Year 11

Year 11 Paper 3: Differentiation and Its Applications

Covers algebraic manipulation, functions and their graphs, coordinate geometry, differentiation, applications of differentiation, and sequences.

20 questions - 80 marks - calculator allowed

Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [2 marks]

Applications of Differentiation

A curve has equation y = 2x^2 - 5x + 3.

Find the gradient of the tangent to the curve at the point where x = 2.

Question 2 [2 marks]

Differentiation

Find dy/dx when y = x^4 - 3x^2 + 7.

Question 3 [2 marks]

Coordinate Geometry

Find the gradient of the straight line joining the points A(-1, 4) and B(3, -8).

Question 4 [3 marks]

Algebraic Manipulation

Expand and simplify (x - 5)(x + 5) - (x - 3)^2.

Question 5 [3 marks]

Sequences

A geometric sequence has first term 3 and common ratio 2.

Find the 8th term of the sequence, and find the sum of the first 8 terms.

Question 6 [3 marks]

Coordinate Geometry

Find the equation of the straight line that passes through the point (-2, 5) and has gradient -4.

Give your answer in the form y = mx + c.

Question 7 [4 marks]

Functions and their Graphs

The function f is defined by f(x) = x^2 - 3x + 1.

Find f(2) and f(-1).

Question 8 [4 marks]

Coordinate Geometry

A circle has equation (x - 2)^2 + (y + 3)^2 = 25.

State the centre and radius of the circle, then determine whether the point (5, 1) lies inside, outside, or on the circle.

Question 9 [4 marks]

Applications of Differentiation

A curve has equation y = x^3 - 3x + 2.

Find the equation of the normal to the curve at the point where x = 0, giving your answer in the form y = mx + c.

Question 10 [4 marks]

Coordinate Geometry

Line L1 has equation y = 2x - 1.

Line L2 is perpendicular to L1 and passes through the point (4, 5).

Find the equation of L2, giving your answer in the form y = mx + c.

Question 11 [4 marks]

Sequences

The first three terms of an arithmetic sequence are 3k - 1, 5k + 2, and 9k - 3, where k is a constant.

Find the value of k, and find the common difference of the sequence.

Question 12 [4 marks]

Coordinate Geometry

A circle has equation (x + 1)^2 + (y - 4)^2 = 20.

State the centre and radius of the circle, then determine whether the point (4, 5) lies inside, outside, or on the circle.

Question 13 [4 marks]

Differentiation

A curve has equation y = x^3 - 12x + 5.

Find the x-coordinates of the stationary points of the curve.

Question 14 [5 marks]

Sequences

A geometric sequence has second term 6 and fifth term 162.

Find the common ratio and the first term of the sequence.

Question 15 [5 marks]

Applications of Differentiation

A curve has equation y = 3x^2 - 2x + 1.

Find the equation of the tangent to the curve at the point where x = -1, giving your answer in the form y = mx + c.

Question 16 [5 marks]

Coordinate Geometry

A circle has centre (1, 2) and passes through the point (4, 6).

Find the equation of the circle.

Then find the equation of the tangent to the circle at the point (4, 6), giving your answer in the form y = mx + c.

Question 17 [6 marks]

Differentiation

A curve has equation y = x^3 - 6x^2 + 3x + 1.

Find the set of values of x for which the curve is decreasing.

Question 18 [5 marks]

Applications of Differentiation

An open-topped box is made from a square sheet of card of side 12 cm by cutting a square of side x cm from each corner and folding up the sides.

Show that the volume V of the box is given by V = x(12 - 2x)^2.

Find the value of x that maximises the volume, and find this maximum volume.

Question 19 [6 marks]

Functions and their Graphs

f(x) = 5/(x - 2) + 1 for x not equal to 2.

Find f^-1(x), the inverse function of f, stating the value excluded from its domain.

Question 20 [5 marks]

Applications of Differentiation

A closed box has a square base of side x cm and height h cm. The total surface area of the box is 150 cm^2.

Show that the volume V cm^3 of the box is given by V = 37.5x - 0.5x^3.

Find the value of x that maximises V, and find this maximum volume.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
differentiating to get dy/dx = 4x - 5 and substituting x = 2	M1
gradient = 3	A1

Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or working	Marks
differentiating each term using the power rule	M1
dy/dx = 4x^3 - 6x	A1

Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or working	Marks
using gradient = (y2 - y1)/(x2 - x1)	M1
gradient = -3	A1

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
expanding (x - 5)(x + 5) to x^2 - 25	M1
expanding (x - 3)^2 to x^2 - 6x + 9	M1
6x - 34	A1

Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or working	Marks
using T8 = ar^7 = 3 x 2^7	M1
T8 = 384	A1
the sum of the first 8 terms = 765	A1
Final answer: T8 = 384; sum of first 8 terms = 765

Mark scheme for Question 6 [3 marks]
Question 6[3 marks]
Answer or working	Marks
using y - 5 = -4(x + 2)	M1
expanding to y = -4x - 8 + 5	M1
y = -4x - 3	A1

Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or working	Marks
substituting x = 2 into f(x) = x^2 - 3x + 1	M1
f(2) = -1	A1
substituting x = -1 into f(x) = x^2 - 3x + 1	M1
f(-1) = 5	A1
Final answer: f(2) = -1, f(-1) = 5

