GCSE Further Maths Short Paper B
Covers Differentiation, Applications of Differentiation, Trigonometric Identities and Equations and 3 more.
Questions
Question 1 [2 marks]
Trigonometric Identities and Equations
Show that sin(x)/cos(x) + cos(x)/sin(x) can be simplified to 1/(sin(x)cos(x)).
Question 2 [4 marks]
Sequences
An arithmetic sequence has first term 8 and common difference 5.
Find the number of terms needed for the sequence to first exceed 100.
Question 3 [4 marks]
Applications of Differentiation
A curve has equation y = x^3 - 2x^2 + 3.
Find the gradient of the tangent to the curve at the point where x = -1, and state the coordinates of this point.
Question 4 [4 marks]
Geometric Proof
O is the centre of a circle, and A and B are points on the circumference.
AB is a chord and M is the midpoint of AB.
Prove that OM is perpendicular to AB.
Question 5 [4 marks]
Differentiation
A curve has equation y = x^3 - 3x^2 - 9x + 2.
Find the coordinates of the stationary points and use the second derivative to determine the nature of each one.
Question 6 [5 marks]
Geometric Proof
In triangle ABC, D and E are points on AB and AC respectively such that DE is parallel to BC.
Prove that triangle ADE is similar to triangle ABC.
Question 7 [5 marks]
Non-Right-Angled Triangle Trigonometry
In triangle LMN, LM = 12 cm, MN = 9 cm, and angle LMN = 55 degrees.
Find the length of LN, giving your answer to 3 significant figures, and hence find the perimeter of triangle LMN.
Question 8 [6 marks]
Sequences
The sum of the first n terms of an arithmetic series is given by Sn = (n/2)(5n + 9).
Find the first term and the common difference of the series, and find the value of n for which Sn = 2180.
Question 9 [6 marks]
Sequences
The sum of the first n terms of an arithmetic series is given by Sn = (n/2)(3n - 1).
Find the first term and the common difference of the series, and find the value of n for which Sn = 330.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| combining the two fractions over a common denominator sin(x)cos(x) | M1 |
| using sin^2(x) + cos^2(x) = 1 to obtain 1/(sin(x)cos(x)) | A1 |
| Final answer: 1/(sin(x)cos(x)), shown using sin^2(x) + cos^2(x) = 1 | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| using Un = a + (n - 1)d to give Un = 5n + 3 | M1 |
| setting up the inequality 5n + 3 > 100 | M1 |
| solving to give n > 19.4 | M1 |
| n = 20 (the 20th term is the first to exceed 100) | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to get dy/dx = 3x^2 - 4x | M1 |
| substituting x = -1 into dy/dx | M1 |
| gradient = 7 | A1 |
| the point (-1, 0) | B1 |
| Final answer: gradient = 7 at the point (-1, 0) | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| stating OA = OB, since both are radii of the circle | B1 |
| stating AM = MB (M is the midpoint of AB) and OM is common to both triangles | B1 |
| concluding triangles OAM and OBM are congruent by SSS, so angle OMA = angle OMB | M1 |
| using angles on a straight line to show each angle is 90 degrees, so OM is perpendicular to AB | A1 |
| Final answer: Triangles OAM and OBM are congruent (SSS), so angle OMA = angle OMB = 90 degrees, hence OM is perpendicular to AB | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating and setting dy/dx = 3x^2 - 6x - 9 = 0 to find x = 3 and x = -1 | M1 |
| the points (3, -25) and (-1, 7) | A1 |
| finding the second derivative d2y/dx2 = 6x - 6 and evaluating it at each x value | M1 |
| identifying (3, -25) as a minimum and (-1, 7) as a maximum | A1 |
| Final answer: (3, -25) is a minimum; (-1, 7) is a maximum | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| stating angle ADE = angle ABC, since DE is parallel to BC (corresponding angles) | B1 |
| stating angle AED = angle ACB, since DE is parallel to BC (corresponding angles) | B1 |
| stating angle DAE, equal to angle BAC, is common to both triangles | B1 |
| concluding triangle ADE is similar to triangle ABC by the AA condition | M1 |
| stating that all three pairs of angles are therefore equal (AAA), confirming the similarity | A1 |
| Final answer: Similar by AA: angle ADE = angle ABC and angle AED = angle ACB (corresponding angles, DE parallel to BC), with angle A common | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| using the cosine rule LN^2 = LM^2 + MN^2 - 2(LM)(MN)cos(55) | M1 |
| substituting to get LN^2 = 144 + 81 - 216cos(55) | M1 |
| LN = 10.1 cm (3 s.f.) | A1 |
| adding LM + MN + LN | M1 |
| perimeter = 31.1 cm (3 s.f.) | A1 |
| Final answer: LN = 10.1 cm (3 s.f.); perimeter = 31.1 cm (3 s.f.) | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| using S1 = a to find a = 7 | M1 |
| finding S2 and using T2 = S2 - S1 to find the common difference d = 5 | M1 |
| a = 7 and d = 5 | A1 |
| forming the equation (n/2)(5n + 9) = 2180 | M1 |
| rearranging to 5n^2 + 9n - 2180 = 0 and solving, e.g. using the quadratic formula | M1 |
| n = 20 | A1 |
| Final answer: first term = 7; common difference = 5; n = 20 | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| using S1 = a to find a = 1 | M1 |
| finding S2 and using T2 = S2 - S1 to find the common difference d = 3 | M1 |
| a = 1 and d = 3 | A1 |
| forming the equation (n/2)(3n - 1) = 330 | M1 |
| rearranging to 3n^2 - n - 660 = 0 and solving, e.g. using the quadratic formula | M1 |
| n = 15 | A1 |
| Final answer: first term = 1; common difference = 3; n = 15 | |