stated) |
| 1 | At x = 2: one-sided limits 2 and 1, so the limit does not exist; not continuous |
| 1 | Names condition 2 as the failure and identifies a jump discontinuity |
A graph or a single substitution without one-sided limits earns no mark for (a) or (b).
Question 6 (constructed response · core)
Let g(x) = (√(x + 4) − 2)/x for x ≠ 0, and g(0) = 1/4.
(a) Find lim (x → 0) g(x), showing your algebra. (b) Is g continuous at x = 0? Justify using the definition. (c) A classmate writes: “Putting x = 0 into the formula gives 0/0, so g is not continuous at 0.” Explain the error. (d) Suppose g(0) were redefined as 0, with the formula unchanged. Which condition of the definition would fail?
Worked solution and suggested Marlbridge rubric
(a) Substitution gives (2 − 2)/0 = 0/0, so rewrite. Multiply top and bottom by the conjugate √(x + 4) + 2. The top becomes (x + 4) − 4 = x. So for x ≠ 0,
g(x) = x/(x(√(x + 4) + 2)) = 1/(√(x + 4) + 2).
So lim (x → 0) g(x) = 1/(√4 + 2) = 1/4.
(b) g(0) = 1/4 exists. lim (x → 0) g(x) = 1/4 exists. They are equal, so g is continuous at x = 0.
(c) The formula is only used for x ≠ 0, so 0/0 says nothing about g(0), which is given as 1/4. And 0/0 does not decide the limit either: after rewriting, the limit is 1/4. Continuity depends on the three conditions, which all hold.
(d) g(0) = 0 would still exist and the limit would still be 1/4, but 1/4 ≠ 0, so condition 3 would fail.
| Point | What earns it |
|---|---|
| 1 | Multiplies by the conjugate and simplifies to 1/(√(x + 4) + 2) for x ≠ 0 |
| 1 | Limit = 1/4 |
| 1 | States g(0) = 1/4 and the limit = 1/4, so g is continuous at 0 (values quoted) |
| 1 | (c) and (d): the formula is not used at x = 0, so 0/0 decides nothing; with g(0) = 0, condition 3 fails because 1/4 ≠ 0 |
Optional check: √4.01 ≈ 2.0025, so g(0.01) ≈ 0.0025/0.01 = 0.25, close to 1/4. A table alone earns no mark for (a).
Question 7 (constructed response · stretch)
Let h(x) = (x² − 3x)/|x − 3| for x ≠ 3, and h(3) = 3.
(a) Find lim (x → 3⁻) h(x) and lim (x → 3⁺) h(x). (b) Is h continuous at x = 3? Justify using the definition. (c) Is h continuous at x = 0? Justify using the definition. (d) A student says: “h(3) = 3 is wrong. If we choose the right value for h(3), h will be continuous at 3.” Is the student correct? Explain.
Worked solution and suggested Marlbridge rubric
(a) Factor the top: x² − 3x = x(x − 3).
- For x < 3, |x − 3| = −(x − 3), so h(x) = x(x − 3)/(−(x − 3)) = −x. So lim (x → 3⁻) h(x) = −3.
- For x > 3, |x − 3| = x − 3, so h(x) = x. So lim (x → 3⁺) h(x) = 3.
(b) h(3) = 3 exists. But the one-sided limits are −3 and 3, so lim (x → 3) h(x) does not exist. Condition 2 fails, so h is not continuous at x = 3. (The right-hand limit happens to equal h(3), but that is not enough: the two-sided limit must exist.)
(c) Near x = 0 we have x < 3, so h(x) = −x. h(0) = 0, and lim (x → 0) h(x) = lim (x → 0) (−x) = 0. They are equal, so h is continuous at x = 0.
(d) The student is not correct. Changing h(3) changes only condition 1 or 3. Condition 2 does not depend on h(3) at all: the one-sided limits are −3 and 3 whatever h(3) is, so the limit never exists and h can never be continuous at 3.
| Point | What earns it |
|---|---|
| 1 | Uses the sign of x − 3 on each side to simplify h to −x (left) and x (right) |
| 1 | One-sided limits −3 and 3, and concludes h is not continuous at 3 because the limit does not exist |
| 1 | At x = 0: h(0) = 0 and lim (x → 0) h(x) = 0, so continuous (values quoted) |
| 1 | Explains that no value of h(3) can help because the limit at 3 does not exist |
Acceptable alternative for (a): a table of values on each side (for example h(2.9) = −2.9 and h(3.1) = 3.1) supports the one-sided limits, but the algebra is needed for full credit.
How did you do?
- Q1 or Q6 wrong: reread “The three-condition definition” and Worked example 2 in the study guide. A formula that gives 0/0 is not the end of the story.
- Q2 or Q5 wrong: redo Worked example 1 and find both one-sided limits at every piecewise boundary.
- Q3 or Q4 wrong: match each row or option to a panel of Figure 1 and name the condition that fails.
- Q7 wrong: see Worked example 3 on absolute values and why no value of h(c) can fix a jump.
Then tick off the topic checklist.