Small-angle approximations Pure

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested learners’ knowledge and understanding of small angle approximations, with learners finding the algebra in part (a) the most demanding aspect. Most attempts began with a substitution of the small angle approximation for cos2x, and stronger responses proceeded efficiently to the given answer with good algebra seen. A common error was not including brackets around the 2x. In some cases, these invisible ( ) 2 2 ; but in both cases a brackets were recovered; in other cases, the omission led to 1 1 x - - maximum score of M1A0* was available. Another common error was to expand the binomial correctly, but to omit the brackets when subtracting from 1, resulting in sign errors. In some cases, the squared sign on the outside of the bracket was omitted completely and this resulted in no marks being scored as they had materially simplified the algebra. Many responses tried to use identities prior to using small angle approximations, making the question significantly more complicated. In some cases, these approaches did not meet the demand of the question “use the small angle approximation for cos theta ”. More successful responses in part (b), made good use of the restart opportunity available here. Most used the small angle approximations and the given answer to part (a) correctly. After substitution into both the numerator and denominator, it was only a very small minority x and 2 x together, rather than who were not successful. Some, for example, tried to add 3 multiply, while others did not deal with the embedded fractions correctly, often dividing by 6 2 2 2 3 x and obtaining the incorrect . - 3

Common mistakes

• cos 2x ≈ 1 − (2x)²/2 with brackets round 2x
• square the whole bracket
• brackets needed when subtracting from 1
• intermediate line 1 − (1 − 4x² + 4x⁴) required
• using identities first does not meet the demand

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested learners’ knowledge and understanding of small angle approximations, More successful responses in part (b), made good use of the restart opportunity available here. Most used the small angle approximations and the given answer to part (a) correctly. 2 24 24 0 x part (b) approached 0. This resulted in many learners setting and solving to - = x = ± . obtain 1 For both B marks follow-through was allowed on their answer to part (b). Most learners did succeed in deducing that the answer to part (c) was their constant term from part (b), and this was sufficient for the first B1 mark. However, a lot of responses did not adequately explain this, so the dB1 mark was frequently not awarded. 2 bx as x tended The most common form of correct answer was to discuss the behaviour of 2 24 24 24 2 x towards zero, for example “as so ”. 0, 24 0 - → x x → - → A significant minority missed the word “hence” and, as a result, some attempted to substitute ( ) 2 1 cos 2 x - 0 or a very small value into the expression , which scored no marks. x x sin tan 3 2

Common mistakes

• use the given answer to (a) over (x/3)(x/2) = x²/6
• multiply, do not add, x/3 and x/2
• 24 − 24x² (dividing by 6 instead of multiplying gives 2/3 − 2/3x²)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested learners’ knowledge and understanding of small angle approximations, Part (c) was answered well by most learners. However, many did not appear to understand “given that x is very small” or did not know how to use this information. There was one dominant misconception, even among those who had been successful in the previous parts of the question, namely that “very small” meant that the expression obtained as an answer to succeed in deducing that the answer to part (c) was their constant term from part (b), and this was sufficient for the first B1 mark. However, a lot of responses did not adequately explain this, so the dB1 mark was frequently not awarded. 2 bx as x tended The most common form of correct answer was to discuss the behaviour of 2 24 24 24 2 x towards zero, for example “as so ”. 0, 24 0 - → x x → - → A significant minority missed the word “hence” and, as a result, some attempted to substitute ( ) 2 1 cos 2 x - 0 or a very small value into the expression , which scored no marks. x x sin tan 3 2

Common mistakes

• approximate value 24 = the constant term from (b)
• reason must refer to the x² term → 0 as x is very small
• setting 24 − 24x² = 0 or substituting x = 0 into the original expression scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was very accessible with many candidates scoring full marks. The most common approach was to directly apply the small angle approximations to the given expression and on the whole, this was successfully simplified to give the correct value. Using the small angle approximation 2 was most successful for the numerator, however there were a few candidates who were unable to process the multiplication. The denominator proved to be trickier to process with errors seen in the bracketing, squaring the angle, using 3cos rather thancos3 or omitting the “1-”. It was very common to see errors such as 2 3 cos3 1 2 = - and 2 2 3 3 1 cos3 1 1 . 2 2 - = -- = - Once the numerator and denominator had been simplified, there was a sizeable minority who struggled with cancelling the fraction. Incorrect answers of 4 3 and 4 3 - were seen frequently, resulting from the poor use of brackets. A significant number of candidates complicated the process by introducing trigonometric identities before applying small angle approximations, presumably not realising they could substitute the appropriate approximations directly. There were many variations in the processing such as double angle formulae and compound angle formulae which all involved a lot more work and inevitably led to complicated expressions with higher order indices in the denominator. Unfortunately, only a few candidates were successful with such approaches as there were often errors in the algebra needed for the simplification or a lack of recognition that they needed to make a limiting argument.

