Pearson Edexcel International Advanced Level Mathematics: Exponentials, logs and numerical methods (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 P3 sections 3 and 6 (topics 3.1 to 3.3 and 6.1 to 6.2) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. P3 is compulsory for IAL Mathematics and Pure Mathematics; its IAS weighting is N/A. The study guide teaches the content and the revision notes summarise it. Marks shown are Marlbridge’s own allocations.

Questions

1. Let f(x) = e^(2 − x) + 3 for real x.

(a) Find the exact y-intercept of the curve y = f(x). [1] (b) Write down the equation of the horizontal asymptote. [1] (c) State the range of f. [1] (d) Explain why the curve is decreasing. [1]

2. Solve each equation. Give the exact answer, then the answer to 3 significant figures.

(a) e^(4x − 3) = 25 [3] (b) ln(7 − 3x) = −2 [3]

3. The function h is given by h(x) = ln(3x − 6).

(a) State the largest possible domain of h and the equation of the vertical asymptote of y = h(x). [2] (b) Find the exact x-coordinate where the curve meets the x-axis. [1] (c) Solve h(x) = 2, giving the exact value of x. [2]

4. Values of x and y are thought to satisfy y = axⁿ. Values of lg x and lg y are shown. The points lie close to a straight line.

lg x	0.176	0.398	0.602	0.778	0.954
lg y	0.093	0.494	0.862	1.179	1.496

(a) Show that plotting lg y against lg x should give a straight line, and state its gradient and intercept in terms of a and n. [2] (b) Take the first and last points to estimate n and a, to 3 significant figures. [3] (c) Use your model to estimate y when x = 20. [2]

5. The number of visits y to a website in week x is modelled by y = kbˣ. The values of lg y for x = 0, 2, 4, 6, 8 are 1.176, 1.498, 1.822, 2.143 and 2.467.

(a) Use the first and last values to find k and b, each to 3 significant figures. [4] (b) Use k = 15.0 and b = 1.45 to find the week in which the model first predicts 500 visits, as a value of x to 3 significant figures. [2]

6. Find the equation of the model in each case, giving constants to 3 significant figures.

(a) The plot of lg y against lg x is a line through (0.2, 0.9) and (1.2, 3.4), and the model is y = axⁿ. [4] (b) The plot of ln y against x is a line through (1, 2.4) and (5, 1.2), and the model is y = kbˣ. [4]

7. Let f(x) = ln(2x) + x − 5, for x > 0.

(a) Show that f(x) = 0 has a root α lying in the interval [3.1, 3.2]. [2] (b) Show that α = 3.16 to 2 decimal places. [2] (c) Let g(x) = 5/(2x − 7). Show that g(3) and g(4) have opposite signs but g(x) = 0 has no root, and explain why. [2]

8. The equation 2x + ln x = 7 has a root α. The iterative formula xₙ₊₁ = (7 − ln xₙ)/2 is used with x₀ = 3.

(a) Find x₁, x₂ and x₃, each to 4 decimal places. [3] (b) Show that if the sequence converges to α, then α satisfies 2α + ln α = 7. [1] (c) Show that α = 2.958 to 3 decimal places. [3]

9. Let f(x) = eˣ − 3x − 2.

(a) Show that a root α of f(x) = 0 lies in [2, 2.2]. [2] (b) Show that f(x) = 0 can be written as x = ln(3x + 2). [2] (c) Starting from x₀ = 2, calculate x₁, x₂, x₃ and x₄ from xₙ₊₁ = ln(3xₙ + 2), giving each to 4 decimal places. [3] (d) By considering f at suitable values, show that α = 2.125 to 3 decimal places. [2] (e) A student uses xₙ₊₁ = (e^(xₙ) − 2)/3 with x₀ = 2. Find x₁ and say what happens to the sequence relative to α. [1]

10. The population N of a colony at time t hours is modelled by N = kbᵗ. The plot of ln N against t is a line through (2, 4.1) and (10, 5.7).

(a) Find k and b to 3 significant figures, and write the model. [5] (b) Find, to 3 significant figures, the time at which the model predicts N = 1000. [3]

11. The point P(2, e²) lies on the curve y = eˣ.

(a) Write down the coordinates of the image of P in the line y = x, and the equation of the curve on which this image lies. [2] (b) The curve y = e^(2x − 3) + 1 meets the line y = 6 at Q. Find the exact x-coordinate of Q. [3]

Answers

1. (a) At x = 0, y = e² + 3 [1] [1] (b) y = 3 [1] [1] (c) f(x) > 3 [1] [1] (d) As x increases, 2 − x decreases, so e^(2 − x) decreases (a = −1 is negative) [1] [1] Examiner insight: Write the asymptote as an equation, y = 3, and the range as an inequality, so the reader sees what you describe.

2. (a) 4x − 3 = ln 25 [1]; x = (3 + ln 25)/4 [1]; x = 1.55 [1] [3] (b) 7 − 3x = e⁻² [1]; x = (7 − e⁻²)/3 [1]; x = 2.29 [1] [3] Examiner insight: Keep the exact form on the page before the decimal, so your method is visible.

3. (a) Domain x > 2 [1]; asymptote x = 2 [1] [2] (b) ln(3x − 6) = 0, so 3x − 6 = 1 and x = 7/3 [1] [1] (c) 3x − 6 = e² [1]; x = (e² + 6)/3 [1] [2] Examiner insight: Give the domain as a strict inequality: x ≥ 2 would include a value where the logarithm does not exist.

