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: every function named has derivatives of all orders on the intervals used; angles in radians; no calculator for Questions 1–4 and 6; in Questions 5 and 7 a calculator may be used for arithmetic, with decimals to at least 4 significant figures. “Pₙ” always means the nth-degree Taylor polynomial about the stated centre. All data are fictional.
Question 1 (multiple choice · foundation)
P₃ is the third-degree Taylor polynomial for f about x = 1. For every z between 1 and 1.2, the size of f⁽⁴⁾(z) is at most 5. Which is the Lagrange error bound for |f(1.2) − P₃(1.2)|?
- (A) 5(0.2)⁴/4!
- (B) 5(0.2)³/3!
- (C) 5(0.2)⁴/3!
- (D) 5(0.2)⁴
Answer and explanation
Answer: (A). With n = 3, the bound is M · |x − a|ⁿ⁺¹/(n + 1)! = 5(0.2)⁴/4! = 5(0.0016)/24 ≈ 0.000333.
- (B) uses n in place of n + 1, both in the power and in the factorial. That is the shape of the last term used, not of the error.
- (C) has the right power but divides by 3! instead of 4!.
- (D) forgets the factorial altogether.
Question 2 (multiple choice · core)
P₂ is the second-degree Maclaurin polynomial for f(x) = e^(2x), used to approximate f(0.1). Which of these can be used as M in the Lagrange error bound?
- (A) 8
- (B) e^0.2
- (C) 4e^0.2
- (D) 8e^0.2
Answer and explanation
Answer: (D). With n = 2 you need f‴(z) = 8e^(2z). On 0 ≤ z ≤ 0.1 this is increasing, so its largest value is 8e^0.2 ≈ 9.77, at z = 0.1. Any M at least this big is valid, and 8e^0.2 is the smallest such choice.
- (A) is f‴ at the centre, z = 0. Because f‴ increases, 8 is smaller than the true maximum, so it is not a valid M.
- (B) forgets the chain-rule factor 2³ = 8.
- (C) is the maximum of f″(z) = 4e^(2z), the wrong derivative for n = 2.
Question 3 (multiple choice · core)
The Maclaurin series for a function g is
1 − x/3 + x²/5 − x³/7 + … + (−1)ᵏ xᵏ/(2k + 1) + …
and it converges to g(x) at x = 0.5. The second-degree Maclaurin polynomial P₂ is used to approximate g(0.5). Using the alternating series error bound, the error is at most
- (A) 0.05
- (B) about 0.0179
- (C) about 0.0069
- (D) about 0.0030
Answer and explanation
Answer: (B). At x = 0.5 the terms 1, 0.5/3, 0.25/5, 0.125/7, … alternate in sign, decrease in size and tend to 0. So the error is at most the first omitted term: 0.5³/7 = 0.125/7 ≈ 0.0179.
- (A) is 0.25/5, the last term used, not the first omitted one.
- (C) is 0.0625/9, the term after the first omitted term.
- (D) divides 0.125/7 by 3!, mixing up the alternating series bound with the factorial in the Lagrange bound.
Question 4 (multiple choice · core)
P₃(1.4) = 2.15 is used to estimate f(1.4), and the Lagrange error bound gives |f(1.4) − P₃(1.4)| ≤ 0.004. Which statement must be true?
- (A) 2.146 ≤ f(1.4) ≤ 2.154
- (B) 2.15 ≤ f(1.4) ≤ 2.154
- (C) The difference between f(1.4) and 2.15 is exactly 0.004.
- (D) f(1.4) = 2.15
Answer and explanation
Answer: (A). The bound limits the size of the error in either direction, so f(1.4) lies in [2.15 − 0.004, 2.15 + 0.004] = [2.146, 2.154].
- (B) assumes the estimate is an underestimate. Nothing in the question gives the sign of the error.
- (C) treats the bound as the actual error. The true error can be anything from 0 up to 0.004.
- (D) treats the estimate as exact.
Question 5 (constructed response · core)
Let f(x) = 1/(1 + x).
(a) Find P₂(x), the second-degree Maclaurin polynomial for f. (b) Use P₂ to approximate f(−0.1). (c) Use the Lagrange error bound to show that |f(−0.1) − P₂(−0.1)| < 0.002.
Worked solution and suggested Marlbridge rubric
(a) f(x) = (1 + x)⁻¹, f′(x) = −(1 + x)⁻², f″(x) = 2(1 + x)⁻³, f‴(x) = −6(1 + x)⁻⁴. f(0) = 1, f′(0) = −1, f″(0) = 2. So P₂(x) = 1 − x + x².
(b) P₂(−0.1) = 1 + 0.1 + 0.01 = 1.11.
(c) n = 2, so use f‴. The size of f‴(z) is 6/(1 + z)⁴. For −0.1 ≤ z ≤ 0, 1 + z is smallest at z = −0.1, so the size is largest at z = −0.1: M = 6/0.9⁴ = 6/0.6561 ≈ 9.145. Bound: M · |−0.1 − 0|³/3! = (6/0.9⁴)(0.001)/6 = 0.001/0.9⁴ ≈ 0.001524. Since 0.001524 < 0.002, |f(−0.1) − P₂(−0.1)| < 0.002.
