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. BC-only material. Assumptions: series are about x = 0 unless a centre is given; angles in radians; no calculator for Questions 1–5; in Questions 6 and 7 a calculator may be used for decimal values, given to 6 decimal places. Notation: “Σ from n = 0 to ∞” is an infinite sum, and “∫ from a to b of f(t) dt” is a definite integral. You may use the series for 1/(1 − x), eˣ, sin x and cos x without proof.
Question 1 (multiple choice · foundation)
Which of the following gives the Maclaurin series for f(x) = 1/(1 + 4x) and the interval on which it represents f?
- (A) Σ from n = 0 to ∞ of (−1)ⁿ 4ⁿ xⁿ, for −1/4 < x < 1/4
- (B) Σ from n = 0 to ∞ of (−1)ⁿ 4ⁿ xⁿ, for −4 < x < 4
- (C) Σ from n = 0 to ∞ of 4ⁿ xⁿ, for −1/4 < x < 1/4
- (D) Σ from n = 0 to ∞ of (−1)ⁿ 4xⁿ, for −1 < x < 1
Answer and explanation
Answer: (A). 1/(1 + 4x) = 1/(1 − (−4x)), a geometric series with a = 1 and r = −4x. So f(x) = Σ (−4x)ⁿ = Σ (−1)ⁿ 4ⁿ xⁿ = 1 − 4x + 16x² − 64x³ + …. It converges when |−4x| < 1, that is |x| < 1/4.
- (B) has the right series but solves |4x| < 1 wrongly, as |x| < 4. Try x = 1: the terms (−4)ⁿ grow, so the series diverges.
- (C) uses r = 4x and loses the minus sign. Its sum is 1/(1 − 4x), a different function.
- (D) puts the power on x only, as if r were −x with a = 4 in each term. (−4x)ⁿ means (−4)ⁿxⁿ.
Question 2 (multiple choice · core)
What is the coefficient of x⁶ in the Maclaurin series for x² cos(2x)?
- (A) 2/3
- (B) −2/3
- (C) 1/24
- (D) −4/45
Answer and explanation
Answer: (A). cos(2x) = 1 − (2x)²/2! + (2x)⁴/4! − (2x)⁶/6! + … = 1 − 2x² + (2/3)x⁴ − (4/45)x⁶ + …. Multiplying by x² gives x² − 2x⁴ + (2/3)x⁶ − (4/45)x⁸ + …. The x⁶ term comes from the x⁴ term of cos(2x): 16/24 = 2/3, with a plus sign.
- (B) uses the wrong sign pattern (− + − instead of + − +): the x⁴ term of cos is positive.
- (C) substitutes into x but not into the power: it uses x⁴/4! instead of (2x)⁴/4!.
- (D) is the coefficient of x⁶ in cos(2x) itself. Multiplying by x² shifts every power up by 2, so that term becomes x⁸.
Question 3 (multiple choice · core)
The function f is defined by f(x) = Σ from n = 1 to ∞ of xⁿ/(n · 5ⁿ) for −5 ≤ x < 5. Which of the following is f′(x) for −5 < x < 5?
- (A) 1/(5 − x)
- (B) 5/(5 − x)
- (C) x/(5 − x)
- (D) 1/(5 − x)²
Answer and explanation
Answer: (A). Differentiate term by term: f′(x) = Σ from n = 1 of n xⁿ⁻¹/(n · 5ⁿ) = Σ from n = 1 of xⁿ⁻¹/5ⁿ = 1/5 + x/25 + x²/125 + …. This is geometric with a = 1/5 and r = x/5, so f′(x) = (1/5)/(1 − x/5) = 1/(5 − x).
- (B) drops the factor 1/5 in the first term: it treats the series as 1 + x/5 + … .
- (C) forgets that differentiating lowers each power by one, so it keeps xⁿ instead of xⁿ⁻¹.
- (D) differentiates a second time, or differentiates 1/(5 − x) instead of recognising it.
Question 4 (multiple choice · stretch)
For −1 < x < 5, Σ from n = 0 to ∞ of (x − 2)ⁿ/3ⁿ⁺¹ = 1/(5 − x). Let G(x) = Σ from n = 0 to ∞ of (x − 2)ⁿ⁺¹/((n + 1) 3ⁿ⁺¹), the series obtained by integrating term by term from 2 to x. What is the interval of convergence of the series for G?
- (A) −1 < x < 5
- (B) −1 ≤ x < 5
- (C) −1 < x ≤ 5
- (D) −1 ≤ x ≤ 5
Answer and explanation
Answer: (B). Integration keeps the radius, 3, so the series converges for −1 < x < 5. Test the endpoints.
