Pearson Edexcel International Advanced Level Mathematics: Exponentials, logarithms and trig equations (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Unit P2 topics 5.1 to 5.3 (exponentials and logarithms) and 6.1 to 6.2 (trigonometric identities and equations) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification, Issue 3, April 2019. P2 is compulsory for IAS Mathematics and IAL Mathematics. Learn the content first in the study guide and the revision notes. Course hub: Edexcel A Level Mathematics. See also the P2 practice questions and the free diagnostics.

Give decimal answers to 3 significant figures, and angles in degrees to 1 decimal place, unless told otherwise. Give exact answers where the question asks for them.

Questions

1. (a) The curve y = (1/4)^x is drawn. State the coordinates of its y-intercept, the equation of its asymptote, and whether the curve rises or falls as x increases. [3] (b) The curve y = a^x, where a > 0, passes through (−3, 125/8). Find the value of a. [2]

2. Show that 3 log_2 6 − log_2 27 = 3. [3]

3. Given that log_a 3 = p and log_a 7 = q, express these in terms of p and q.

(a) log_a 63 [1] (b) log_a (a/49) [2] (c) log_a √21 [1]

4. Solve each equation, giving your answer to 3 significant figures.

(a) 4^x = 30 [2] (b) 3^(2x − 1) = 100 [3]

5. (a) Use the change of base formula to find log_9 27 as an exact fraction. [2] (b) Show that log_a b × log_b a = 1, where a, b > 0 and a, b ≠ 1. [2]

6. The value, V dollars, of a machine t years after it is bought is modelled by V = 24000 × 0.85^t.

(a) Find the value of the machine after 3 years. [1] (b) Find the smallest whole number of years after which the machine is worth less than 5000 dollars. [5] (c) Find the value of the machine at that time, to the nearest dollar. [1] (d) Explain why this model never gives a value of zero. [1]

7. (a) Given that cos θ = −5/13 and 180° < θ < 270°, find the exact values of sin θ and tan θ. [3] (b) Prove that sin θ/(1 + cos θ) + (1 + cos θ)/sin θ ≡ 2/sin θ. [4]

8. Solve each equation in the given interval.

(a) sin(x − 20°) = −0.7 for 0° ≤ x < 360° [4] (b) tan 2x = −1.5 for −90° < x < 90° [4]

9. Solve cos(x + π/4) = −√3/2 for 0 ≤ x < 2π. Give your answers as exact multiples of π. [4]

10. Solve 2 sin²x + 5 cos x + 1 = 0 for 0° ≤ x < 360°. [6]

11. A student is solving 3 sin x cos x = 2 sin x for 0° ≤ x < 360° and divides both sides by sin x.

(a) Explain what is wrong with this step. [1] (b) Solve the equation correctly, giving angles to 1 decimal place where they are not exact. [4]

12. Solve sin 2(x + 15°) = −1/2 for −180° ≤ x < 180°. [6]

Answers

1. (a) The curve crosses the y-axis at (0, 1). [1] The asymptote is y = 0. [1] The base is less than 1, so the curve falls as x increases. [1] (b) a⁻³ = 125/8, so a³ = 8/125. [1] a = 2/5. [1] [5] Examiner insight: For a “state” part, brief correct statements are enough. Give the point as a coordinate pair and name the asymptote as an equation, since “the x-axis” in words is less precise than y = 0.

2. 3 log_2 6 = log_2 216. [1] log_2 216 − log_2 27 = log_2 8. [1] log_2 8 = 3 because 2³ = 8, so the result is 3. [1] [3] Examiner insight: On a “show that” question, write every law you use as a line of working, because the answer is given and only the reasoning can earn credit.

3. (a) 63 = 3² × 7, so log_a 63 = 2p + q. [1] (b) log_a (a/49) = log_a a − log_a 49. [1] = 1 − 2q. [1] (c) log_a √21 = ½ log_a (3 × 7) = ½(p + q). [1] [4] Examiner insight: Factorise numbers into the values you were given before applying any law. The constant 1 from log_a a is easy to drop, so write it explicitly.

4. (a) x ln 4 = ln 30. [1] x = ln 30 / ln 4 = 2.45. [1] (b) (2x − 1) ln 3 = ln 100, so 2x − 1 = ln 100 / ln 3 = 4.1918. [1] 2x = 5.1918. [1] x = 2.60. [1] [5] Examiner insight: Keep the full quotient until the final line and round once. Solving a linear equation with an early rounded value can shift the third significant figure.

5. (a) log_9 27 = log_3 27 / log_3 9. [1] = 3/2, so the answer is 3/2. [1] (b) Using any common base c: log_a b = log_c b / log_c a and log_b a = log_c a / log_c b. [1] Multiplying, the numerators and denominators cancel, so the product is 1. [1] [4] Examiner insight: For an algebraic proof, state the base you change to and show the cancellation. Putting numerical values in for a and b would not prove a general statement.