Why the endpoint matters: taking M = 6 (the value at the centre) gives 0.001, but the actual error is 1/0.9 − 1.11 ≈ 0.001111, which is bigger than 0.001. A wrong M can give a “bound” that is false.
| Point | What earns it |
|---|---|
| 1 | P₂(x) = 1 − x + x² |
| 1 | Estimate 1.11 |
| 1 | Uses the third derivative and a valid M, justified (largest at z = −0.1) |
| 1 | Bound about 0.001524 (or 0.001/0.9⁴) |
| 1 | Explicit comparison: bound < 0.002 |
Total: 5 points. Any larger valid M is accepted if the bound is still shown to be less than 0.002; for example M = 10 gives 10(0.001)/6 ≈ 0.001667 < 0.002. M = 6 does not earn the M point.
Question 6 (constructed response · core)
A function g has the following values at x = 2. For all x in the interval 2 ≤ x ≤ 2.6, the size of g⁽⁴⁾(x) is at most 15.
| x | g(x) | g′(x) | g″(x) | g‴(x) |
|---|---|---|---|---|
| 2 | −1 | 3 | −4 | 9 |
(a) Write the third-degree Taylor polynomial P₃ for g about x = 2. (b) Use P₃ to approximate g(2.4). (c) Use the Lagrange error bound to show that |g(2.4) − P₃(2.4)| < 0.02. (d) Find an interval that must contain g(2.4). Can you conclude that g(2.4) is negative? Explain.
Worked solution and suggested Marlbridge rubric
(a) Coefficients: −1, 3, −4/2 = −2, 9/6 = 1.5. P₃(x) = −1 + 3(x − 2) − 2(x − 2)² + 1.5(x − 2)³
(b) x − 2 = 0.4: P₃(2.4) = −1 + 1.2 − 0.32 + 0.096 = −0.024.
(c) n = 3, M = 15: bound = 15(0.4)⁴/4! = 15(0.0256)/24 = 0.016. Since 0.016 < 0.02, |g(2.4) − P₃(2.4)| < 0.02.
(d) g(2.4) lies in [−0.024 − 0.016, −0.024 + 0.016] = [−0.04, −0.008]. Every number in this interval is negative, so yes, g(2.4) < 0.
| Point | What earns it |
|---|---|
| 1 | Coefficients −2 and 1.5 (dividing by 2! and 3!), in powers of (x − 2) |
| 1 | Estimate −0.024 |
| 1 | Bound 15(0.4)⁴/4! = 0.016 |
| 1 | Explicit comparison 0.016 < 0.02 |
| 1 | Interval [−0.04, −0.008] and the conclusion g(2.4) < 0 because the whole interval is below 0 |
Total: 5 points. In (d), using 0.02 in place of 0.016 gives [−0.044, −0.004], which also leads to the correct conclusion and is accepted. Saying “yes, because P₃(2.4) is negative” without the interval does not earn the point: the estimate alone does not prove the sign.
Question 7 (constructed response · stretch)
The Maclaurin polynomials for cos x are used to approximate cos 0.5.
(a) Using M = 1, find the smallest n for which the Lagrange error bound for |cos 0.5 − Pₙ(0.5)| is less than 0.0001. (b) Explain why P₄(0.5) already meets the accuracy in (a), even though your answer to (a) may be larger than 4. (c) Use the alternating series error bound to bound the error of P₄(0.5) = 1 − 0.5²/2 + 0.5⁴/24, and decide whether P₄(0.5) is an overestimate or an underestimate.
Worked solution and suggested Marlbridge rubric
(a) Every derivative of cos x is ±sin x or ±cos x, so M = 1 is valid. Bound: 0.5ⁿ⁺¹/(n + 1)!.
- n = 3: 0.5⁴/24 ≈ 0.002604 (too big)
- n = 4: 0.5⁵/120 ≈ 0.0002604 (too big)
- n = 5: 0.5⁶/720 ≈ 0.0000217 (less than 0.0001)
So n = 5.
(b) The x⁵ coefficient of cos x is 0 (all odd coefficients are 0), so P₅ = P₄. The n = 5 bound therefore applies to P₄(0.5) as well, so P₄(0.5) is within about 0.0000217 < 0.0001.
(c) The series 1 − x²/2! + x⁴/4! − x⁶/6! + … has terms that alternate, decrease in size at x = 0.5 and tend to 0. The first omitted term is −0.5⁶/720 ≈ −0.0000217, so the error is at most 0.0000217. That term is negative, so the true value is below P₄(0.5): P₄(0.5) ≈ 0.877604 is an overestimate. (Calculator check: cos 0.5 ≈ 0.877583, so P₄(0.5) is too big by about 0.0000216.)
| Point | What earns it |
|---|---|
| 1 | Correct bound expression 0.5ⁿ⁺¹/(n + 1)! with M = 1 justified |
| 1 | n = 5, with the n = 4 value shown to be too big |
| 1 | P₅ = P₄ because the x⁵ coefficient is 0 |
| 1 | Alternating series bound 0.5⁶/720 ≈ 0.0000217, with the conditions stated |
| 1 | Overestimate, because the first omitted term is negative |
Total: 5 points. In (c), a correct Lagrange argument for the sign (for example using the sign of the next derivative) is also accepted.
How did you do?
- Q1 or Q6(c) wrong: recheck the formula: f⁽ⁿ⁺¹⁾, power n + 1 and (n + 1)!. See “The Lagrange error bound” in the study guide.
- Q2 or Q5(c) wrong: reread “Choosing M”, and check both ends of the interval.
- Q4 or Q6(d) wrong: look again at the number line in Worked example 1.
- Q3 or Q7(c) wrong: revisit “The alternating series error bound as an alternative”.
- Q7(a)–(b) wrong: work through Worked example 2, choosing the degree.
Then tick off the topic checklist.