OCR A Level Mathematics A: Applications of differentiation: tangents, normals, stationary points, optimisation and increasing/decreasing functions (H240) – 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 the applications of differentiation in OCR A Level Mathematics A (H240), specification Version 3.1 (August 2026): references 1.07m, 1.07n and 1.07o (Stage 1) and 1.07p (A Level only). Question 6 and part (c) of question 8 use inflection, which is A Level only.

Review the ideas in the study guide or the revision notes. See also the course hub and the checklist.

Questions

1. Find the equation of the tangent to y = x² − 6√x at the point where x = 4, in the form ax + by + c = 0 with integer coefficients. [4]

2. The curve y = 12/x + x passes through the point where x = 3.

(a) Find the equation of the normal at this point. [4] (b) Find the area of the triangle bounded by this normal and the two coordinate axes. [2]

3. Find the coordinates of the points on y = x³ − 3x² + 2x where the gradient is 2. [4]

4. Find the coordinates of the stationary points of y = x³ − 3x² − 24x + 5 and determine their nature. [6]

5. (a) Find the values of x for which f(x) = 2x³ + 3x² − 12x is decreasing. [3] (b) Show that g(x) = x³ − 3x² + 4x − 7 is increasing for all real x. [3]

6. (A Level only) A curve has equation y = 3x⁴ − 8x³ + 6x².

(a) Find the stationary points and show that one of them is a point of inflection. [7] (b) Show that the curve has a second point of inflection, and state, with a reason, whether it is stationary. [2]

7. An open cylindrical tank, with a base but no lid, has volume 64π cm³. Its base radius is r cm. Find the value of r for which the total outer surface area is least, justify that it is least, and find that area. [7]

8. f(x) = (x − 1)(x + 2)².

(a) Show that f(x) = x³ + 3x² − 4. [1] (b) Find the stationary points and determine their nature. [4] (c) (A Level only) Find the point of inflection. [1] (d) Sketch the curve, labelling intercepts and the points found. [2]

9. A coach company models the cost, C pounds, of a journey at a steady speed v km/h by C = 3200/v + 0.025v², for 20 ≤ v ≤ 90.

(a) Find the speed that gives the least cost and show that it is a minimum. [5] (b) Find this least cost. [1] (c) Check that this is the least cost over the whole range. [2]

10. The curve y = x³ + px² + qx has stationary points at x = 1 and x = −3. Find p and q, and show that the point where x = −3 is a maximum. [5]

11. The normal to y = x² − 4x + 5 at the point where x = 3 meets the curve again at the point P. Find the coordinates of P. [6]

12. The function f(x) = x + k/x², for x > 0, has a stationary point at x = 2.

(a) Find k. [2] (b) Find the nature of the stationary point and the value of f there. [3] (c) Show that f is increasing for x > 2. [2]

Answers

1. y = 4 when x = 4 (16 − 12) [1]. dy/dx = 2x − 3x^(−1/2) [1]. At x = 4 the gradient is 8 − 3/2 = 13/2 [1]. y − 4 = (13/2)(x − 4), so 13x − 2y − 44 = 0 [1]. [4] Examiner insight: Work out y from the curve and the gradient from the derivative separately; mixing them up is a common way to lose credit, and a bare final line without the derivative gives little to reward if it is wrong.

2. (a) dy/dx = −12/x² + 1 [1]. At x = 3 this is −1/3, and y = 4 + 3 = 7 [1]. Normal gradient 3 [1]. y − 7 = 3(x − 3), so y = 3x − 2 [1]. (b) The normal meets the axes at (2/3, 0) and (0, −2) [1]. Area = ½ × 2/3 × 2 = 2/3 [1]. [6] Examiner insight: Lengths are positive, so use the size of the negative intercept in the area; write the intercepts down before multiplying so a sign slip is easy to spot.

3. dy/dx = 3x² − 6x + 2 [1]. Set equal to 2: 3x² − 6x = 0 [1]. x = 0 or x = 2 [1]. Points (0, 0) and (2, 0) [1]. [4] Examiner insight: Give coordinates, not just x-values, when the question asks for points; the second coordinate comes from the curve, not from the gradient.

4. dy/dx = 3x² − 6x − 24 = 3(x − 4)(x + 2) [1]. x = 4 or x = −2 [1]. y(4) = −75 and y(−2) = 33 [1]. d²y/dx² = 6x − 6 [1]. At x = 4 it is 18 > 0, so minimum at (4, −75) [1]. At x = −2 it is −18 < 0, so maximum at (−2, 33) [1]. [6] Examiner insight: A conclusion that quotes the sign of d²y/dx² at each point is the safest way to support “maximum” or “minimum”; a local maximum need not be the greatest value of the function.

5. (a) f′(x) = 6x² + 6x − 12 = 6(x + 2)(x − 1) [1]. The roots are −2 and 1, and the parabola opens upward, so f′ is negative between them [1]. −2 < x < 1 [1]. (b) g′(x) = 3x² − 6x + 4 [1] = 3(x − 1)² + 1 [1]. This is at least 1, so positive for all x, and g is increasing everywhere [1]. [6] Examiner insight: Give an interval with inequalities, not the roots alone; for a “show that” about all x, completing the square and stating why the result is positive is the complete argument.