-
At x = 5: (x − 2)ⁿ⁺¹ = 3ⁿ⁺¹, so the terms are 1/(n + 1): the harmonic series, which diverges.
-
At x = −1: (x − 2)ⁿ⁺¹ = (−3)ⁿ⁺¹, so the terms are (−1)ⁿ⁺¹/(n + 1): −1 + 1/2 − 1/3 + …, which converges by the alternating series test.
-
(A) assumes the endpoints are the same as for the original geometric series. Only the radius is guaranteed.
-
(C) mixes up the endpoints: x = 5 gives the harmonic series, not the alternating one.
-
(D) assumes integration makes both endpoints converge.
Question 5 (constructed response · core)
(a) Write the first four terms and the general term of the Maclaurin series for 1/(1 + x). State its interval of convergence. (b) Use (a) to find the first four terms and the general term of the Maclaurin series for 1/(1 + x)². Explain your method. (c) Use (b) to find the exact value of Σ from n = 1 to ∞ of (−1)ⁿ⁺¹ n/3ⁿ⁻¹ = 1 − 2/3 + 3/9 − 4/27 + ….
Worked solution and suggested Marlbridge rubric
(a) 1/(1 + x) = 1/(1 − (−x)) = 1 − x + x² − x³ + … = Σ from n = 0 of (−1)ⁿ xⁿ, for −1 < x < 1 (a geometric series diverges at both endpoints).
(b) d/dx [1/(1 + x)] = −1/(1 + x)². So 1/(1 + x)² = −d/dx [1 − x + x² − x³ + x⁴ − …] = −(−1 + 2x − 3x² + 4x³ − …).
1/(1 + x)² = 1 − 2x + 3x² − 4x³ + … = Σ from n = 1 of (−1)ⁿ⁺¹ n xⁿ⁻¹, for −1 < x < 1.
(c) The given series is the series from (b) at x = 1/3, which is inside the interval: (−1)ⁿ⁺¹ n (1/3)ⁿ⁻¹ = (−1)ⁿ⁺¹ n/3ⁿ⁻¹. So the sum is 1/(1 + 1/3)² = 1/(16/9) = 9/16.
Check: the partial sums 1, 0.333, 0.667, 0.519, … settle towards 0.5625. ✓
| Point | What earns it |
|---|---|
| 1 | Correct series and general term in (a), with r = −x |
| 1 | Interval −1 < x < 1 in (a) |
| 1 | Differentiates term by term and deals with the minus sign from d/dx [1/(1 + x)] = −1/(1 + x)² |
| 1 | Correct terms and general term in (b), starting at n = 1 |
| 1 | Identifies x = 1/3, notes that it is inside the interval, and gets 9/16 |
Total: 5 points. Acceptable alternative for (b): square the series from (a) and collect terms up to x³; the general term then needs the pattern to be stated. A decimal alone (0.5625) for (c) does not earn the last point, because the question asks for an exact value.
Question 6 (constructed response · stretch)
Let F(x) = ∫ from 0 to x of (1 − e^(−t))/t dt, where the integrand is taken to be 1 at t = 0 (its limit there).
(a) Write the first four nonzero terms and the general term of the Maclaurin series for (1 − e^(−t))/t. (b) Write the first four nonzero terms and the general term of the Maclaurin series for F(x). For which x does it converge? (c) Use the first three nonzero terms of (b) to estimate F(1/2). Show that the estimate differs from F(1/2) by less than 0.001. (d) Use your series to find F″(0).
Worked solution and suggested Marlbridge rubric
(a) Substitute −t into eˣ: e^(−t) = 1 − t + t²/2! − t³/3! + t⁴/4! − …. 1 − e^(−t) = t − t²/2! + t³/3! − t⁴/4! + …. Divide by t: (1 − e^(−t))/t = 1 − t/2 + t²/6 − t³/24 + … = Σ from n = 0 of (−1)ⁿ tⁿ/(n + 1)!
(b) Integrate term by term from 0 to x (no extra constant, because F(0) = 0): F(x) = x − x²/4 + x³/18 − x⁴/96 + … = Σ from n = 0 of (−1)ⁿ xⁿ⁺¹/((n + 1)(n + 1)!) The series for eˣ converges for all x, and substitution, division by t and integration keep an infinite radius, so the series converges for all real x.