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
centre (2, -3)	B1
radius 5	B1
calculating the distance from (5, 1) to the centre as sqrt(3^2 + 4^2)	M1
concluding the point lies on the circle since this distance equals 5	A1
Final answer: Centre (2, -3), radius 5; the point (5, 1) lies on the circle

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
finding y = 2 and dy/dx = -3 at x = 0	M1
using the tangent gradient to find the normal gradient as 1/3	M1
using y - 2 = (1/3)(x - 0)	M1
y = (1/3)x + 2	A1
Final answer: y = x/3 + 2

Mark scheme for Question 10 [4 marks]
Question 10[4 marks]
Answer or working	Marks
identifying the gradient of L1 as 2	M1
using the perpendicular gradient rule to find gradient -1/2	M1
using y - 5 = -0.5(x - 4)	M1
y = -0.5x + 7	A1

Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or working	Marks
setting up the equation (T2 - T1) = (T3 - T2)	M1
forming 2k + 3 = 4k - 5	M1
k = 4	A1
common difference = 11	A1
Final answer: k = 4; common difference = 11

Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or working	Marks
centre (-1, 4)	B1
radius sqrt(20), i.e. 2sqrt(5)	B1
calculating the distance from (4, 5) to the centre as sqrt(5^2 + 1^2)	M1
concluding the point lies outside the circle, since sqrt(26) > sqrt(20)	A1
Final answer: Centre (-1, 4), radius 2sqrt(5); the point (4, 5) lies outside the circle

Mark scheme for Question 13 [4 marks]
Question 13[4 marks]
Answer or working	Marks
differentiating to get dy/dx = 3x^2 - 12	M1
setting dy/dx = 0	M1
solving to get x^2 = 4	M1
x = 2 and x = -2	A1

Mark scheme for Question 14 [5 marks]
Question 14[5 marks]
Answer or working	Marks
writing T2 = ar = 6 and T5 = ar^4 = 162	M1
dividing to eliminate a, giving r^3 = 27	M1
r = 3	A1
substituting back into ar = 6 to find a	M1
a = 2	A1
Final answer: common ratio r = 3; first term a = 2

Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or working	Marks
finding y = 6 at x = -1	M1
differentiating to get dy/dx = 6x - 2	M1
substituting x = -1 to find the gradient = -8	M1
using y - 6 = -8(x + 1)	M1
y = -8x - 2	A1

Mark scheme for Question 16 [5 marks]
Question 16[5 marks]
Answer or working	Marks
finding the radius as sqrt(3^2 + 4^2) = 5	M1
the circle equation (x - 1)^2 + (y - 2)^2 = 25	A1
finding the gradient of the radius to (4, 6) as 4/3	M1
using the perpendicular gradient -3/4 with the point (4, 6)	M1
the tangent y = -0.75x + 9	A1
Final answer: Circle: (x - 1)^2 + (y - 2)^2 = 25; tangent: y = -0.75x + 9

Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or working	Marks
differentiating to get dy/dx = 3x^2 - 12x + 3	M1
setting dy/dx = 0 and dividing by 3 to get x^2 - 4x + 1 = 0	M1
using the quadratic formula (or completing the square) to solve	M1
x = 2 + sqrt(3) or x = 2 - sqrt(3)	A1
reasoning that dy/dx < 0 between the roots, since the coefficient of x^2 is positive	M1
2 - sqrt(3) < x < 2 + sqrt(3)	A1

Mark scheme for Question 18 [5 marks]
Question 18[5 marks]
Answer or working	Marks
expanding (12 - 2x)^2 and multiplying by x to show V = 144x - 48x^2 + 4x^3	M1
differentiating to get dV/dx = 144 - 96x + 12x^2	M1
solving dV/dx = 0 to find x = 2, rejecting x = 6 as it gives a degenerate box	M1
x = 2 cm	A1
maximum volume V = 128 cm^3	A1
Final answer: x = 2 cm; maximum volume = 128 cm^3

Mark scheme for Question 19 [6 marks]
Question 19[6 marks]
Answer or working	Marks
subtracting 1 from both sides to give y - 1 = 5/(x - 2)	M1
taking reciprocals of both sides to give 1/(y - 1) = (x - 2)/5	M1
multiplying both sides by 5 to give 5/(y - 1) = x - 2	M1
adding 2 to both sides to give x = 2 + 5/(y - 1)	M1
swapping x and y to give f^-1(x) = 2 + 5/(x - 1)	A1
stating x = 1 is excluded from the domain of f^-1	B1
Final answer: f^-1(x) = 2 + 5/(x - 1), x not equal to 1

Mark scheme for Question 20 [5 marks]
Question 20[5 marks]
Answer or working	Marks
writing the surface area equation 2x^2 + 4xh = 150 and rearranging to h = 37.5/x - x/2	M1
substituting into V = x^2h to show V = 37.5x - 0.5x^3	M1
differentiating to get dV/dx = 37.5 - 1.5x^2 and solving dV/dx = 0 to find x = 5 (rejecting the negative root)	M1
x = 5 cm	A1
maximum volume V = 125 cm^3	A1
Final answer: x = 5 cm; maximum volume = 125 cm^3