6. (a) dy/dx = 12x³ − 24x² + 12x = 12x(x − 1)² [1]. Stationary at x = 0 and x = 1, giving (0, 0) and (1, 1) [1]. d²y/dx² = 36x² − 48x + 12 = 12(3x − 1)(x − 1) [1]. The second derivative is 12 > 0 at x = 0, so (0, 0) is a minimum [1]. At x = 1 it is 0, so the test cannot decide [1]. Just below 1 (x = ½) d²y/dx² = −3, and above 1 (x = 2) it is 60, so it changes sign [1]. (1, 1) is a stationary point of inflection [1]. (b) d²y/dx² is also zero at x = 1/3, positive at x = 0 and negative at x = ½, so the sign changes: an inflection at (1/3, 11/27) [1]. dy/dx there is 12(1/3)(2/3)² = 16/9 ≠ 0, so it is not stationary [1]. [9] Examiner insight: For inflection, show the second derivative is zero and show the sign change with two test values; a zero alone is not enough, and naming the point as stationary or not completes the answer.

7. h = 64/r² from the volume πr²h = 64π [1]. A = πr² + 2πrh = πr² + 128π/r [1]. dA/dr = 2πr − 128π/r² [1]. Set to zero: r³ = 64, so r = 4 [1]. d²A/dr² = 2π + 256π/r³ = 6π > 0, so a minimum [1]. h = 4 [1]. Least area = 16π + 32π = 48π cm² [1]. [7] Examiner insight: Show how the second variable is eliminated, and justify the minimum rather than assume it; the answer should state the area with its unit.

8. (a) (x − 1)(x² + 4x + 4) = x³ + 4x² + 4x − x² − 4x − 4 = x³ + 3x² − 4 [1]. (b) f′(x) = 3x² + 6x = 3x(x + 2) [1]. Stationary points (0, −4) and (−2, 0) [1]. f″(x) = 6x + 6 [1]. f″(0) = 6, so minimum at (0, −4); f″(−2) = −6, so maximum at (−2, 0) [1]. (c) f″(x) = 0 at x = −1, with sign change, and f(−1) = −2, so (−1, −2) [1]. (d) Axis crossing at x = 1, a touch at x = −2, and y-intercept −4 [1]. Cubic shape rising to the maximum (−2, 0) on the x-axis, down through (−1, −2) to the minimum (0, −4), then rising through (1, 0), with every point labelled [1]. [8] Examiner insight: A “show that” expects every step of the expansion written out; in the sketch, the touching of the axis at the maximum and the labelled coordinates are what the sketch needs to display.

9. (a) dC/dv = −3200/v² + 0.05v [1]. Set to zero: v³ = 64 000 [1]. v = 40 km/h [1]. d²C/dv² = 6400/v³ + 0.05 [1] = 0.15 > 0 at v = 40, so a minimum [1]. (b) C = 80 + 40 = £120 [1]. (c) C(20) = 170 and C(90) = 238.06 (2 d.p.) [1]. Both exceed 120, so the least cost over 20 ≤ v ≤ 90 is 120 [1]. [8] Examiner insight: When a model has a restricted range, checking the end values closes the argument; also include units, here km/h and pounds.

10. dy/dx = 3x² + 2px + q [1]. Substituting gives 3 + 2p + q = 0 and 27 − 6p + q = 0 [1]. Subtracting: 24 − 8p = 0 [1]. So p = 3 and q = −9 [1]. d²y/dx² = 6x + 6 = −12 < 0 at x = −3, so it is a maximum [1]. [5] Examiner insight: Both conditions come from dy/dx = 0; write the two equations down explicitly before eliminating, and finish by quoting the sign of the second derivative.

11. At x = 3, y = 2 [1]. dy/dx = 2x − 4 = 2 [1]. Normal: y − 2 = −½(x − 3) [1]. Equate: x² − 4x + 5 = −x/2 + 7/2, so 2x² − 7x + 3 = 0 [1]. (2x − 1)(x − 3) = 0 [1]. x = 3 is the known point, so x = ½ and P = (1/2, 13/4) [1]. [6] Examiner insight: One root of the intersection equation is the point you started from, which is a built-in check on your algebra; use it to confirm before finding the other.

12. (a) f′(x) = 1 − 2k/x³ [1]. At x = 2: 1 − k/4 = 0, so k = 4 [1]. (b) f″(x) = 24/x⁴ [1]. At x = 2 it is 3/2 > 0, so a minimum [1]. f(2) = 2 + 1 = 3 [1]. (c) f′(x) = 1 − 8/x³, and for x > 2 we have x³ > 8, so 8/x³ < 1 [1]. Hence f′(x) > 0 and f is increasing [1]. [7] Examiner insight: In part (c), link the inequality on x directly to the sign of f′; stating “f′ > 0, so increasing” as the final line shows the reasoning has landed.

Where marks are usually lost

• Using the derivative value as a y-coordinate rather than substituting into the curve.
• Writing the normal gradient as −m or 1/m.
• Not rearranging an equation into the form the question specifies.
• Asserting “maximum” or “minimum” with no sign evidence.
• Calling a point an inflection from d²y/dx² = 0 alone, with no sign check on either side.
• Answering a “where is it increasing” question with roots instead of intervals.
• Leaving out units, or giving an optimum without checking it lies in the allowed range.
• Skipping the algebra of “show that” and writing only the target.

Next steps

• Re-read the revision notes and the study guide.
• Go to the course hub and tick off the checklist.
• Try the free 10-minute diagnostics to see which topic to revise next.
• Book a free trial class for help with any question you got stuck on.

Official syllabus

Cambridge OCR Level 3 Advanced GCE in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by Cambridge OCR: section 1.07, Differentiation, references 1.07m to 1.07p.