AS Paper 2: Calculus and Vectors
Covers proof and algebraic methods, trigonometry, differentiation, integration and vectors.
Questions
Question 1 [2 marks]
Proof and Algebraic Methods
Without using a calculator, simplify fully (3x^2 - 12) / (x^2 + x - 6).
Question 2 [3 marks]
Trigonometry
Without using a calculator, find the exact value of sin(45 deg) tan(60 deg) + cos(45 deg), giving your answer in the form (sqrt(a) + sqrt(b))/2.
Question 3 [3 marks]
Differentiation
Without using a calculator, find dy/dx for y = 6*sqrt(x) + 5x, writing sqrt(x) as x^(1/2) before differentiating.
Question 4 [3 marks]
Vectors
Find the magnitude of the vector v = 5i - 12j, and find a unit vector in the same direction as v.
Question 5 [3 marks]
Trigonometry
Solve 2 sin(x) = 1 for 0 <= x <= 360 deg, giving all solutions.
Question 6 [4 marks]
Integration
Without using a calculator, find the integral of (8x^3 - 6x + 5/x^2) with respect to x, writing 5/x^2 as 5x^-2 before integrating.
Question 7 [3 marks]
Proof and Algebraic Methods
The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.
Find f^-1(x) and state its domain.
Question 8 [4 marks]
Differentiation
A curve has equation y = (4x - 1)/(x + 2).
Find dy/dx using the quotient rule, and find the gradient of the curve at the point where x = 1.
Question 9 [4 marks]
Trigonometry
Solve the equation 3 tan(x) - 2 = 4 for 0 <= x <= 360 deg.
Question 10 [4 marks]
Integration
Without using a calculator, find the integral of (2x - 1)^5 with respect to x, using the substitution u = 2x - 1 or otherwise.
Question 11 [5 marks]
Vectors
Find the angle between the vectors u = 3i + 4j and v = -i + 2j, giving your answer to 1 decimal place.
Question 12 [4 marks]
Proof and Algebraic Methods
Without using a calculator, prove by contradiction that sqrt(3) is an irrational number.
Question 13 [5 marks]
Vectors
Vectors a and b are such that |a| = 5, |b| = 3 and the angle between a and b is 60 degrees.
Find a.b, and find |a + b|.
Question 14 [5 marks]
Proof and Algebraic Methods
Without using a calculator, expand and simplify (3 + sqrt(5))(4 - sqrt(5)), giving your answer in the form a + b*sqrt(5), stating the values of a and b.
Question 15 [5 marks]
Vectors
The points A, B and C have position vectors a = i + 2j - k, b = 3i - j + 3k and c = 5i + tj + 7k respectively, relative to a fixed origin O, where t is a constant.
Given that A, B and C are collinear, find the value of t.
Question 16 [5 marks]
Proof and Algebraic Methods
Use proof by contradiction to show that there is no largest even integer.
Question 17 [6 marks]
Differentiation
A curve is defined implicitly by x*y^2 - x^2 = 8. The point (1, 3) lies on the curve.
Find dy/dx at this point, and hence find the equation of the tangent to the curve at (1, 3), giving your answer in the form y = mx + c.
Question 18 [6 marks]
Proof and Algebraic Methods
The cubic polynomial f(x) = x^3 + ax^2 - 7x + 10 has (x - 1) as a factor.
Find the value of a, and hence find all three roots of the equation f(x) = 0.
Question 19 [6 marks]
Integration
Find the area of the region enclosed between the curves y = x^2 and y = 8 - x^2.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the numerator as 3(x - 2)(x + 2) and the denominator as (x + 3)(x - 2) | M1 |
| cancelling the common factor (x - 2) to give 3(x + 2)/(x + 3) | A1 |
| Final answer: 3(x + 2)/(x + 3) | |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| stating sin(45 deg) = sqrt(2)/2, tan(60 deg) = sqrt(3) and cos(45 deg) = sqrt(2)/2 | M1 |
| computing sin(45 deg) tan(60 deg) = sqrt(6)/2 | M1 |
| the total (sqrt(6) + sqrt(2))/2 | A1 |
| Final answer: (sqrt(6) + sqrt(2))/2 | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| rewriting 6*sqrt(x) as 6x^(1/2) | M1 |
| differentiating each term | M1 |
| dy/dx = 3/sqrt(x) + 5 | A1 |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| |v| = sqrt(5^2 + 12^2) | M1 |
| |v| = 13 | A1 |
| the unit vector (5/13)i - (12/13)j | A1 |
| Final answer: |v| = 13; unit vector = (5/13)i - (12/13)j | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| sin(x) = 0.5 | M1 |
| x = 30 deg | A1 |
| x = 150 deg | A1 |
| Final answer: x = 30 deg or x = 150 deg | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| rewriting 5/x^2 as 5x^-2 | M1 |
| integrating each term | M1 |
| the 2x^4 - 3x^2 terms | A1 |
| 2x^4 - 3x^2 - 5/x + c | A1 |
| Question 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1 | M1 |
| collecting x terms to give x(y - 2) = -1 - 3y | M1 |
| f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2 | A1 |
| Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| applying the quotient rule with u' = 4 and v' = 1 | M1 |