4. (a) lg y = n lg x + lg a [1]; this is a straight line with gradient n and intercept lg a [1] [2] (b) Gradient = (1.496 − 0.093)/(0.954 − 0.176) = 1.803, so n = 1.80 [1]; intercept = 0.093 − 1.803 × 0.176 = −0.224 [1]; a = 10^(−0.224) = 0.597 [1] [3] (c) y = 0.597 × 20^1.80 [1]; y = 131 (accept values close to this) [1] [2] Examiner insight: Show the gradient as a fraction with the numbers in it, so a slip in one figure leaves your method visible. Remember a is 10 to the power of the intercept.

5. (a) Gradient = (2.467 − 1.176)/8 = 0.1614 [1]; b = 10^0.1614 = 1.45 [1]; lg k = 1.176 [1]; k = 15.0 [1] [4] (b) 15.0 × 1.45ˣ = 500, so 1.45ˣ = 33.3 [1]; x = lg(33.3)/lg(1.45) = 9.44 [1] [2] Examiner insight: With x on the horizontal axis, the intercept is at x = 0, so lg k is the first value. Check your model against x = 8.

6. (a) Gradient = (3.4 − 0.9)/(1.2 − 0.2) = 2.5, so n = 2.5 [1]; intercept = 0.9 − 2.5 × 0.2 = 0.4 [1]; a = 10^0.4 = 2.51 [1]; y = 2.51x^2.5 [1] [4] (b) Gradient = (1.2 − 2.4)/(5 − 1) = −0.3 [1]; ln k = 2.4 − (−0.3)(1) = 2.7 [1]; k = e^2.7 = 14.9 [1]; b = e^(−0.3) = 0.741 so y = 14.9 × 0.741ˣ [1] [4] Examiner insight: In part (a) the horizontal axis is lg x, so the intercept is where x = 1. In both parts, convert the constants back from logs.

7. (a) f(3.1) = −0.0755 [1]; f(3.2) = 0.0563, and f is continuous on the interval, so there is a sign change and a root in [3.1, 3.2] [1] [2] (b) f(3.155) = −0.0029 [1]; f(3.165) = 0.0103, a sign change, so α = 3.16 to 2 decimal places [1] [2] (c) g(3) = −5 and g(4) = 5, but g is not continuous at x = 3.5 [1]; a fraction with numerator 5 is never zero [1] [2] Examiner insight: For a sign-change argument, give the values of f, state that f is continuous, and write a conclusion.

8. (a) x₁ = (7 − ln 3)/2 = 2.9507 [1]; x₂ = 2.9590 [1]; x₃ = 2.9576 [1] [3] (b) If xₙ → α then xₙ₊₁ → α as well, so α = (7 − ln α)/2, which rearranges to 2α + ln α = 7 [1] [1] (c) Let f(x) = 2x + ln x − 7. f(2.9575) = −0.0007 [1]; f(2.9585) = 0.0017 [1]; f is continuous and changes sign, so α = 2.958 to 3 decimal places [1] [3] Examiner insight: Round only when you write each term down. Rounding early can change the fourth decimal place.

9. (a) f(2) = e² − 8 = −0.611 [1]; f(2.2) = 0.425, and with f continuous the sign change gives a root in [2, 2.2] [1] [2] (b) eˣ = 3x + 2 [1]; taking natural logs, x = ln(3x + 2) [1] [2] (c) x₁ = 2.0794 [1]; x₂ = 2.1088 [1]; x₃ = 2.1194 and x₄ = 2.1233 [1] [3] (d) f(2.1245) = −0.0048 and f(2.1255) = 0.0006 [1]; the sign change in a continuous function gives α = 2.125 to 3 decimal places [1] [2] (e) x₁ = (e² − 2)/3 = 1.7964, which is moving away from α (the next term is 1.3425), so this formula does not converge to α from x₀ = 2 [1] [1] Examiner insight: In a “show that”, write each line so the reader sees how you reach the printed form.

10. (a) Gradient = (5.7 − 4.1)/(10 − 2) = 0.2, so ln b = 0.2 [1]; b = e^0.2 = 1.22 [1]; ln k = 4.1 − 0.2 × 2 = 3.7 [1]; k = e^3.7 = 40.4 [1]; N = 40.4 × 1.22ᵗ [1] [5] (b) ln 1000 = 3.7 + 0.2t [1]; t = (ln 1000 − 3.7)/0.2 [1]; t = 16.0 hours [1] [3] Examiner insight: Using ln N = 3.7 + 0.2t in part (b) avoids rounding errors from k and b. Include the units, hours.

11. (a) The image is (e², 2) [1]; it lies on y = ln x, the reflection of y = eˣ in y = x [1] [2] (b) e^(2x − 3) + 1 = 6, so e^(2x − 3) = 5 [1]; 2x − 3 = ln 5 [1]; x = (3 + ln 5)/2 [1] [3] Examiner insight: Subtract the constant before taking logs, because ln(e^(2x − 3) + 1) does not simplify.

Where marks are usually lost

• Taking ln of a sum term by term, or failing to isolate the exponential first.
• Forgetting that the asymptote of e^(ax + b) + c is y = c, not y = 0.
• Using the intercept itself as a or k, instead of 10 or e to that power.
• Using log x on the horizontal axis for an exponential law, or x for a power law.
• Giving a sign-change argument without values, continuity or a conclusion.
• Quoting a root to the wrong accuracy because the interval tested is too wide.
• Iterating with too few decimal places.

Next steps

• Recall methods with the revision notes.
• Re-read the study guide.
• Use the Edexcel A Level Mathematics hub and the printable checklist.
• Try all 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, Unit P3, sections 3 and 6 (Pearson Education Limited).