A Level Paper 2: Calculus and Vectors
Covers proof and algebraic methods, coordinate geometry, differentiation, integration and vectors.
Questions
Question 1 [2 marks]
Proof and Algebraic Methods
Without using a calculator, show that sqrt(45) + sqrt(20) can be written as k*sqrt(5), stating the value of k.
Question 2 [2 marks]
Integration
Without using a calculator, find the integral of (6x^2 - 4x + 3) with respect to x.
Question 3 [3 marks]
Coordinate Geometry
The points A(-3, 7) and B(5, -1) are given.
Find the distance AB, and find the midpoint of AB.
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 [4 marks]
Differentiation
Using differentiation from first principles, show that the derivative of x^2 is 2x.
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 [4 marks]
Coordinate Geometry
The line l1 has equation x + 2y = 16.
The line l2 is perpendicular to l1 and passes through the point (4, 1).
Find the equation of l2 in the form y = mx + c, and find the coordinates of the point where l1 and l2 intersect.
Question 8 [4 marks]
Integration
Evaluate the definite integral of sin(2x) with respect to x between x = 0 and x = pi/3.
Question 9 [4 marks]
Differentiation
Without using a calculator, find dy/dx for y = sin(3x) + cos(2x), and find the exact gradient of the curve at x = pi/6.
Question 10 [5 marks]
Integration
Without using a calculator, find the integral of sqrt(3x + 1) with respect to x, using the substitution u = 3x + 1 or otherwise.
Question 11 [5 marks]
Differentiation
A curve is defined implicitly by x^2 + y^2 - 4x + 6y = 12. The point (5, 1) lies on the curve.
Using implicit differentiation, find dy/dx in terms of x and y, and find the gradient of the curve at (5, 1).
Question 12 [5 marks]
Integration
The curve y = 6x - x^2 meets the x-axis at the origin and at the point (6, 0).
Find the area enclosed between the curve and the x-axis.
Question 13 [6 marks]
Coordinate Geometry
A circle has equation x^2 + y^2 + 8x - 2y + 8 = 0.
Find the centre and radius of the circle, and determine whether the point (0, 5) lies inside, on, or outside the circle.
Question 14 [5 marks]
Proof and Algebraic Methods
Prove that (2n + 1)^2 - (2n - 1)^2 is a multiple of 8 for every integer n.
Question 15 [6 marks]
Differentiation
A curve has equation y = x^3 - 6x^2 + 9x + 2.
Find the coordinates of the point of inflection on the curve, justifying that it is a point of inflection.
Question 16 [6 marks]
Coordinate Geometry
The circle C has equation x^2 + y^2 = 25.
The line l has equation y = x + c.
Find the two values of c for which l is a tangent to C.
Question 17 [6 marks]
Vectors
Relative to a fixed origin O, points A and B have position vectors 2i + 3j - k and 8i - 3j + 5k.
The point C divides AB in the ratio 2:1 (so that AC:CB = 2:1). Find the position vector of C.
Question 18 [6 marks]
Coordinate Geometry
A circle C has centre (1, 2) and radius 10.
The line l has equation 3x - 4y - 5 = 0.
Find the length of the chord cut off on C by l.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| simplifying sqrt(45) = 3*sqrt(5) and sqrt(20) = 2*sqrt(5) | M1 |
| k = 5 | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| integrating each term | M1 |
| 2x^3 - 2x^2 + 3x + c | A1 |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| using the distance formula sqrt((5 - (-3))^2 + (-1 - 7)^2) | M1 |
| AB = 8*sqrt(2) (or 11.3 to 3 sf) | A1 |
| the midpoint (1, 3) | A1 |
| Final answer: AB = 8*sqrt(2); midpoint = (1, 3) | |
| 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[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing f'(x) = lim as h tends to 0 of [(x + h)^2 - x^2]/h | M1 |
| expanding (x + h)^2 = x^2 + 2xh + h^2 | M1 |
| simplifying the difference quotient to 2x + h | A1 |
| taking the limit as h tends to 0 to give f'(x) = 2x, completing the proof | A1 |
| Final answer: Proof: f'(x) = lim (h -> 0) of (2x + h) = 2x | |
| 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[4 marks] | |
|---|---|
| Answer or working | Marks |
| gradient of l1 = -1/2, so gradient of l2 = 2 | M1 |
| forming l2 as y - 1 = 2(x - 4) | M1 |
| l2: y = 2x - 7 | A1 |
| the intersection point (6, 5) | A1 |