Common mistakes

• tan 2θ ≈ 2θ
• cos 3θ ≈ 1 − (3θ)²/2 with brackets (not 1 − 3θ²/2)
• keep the "1 −" in the denominator
• answer 4/9 (not ±4/3 or 0.444)
• using identities first makes it much harder

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question testing the small angle approximation for cosine and finding the equation of a straight line was accessible to many candidates with a large number scoring full marks. In part (a), some candidates formed and solved the equation but were not able to use the information that x was small to choose the correct solution. There was a very small minority of candidates who did not quote the correct small angle approximation, despite this being in the formula book. A slightly larger minority misunderstood that x being small meant that x was 0 and therefore did not include the 2x term in their solution attempt. Candidates should also pay closer attention to the question as several did not round to the required decimal places. It was also notable that some candidates at this level found it difficult to expand 2 1 1 2 2 x - when multiplying the fractions together.

Common mistakes

• substitute cos x ≈ 1 − x²/2 into f′(x) = 0 and keep the 2x term (x small is not x = 0)
• choose the small root only
• 3 d.p.

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The first method mark in this question was often gained for sight of but the second mark proved more challenging, with errors seen in the correct application of the small-angle approximation. Some candidates missed out the square completely and just wrote . Many of those who did attempt to expand made mistakes with fractions and powers, with 2 2 3 1 4 theta theta - + occasionally seen. Candidates who realised they could use and then use the approximation fared better. In contrast, those using the double angle formula performed less well, with many fraction and sign errors seen with this approach. It was not uncommon to see candidates who followed the main method ‘multiply everything by 4’ at the end leading to an incorrect final expression of , forfeiting the final A mark.

Common mistakes

• square the whole bracket 3(1 − θ²/2)² (do not drop the square)
• sin(θ/2) ≈ θ/2
• ignore θ³ and higher terms
• do not multiply everything by 4 at the end
• a = 3, b = 2, c = −3

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved to be more challenging for a number of candidates although many were still able to score full marks. In part (b) almost all candidates were able to write down the small angle approximation for cosx even if they were unable to use it in order to answer the question. Most candidates did substitute correctly into the equation in order to gain a quadratic equation, although some did contain a mixture of variables. Sign errors sometimes occurred at this stage and the weakest candidates were unable to deal correctly with resulting fractions in order to simplify the equation. Many candidates appear to be using calculators in order to solve quadratics and in this question, this usually resulted in full marks. The final mark was lost by some candidates who correctly solved the quadratic equation but gave both roots as their final answer.

Common mistakes

• cos x ≈ 1 − x²/2
• x² + 4x − 1 = 0
• give only the small root −2 + √5 ≈ 0.236, not both roots

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A number of candidates did not attempt this question. Whether this was due to lack of understanding or time constraints is unknown. The small angle approximation in (b) part (i) was also well handled. Most settled for the form 8 x y although some rearranged to y = . Marks were generally lost when sin 2y was replaced by 2 instead of 2y. Many candidates failed to see the link between (a) and (b)(i) even though they had both answers correct. They were unable to see that the gradients were the same or else failed to explain themselves coherently.

Common mistakes

• sin 2y ≈ 2y (not 2θ) gives x = 8y
• the gradient of x = 8y is 1/8, the same as (a)
• "8 and 1/8 are reciprocals" alone not enough

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was accessible to most students and many gave fully correct solutions. Almost all students were able to score a mark for sin3 3 but there were a number who simplified the denominator to 6 instead of 2 6 resulting in a final solution of 4 . 3 Replacing the cos 4𝜃 proved more problematic with common incorrect attempts on the numerator including 2 1 4 1 2 and more frequently 2 4 1 1 2 where the 4 was not squared leading to an incorrect answer of 1/3. Other common incorrect attempts involved simplifying 1 cos4 to 2 8 rather than 2 8 leading to the incorrect answer of - 4/3. Even in cases where both substitutions were performed correctly many careless errors were made when simplifying. There were a small number of students who tried to expand and simplify the expression using the double angle formula for cos4 , but in most cases this resulted in an incorrect expression which did not lead to the correct answer.

Common mistakes

• sin3θ ≈ 3θ and cos4θ ≈ 1 − (4θ)²/2
• square the 4θ (not squaring gives 1/3)
• 1 − cos4θ = +8θ² not −8θ² (gives −4/3)
• denominator 2θ × 3θ = 6θ², not 6θ