(c) F(1/2) ≈ 1/2 − 1/16 + 1/144 = 4/9 ≈ 0.444444. At x = 1/2 the series alternates and its terms decrease in size to 0, so the error is less than the size of the next term: (1/2)⁴/96 = 1/1536 ≈ 0.000651 < 0.001. ✓ (A calculator gives F(1/2) ≈ 0.443842, so the actual error is about 0.000602.)
(d) The coefficient of x² is −1/4 = F″(0)/2!, so F″(0) = −1/2.
| Point | What earns it |
|---|---|
| 1 | Correct substitution into the series for eˣ |
| 1 | Correct series for the integrand, including the general term |
| 1 | Correct series for F with no extra constant, and general term |
| 1 | Converges for all real x, with a reason |
| 1 | Estimate 0.444444 (or 4/9) |
| 1 | Error bound 1/1536 from the alternating series error bound, with the conditions stated |
| 1 | F″(0) = −1/2 from the coefficient |
Total: 7 points. The error-bound point needs a reason (alternating, decreasing terms, limit 0); a calculator comparison alone does not earn it.
Question 7 (constructed response · stretch)
Let h(x) = ∫ from 0 to x of 1/(1 + t⁴) dt.
(a) Write the first four nonzero terms and the general term of the Maclaurin series for 1/(1 + t⁴). For which t does it converge? (b) Find the Maclaurin series for h(x): give its first four nonzero terms and its general term. (c) Find the interval of convergence of the series in (b). Justify your answer at each endpoint. (d) Find h⁽⁹⁾(0). (e) Use the first two nonzero terms of (b) to approximate h(1/2), and give a bound on the error.
Worked solution and suggested Marlbridge rubric
(a) Replace x by −t⁴ in 1/(1 − x): 1/(1 + t⁴) = 1 − t⁴ + t⁸ − t¹² + … = Σ from n = 0 of (−1)ⁿ t^(4n), for |t⁴| < 1, that is −1 < t < 1.
(b) Integrate from 0 to x (h(0) = 0, so no constant): h(x) = x − x⁵/5 + x⁹/9 − x¹³/13 + … = Σ from n = 0 of (−1)ⁿ x^(4n+1)/(4n + 1)
(c) The radius is still 1. Test the endpoints.
- x = 1: Σ (−1)ⁿ/(4n + 1) = 1 − 1/5 + 1/9 − …. The terms alternate, decrease in size and tend to 0, so it converges (alternating series test).
- x = −1: (−1)^(4n+1) = −1, so the terms are (−1)ⁿ⁺¹/(4n + 1) = −1 + 1/5 − 1/9 + …, the same series multiplied by −1, so it converges.
Interval of convergence: −1 ≤ x ≤ 1. Both endpoints were gained by integrating.
(d) The x⁹ term is +x⁹/9, and its coefficient equals h⁽⁹⁾(0)/9!. So h⁽⁹⁾(0) = 9!/9 = 8! = 40 320.
(e) h(1/2) ≈ 1/2 − (1/2)⁵/5 = 1/2 − 1/160 = 0.49375. The series is alternating with decreasing terms, so the error is less than the next term, (1/2)⁹/9 = 1/4608 ≈ 0.000217. (A calculator gives h(1/2) ≈ 0.493958, an error of about 0.000208.)
| Point | What earns it |
|---|---|
| 1 | Correct series for 1/(1 + t⁴) with general term |
| 1 | Convergence condition −1 < t < 1 |
| 1 | Correct series for h with general term and no extra constant |
| 1 | Radius 1 and convergence at x = 1, with the test named |
| 1 | Convergence at x = −1 with a reason, and the interval −1 ≤ x ≤ 1 |
| 1 | h⁽⁹⁾(0) = 8! = 40 320 from the coefficient |
| 1 | 0.49375 with error bound 1/4608 (about 0.000217), justified |
Total: 7 points. Acceptable alternatives: in (c), x = −1 may be handled by noting that h is an odd function, so the series at −1 is minus the series at 1; in (e), an error bound left as (1/2)⁹/9 earns the point.
How did you do?
- Q1 or Q5(a) wrong: reread “Rewriting into geometric form” in the study guide.
- Q2 or Q6(a) wrong: practise substitution into every power, and multiplying or dividing by a power of x.
- Q3 or Q5(b) wrong: revisit “Term-by-term differentiation and integration”, including where the sum starts.
- Q4 or Q7(c) wrong: work through Worked example 2, part (b), on testing endpoints again.
- Q6(b), Q6(d), Q7(b) or Q7(d) wrong: check the constant after integrating and the rule f⁽ⁿ⁾(0) = n! × coefficient.
Then tick off the topic checklist.