| Final answer: l2: y = 2x - 7; intersection (6, 5) | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative -1/2 cos(2x) | M1 |
| substituting the limits x = pi/3 and x = 0 | M1 |
| -1/2 cos(2pi/3) - (-1/2 cos(0)) | A1 |
| the value 3/4 | A1 |
| Final answer: 3/4 | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating sin(3x) using the chain rule to give 3cos(3x) | M1 |
| differentiating cos(2x) using the chain rule to give -2sin(2x) | M1 |
| dy/dx = 3cos(3x) - 2sin(2x) | A1 |
| substituting x = pi/6 to give the gradient = 3cos(pi/2) - 2sin(pi/3) = 0 - sqrt(3) = -sqrt(3) | A1 |
| Final answer: dy/dx = 3cos(3x) - 2sin(2x); gradient at x = pi/6 is -sqrt(3) | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the reverse chain rule, using u = 3x + 1 | M1 |
| rewriting the integrand as u^(1/2) and integrating to obtain (2/3)u^(3/2) | M1 |
| dividing by the derivative factor 3, giving (2/9)u^(3/2) | A1 |
| substituting back to (2/9)(3x + 1)^(3/2) | A1 |
| including the constant of integration + c | A1 |
| Final answer: (2/9)(3x + 1)^(3/2) + c | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term implicitly: 2x + 2y(dy/dx) - 4 + 6(dy/dx) = 0 | M1 |
| collecting the dy/dx terms: dy/dx (2y + 6) = 4 - 2x | M1 |
| dy/dx = (2 - x)/(y + 3) | A1 |
| substituting the point (5, 1) | M1 |
| the gradient = -3/4 | A1 |
| Final answer: dy/dx = (2 - x)/(y + 3); gradient at (5, 1) is -3/4 | |
| Question 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the limits x = 0 and x = 6, from solving 6x - x^2 = 0 | M1 |
| finding the antiderivative 3x^2 - x^3/3 | M1 |
| substituting x = 6 to get 108 - 72 = 36 | M1 |
| substituting x = 0 to get 0 | A1 |
| area = 36 square units | A1 |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| completing the square on the x terms, (x + 4)^2 - 16 | M1 |
| completing the square on the y terms, (y - 1)^2 - 1 | M1 |
| centre (-4, 1) | A1 |
| radius 3 | A1 |
| finding the distance from (-4, 1) to (0, 5) as sqrt(4^2 + 4^2) = sqrt(32) | M1 |
| concluding the point lies outside the circle, since sqrt(32) (approx 5.66) > 3 | A1 |
| Final answer: centre (-4, 1), radius 3; (0, 5) lies outside the circle | |
| Question 14[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing (2n + 1)^2 - (2n - 1)^2 as a difference of two squares | B1 |
| identifying the two factors as (2n + 1) - (2n - 1) and (2n + 1) + (2n - 1) | M1 |
| simplifying the factors to 2 and 4n respectively | A1 |
| multiplying the factors to obtain 8n | M1 |
| concluding that 8n is a multiple of 8 for every integer n, completing the proof | A1 |
| Final answer: Proof: (2n + 1)^2 - (2n - 1)^2 = 2 x 4n = 8n, a multiple of 8 for every integer n. | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to find dy/dx = 3x^2 - 12x + 9 | M1 |
| differentiating again to find d^2y/dx^2 = 6x - 12 | M1 |
| setting d^2y/dx^2 = 0 to find x = 2 | M1 |
| confirming d^2y/dx^2 changes sign either side of x = 2 (negative for x < 2, positive for x > 2) | A1 |
| the y-coordinate y = 4 | A1 |
| the point of inflection (2, 4) | A1 |
| Final answer: Point of inflection = (2, 4) | |
| Question 16[6 marks] | |
|---|---|
| Answer or working | Marks |
| substituting y = x + c into the equation of C | M1 |
| expanding and simplifying to the quadratic 2x^2 + 2cx + (c^2 - 25) = 0 | M1 |
| using the tangency condition that the discriminant equals zero | M1 |
| forming the equation 4c^2 - 8(c^2 - 25) = 0 | A1 |
| simplifying to c^2 = 50 | M1 |
| c = 5*sqrt(2) or c = -5*sqrt(2) | A1 |
| Question 17[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding AB = b - a | M1 |
| AB = 6i - 6j + 6k | A1 |
| using the ratio to find (2/3)AB | M1 |
| (2/3)AB = 4i - 4j + 4k | A1 |
| forming C = a + (2/3)AB | M1 |
| C = 6i - j + 3k | A1 |
| Question 18[6 marks] | |
|---|---|
| Answer or working | Marks |
| using the perpendicular distance formula |3(1) - 4(2) - 5| / sqrt(3^2 + (-4)^2) | M1 |
| the perpendicular distance = 2 | A1 |
| using half the chord length = sqrt(r^2 - d^2) | M1 |
| half the chord length = sqrt(100 - 4) = sqrt(96) = 4*sqrt(6) | A1 |
| doubling to find the full chord length | M1 |
| the chord length = 8*sqrt(6) (or 19.6 to 3 sf) | A1 |
| Final answer: Chord length = 8*sqrt(6) | |