| dy/dx = 9/(x + 2)^2 after simplification | A1 |
| substituting x = 1 | M1 |
| gradient = 1 | A1 |
| Final answer: dy/dx = 9/(x + 2)^2; gradient at x = 1 is 1 | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to tan(x) = 2 | M1 |
| x = 63.4 deg (1 dp) | A1 |
| using the periodicity of tan to find a second solution x + 180 deg | M1 |
| x = 243.4 deg (1 dp) | A1 |
| Final answer: x = 63.4 deg or x = 243.4 deg (1 dp) | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the reverse chain rule, using u = 2x - 1 | M1 |
| integrating to obtain u^6/6 | M1 |
| dividing by the derivative factor 2, giving (2x - 1)^6 / 12 | A1 |
| including the constant of integration + c | A1 |
| Final answer: (2x - 1)^6 / 12 + c | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| computing u.v = (3)(-1) + (4)(2) = 5 | M1 |
| computing |u| = 5 and |v| = sqrt(5) | M1 |
| cos(theta) = 5/(5*sqrt(5)) = 1/sqrt(5) | A1 |
| using the inverse cosine | M1 |
| theta = 63.4 deg (1 dp) | A1 |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| assuming, for contradiction, that sqrt(3) is rational, so sqrt(3) = p/q where p and q are integers with no common factor | B1 |
| squaring to give p^2 = 3q^2 and deducing that p must be a multiple of 3 | M1 |
| substituting p = 3k to give q^2 = 3k^2 and deducing that q must also be a multiple of 3 | M1 |
| identifying that this contradicts p/q being in its lowest terms, so sqrt(3) is irrational | A1 |
| Final answer: Proof: assuming sqrt(3) = p/q in lowest terms forces both p and q to be multiples of 3, a contradiction, so sqrt(3) is irrational. | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| using a.b = |a||b|cos(theta) | M1 |
| a.b = 7.5 | A1 |
| using |a + b|^2 = |a|^2 + 2(a.b) + |b|^2 | M1 |
| |a + b|^2 = 25 + 15 + 9 = 49 | A1 |
| |a + b| = 7 | A1 |
| Final answer: a.b = 7.5, |a + b| = 7 | |
| Question 14[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying out to give the four terms 12 - 3*sqrt(5) + 4*sqrt(5) - 5 | M1 |
| simplifying sqrt(5) x sqrt(5) to 5 | M1 |
| combining the constant terms to 7 | A1 |
| combining the surd terms to sqrt(5) | A1 |
| the final answer 7 + sqrt(5), so a = 7 and b = 1 | A1 |
| Final answer: 7 + sqrt(5), a = 7, b = 1 | |
| Question 15[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding AB = b - a = 2i - 3j + 4k | M1 |
| finding AC = c - a = 4i + (t - 2)j + 8k | M1 |
| recognising AC is a scalar multiple of AB and using the i (or k) components to find the scale factor = 2 | M1 |
| forming the equation t - 2 = 2 x (-3) using the j-components | A1 |
| t = -4 | A1 |
| Question 16[5 marks] | |
|---|---|
| Answer or working | Marks |
| assuming, for contradiction, that there is a largest even integer, N | B1 |
| writing N = 2k for some integer k, since N is even | M1 |
| considering N + 2 = 2k + 2 = 2(k + 1) | M1 |
| recognising 2(k + 1) is also an even integer | A1 |
| identifying that N + 2 > N, which contradicts N being the largest even integer, so no largest even integer exists | A1 |
| Final answer: Proof: assuming a largest even integer N = 2k exists, N + 2 = 2(k + 1) is also even and greater than N, a contradiction, so no largest even integer exists. | |
| Question 17[6 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating x*y^2 using the product rule to give y^2 + 2xy(dy/dx) | M1 |
| differentiating -x^2 to give -2x, with the derivative of the constant 8 equal to 0 | M1 |
| the equation y^2 + 2xy(dy/dx) - 2x = 0 | A1 |
| substituting x = 1, y = 3 and solving for dy/dx | M1 |
| dy/dx = -7/6 | A1 |
| the tangent y = -7/6 x + 25/6, using the point (1, 3) | A1 |
| Final answer: dy/dx = -7/6; tangent: y = -7/6 x + 25/6 | |
| Question 18[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting x = 1 into f(x) and setting f(1) = 0 | M1 |
| forming the equation a + 4 = 0, giving a = -4 | A1 |
| dividing f(x) = x^3 - 4x^2 - 7x + 10 by (x - 1) | M1 |
| the quotient x^2 - 3x - 10 | A1 |
| factorising the quotient as (x - 5)(x + 2) | M1 |
| the full solution set x = 1, x = 5, x = -2, stated together | A1 |
| Final answer: a = -4; x = 1, x = 5, x = -2 | |
| Question 19[6 marks] | |
|---|---|
| Answer or working | Marks |
| setting x^2 = 8 - x^2 to find the intersections | M1 |
| x = -2 and x = 2 | A1 |
| recognising y = 8 - x^2 lies above y = x^2 between the intersections and forming the integral of ((8 - x^2) - x^2) dx | M1 |
| finding the antiderivative 8x - (2/3)x^3 | M1 |
| substituting the limits to obtain 64/3 | A1 |
| area = 64/3 square units, stated as an exact answer | A1 |
| Final answer: Area = 64/3 square units | |