6. (a) V = 24000 × 0.85³ = 14 739 dollars. [1] (b) 24000 × 0.85^t < 5000. [1] 0.85^t < 5/24. [1] t ln 0.85 < ln(5/24). [1] ln 0.85 is negative, so dividing reverses the inequality: t > ln(5/24) / ln 0.85 = 9.65. [1] The smallest whole number is t = 10. [1] (c) V = 24000 × 0.85¹⁰ = 4725 dollars (to the nearest dollar). [1] (d) 0.85^t > 0 for every t, so V is never zero. [1] [8] Examiner insight: Write the sign of ln 0.85 when you divide by it. An answer of 9.65 should be rounded up to 10 in the context, and 9 or 9.65 would not answer a “whole number of years” question.

7. (a) sin²θ = 1 − 25/169 = 144/169. [1] In the third quadrant sine is negative, so sin θ = −12/13. [1] tan θ = (−12/13) ÷ (−5/13) = 12/5. [1] (b) Combine over a common denominator: [sin²θ + (1 + cos θ)²] / [sin θ (1 + cos θ)]. [1] Expand the numerator: sin²θ + 1 + 2 cos θ + cos²θ. [1] Use sin²θ + cos²θ = 1 to get 2 + 2 cos θ = 2(1 + cos θ). [1] Cancel (1 + cos θ) to leave 2/sin θ. [1] [7] Examiner insight: Show the choice of sign with a reason (the quadrant). In a proof, work on one side only and make the use of sin²θ + cos²θ = 1 visible.

8. (a) Let u = x − 20°. The interval is −20° ≤ u < 340°. [1] arcsin(−0.7) = −44.4°, which is outside the interval. [1] Use u = 180° + 44.4° = 224.4° and u = 360° − 44.4° = 315.6°. [1] x = u + 20°, so x = 244.4° or 335.6°. [1] (b) Let u = 2x. The interval is −180° < u < 180°. [1] arctan(−1.5) = −56.3°. [1] Adding 180° gives u = 123.7°. [1] x = −28.2° or 61.8°. [1] [8] Examiner insight: Write the transformed interval before listing angles. A principal value outside the interval has to be converted, and for tan the second value comes from adding 180°.

9. The interval for u = x + π/4 is π/4 ≤ u < 9π/4. [1] cos u = −√3/2 gives u = 5π/6 or 7π/6. [1] Neither 5π/6 + 2π nor 7π/6 + 2π is below 9π/4, so no more values. [1] x = u − π/4, so x = 7π/12 or 11π/12. [1] [4] Examiner insight: An exact-value request means a decimal answer would not be what was asked. Keep the answers in terms of π and check each against the original interval.

10. Replace sin²x by 1 − cos²x: 2 − 2 cos²x + 5 cos x + 1 = 0. [1] Rearrange: 2 cos²x − 5 cos x − 3 = 0. [1] Factorise: (2 cos x + 1)(cos x − 3) = 0. [1] cos x = 3 is impossible because cos x ≤ 1. [1] cos x = −1/2 gives x = 120° [1] and x = 240°. [1] [6] Examiner insight: Say why you reject a root. Leaving cos x = 3 in the working without comment, or trying to find an angle for it, loses the argument.

11. (a) sin x could be zero, and dividing by it removes the solutions where sin x = 0. [1] (b) 3 sin x cos x − 2 sin x = 0, so sin x (3 cos x − 2) = 0. [2] sin x = 0 gives x = 0°, 180°. [1] cos x = 2/3 gives x = 48.2°, 311.8°. [1] [5] Examiner insight: Collect everything on one side and factorise. A solution set that stops at 48.2° and 311.8° is a typical result of dividing by a function.

12. Let u = 2(x + 15°) = 2x + 30°. The interval is −330° ≤ u < 390°. [1] sin u = −1/2 has a principal value u = −30°. [1] The values in the interval are u = −150°, −30°, 210° and 330°. [2] x = (u − 30°)/2 gives x = −90°, −30°, 90°, 150°. [2] [6] Examiner insight: Work out the new interval first, since the factor 2 makes it twice as wide. List the values systematically by adding and subtracting 360° and check both end points.

Where marks are usually lost

• Writing log_a(x + y) = log_a x + log_a y, or cancelling logarithms as if they were factors.
• Leaving out log_a a = 1 when combining logarithms with a constant.
• Rounding intermediate values when solving equations with a power, then giving a final answer that is slightly out.
• Not reversing the inequality when dividing by the logarithm of a base between 0 and 1.
• Giving a decimal answer when the question asks for a whole number of years or an exact form.
• Using the interval for x on the whole angle inside the bracket, so missing solutions.
• Dividing by sin x or cos x instead of factorising.
• Not rejecting an impossible value such as cos x = 3 with a clear reason.
• Mixing degrees and radians, or forgetting to subtract the shift at the end.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Return to the Edexcel A Level Mathematics hub or the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019 (Pearson Education Limited), Unit P2 topics 5.1 to 5.3 and 6.1 to 6.2.