Edexcel A Level Mathematics: Pure Mathematics 1 — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Pure Mathematics 1 revision notes

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Section A

1. Express 2x² − 12x + 23 in the form a(x + b)² + c. [3]

2. Hence state the coordinates of the turning point and explain why the curve has no real roots. [3]

Section B

3. The line L passes through A(1, 5) and B(7, −7).

(a) Find the equation of L in the form ax + by + c = 0. [3] (b) Find the equation of the perpendicular bisector of AB. [4]

4. Line L1 has equation 3x − 2y + 4 = 0. Find the equation of the line L2 that is parallel to L1 and passes through the point (2, 1), giving your answer in the form ax + by + c = 0. [3]

5. A curve has equation y = x³ − 4x + 1. Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form y = mx + c. [5]

6. A curve satisfies dy/dx = 6x² − 4x + 1 and passes through the point (1, 3). Find y in terms of x. [4]

7. A sector of a circle has radius 8 cm and angle 1.2 radians. Find (a) the arc length and (b) the area of the sector. [5]

8. Rationalise the denominator of 5/(3 + √2), giving your answer in the form (a + b√2)/c. [3]

9. Find the set of values of k for which kx² + 4x + 1 is positive for all real x. [3]

10. A triangle has a = 8, angle A = 30°, and side b = 12. Use the sine rule to find the two possible values of angle B, and explain why both are valid. [4]

11. State the derivative of f(x) from first principles, using limit notation. [2]

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Answers

1. 2(x² − 6x) + 23 [1] = 2[(x − 3)² − 9] + 23 [1] = 2(x − 3)² + 5 [1].

2. Turning point at (3, 5) [1]. The minimum value of 2(x − 3)² is 0, so the minimum value of the expression is 5 [1], which is positive, so the curve never crosses the x-axis [1].

3. (a) Gradient = (−7 − 5) ÷ (7 − 1) = −2 [1]; y − 5 = −2(x − 1) [1]; 2x + y − 7 = 0 [1]. (b) Midpoint = (4, −1) [1]; perpendicular gradient = ½ [1]; y + 1 = ½(x − 4) [1]; y = ½x − 3 (or x − 2y − 6 = 0) [1].

4. L1: 3x − 2y + 4 = 0 rearranges to y = (3/2)x + 2, so its gradient is 3/2 [1]. A parallel line has the same gradient, so through (2, 1): y − 1 = (3/2)(x − 2) [1], which rearranges to 3x − 2y − 4 = 0 [1].

5. At x = 2, y = 8 − 8 + 1 = 1, so the point is (2, 1) [1]. dy/dx = 3x² − 4 [1], so at x = 2 the gradient of the tangent is 3(4) − 4 = 8 [1], and the gradient of the normal is the negative reciprocal, −⅛ [1]. y − 1 = −⅛(x − 2), giving y = −⅛x + 5/4 [1].

6. Integrating term by term: y = 2x³ [1] − 2x² + x + c [1]. Substituting the point (1, 3): 2 − 2 + 1 + c = 3 [1], so c = 2, giving y = 2x³ − 2x² + x + 2 [1].

7. (a) Arc length s = rθ = 8 × 1.2 = 9.6 cm [2]. (b) Area of sector A = ½r²θ = ½ × 64 × 1.2 = 38.4 cm² [3].

8. Multiply by the conjugate: 5/(3 + √2) × (3 − √2)/(3 − √2) [1] = 5(3 − √2) ÷ (9 − 2) [1] = (15 − 5√2)/7 [1].

9. Requires a > 0 (k > 0) [1] and discriminant < 0: 4² − 4(k)(1) < 0 [1], so 16 − 4k < 0, giving k > 4 [1].

10. sin B / b = sin A / a, so sin B = (12 × sin 30°) ÷ 8 = 6 ÷ 8 = 0.75 [1]. B = sin⁻¹(0.75) = 48.6° [1] or the ambiguous-case alternative B = 180° − 48.6° = 131.4° [1]. Both are valid since A + B stays below 180° in each case (30 + 48.6 and 30 + 131.4 are both under 180°) [1].

11. f′(x) = lim(h→0) [f(x + h) − f(x)] / h [2].

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Where marks are usually lost

• Forgetting to factor out the 2 before completing the square.
• Using the original gradient rather than its negative reciprocal for a normal — or its negative reciprocal instead of the same gradient for a parallel line.
• Omitting the constant of integration in an indefinite integral, or forgetting to use the given point to find its value.
• Giving only the discriminant condition for “always positive”, forgetting a > 0 is also required.
• Missing the second, obtuse solution in an ambiguous-case sine rule question.
• Mixing degree and radian mode when using the arc-length or sector-area formulas.
• Writing the first-principles derivative without the limit notation.