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INSTRUCTIONS TO CANDIDATES

• Answer all the questions.

• Give non-exact numerical answers correct to 3 significant figures, unless a different

degree of accuracy is specified in the question or is clearly appropriate.

• You are not permitted to use a calculator in this paper.

INFORMATION FOR CANDIDATES

• The number of marks is given in brackets [ ] at the end of each question or part

question.

• You are reminded of the need for clear presentation in your answers.

GCE Examinations

Advanced / Advanced Subsidiary

Core Mathematics C1 Paper 1

Time: 1 hour 30 minutes

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x

y

1. Express

( )5

6

3

1

x

in the form xn, stating the value of n. [4]

2. Factorise 21x2 + 4x − 1. [2]

Hence, or otherwise, solve the equation

0 1 4 21 3

1

3

2

=−+ yy .

Give your answers as fractions. [6]

3. The straight line p is perpendicular to the line with equation x + 2y = 1 and passes

through the point A(a, 2). Find, in terms of the constant a, an equation for the

line p. [6]

Given that the line p crosses the y-axis at the point B(0, 3), find the value of a. [3]

4. Solve the simultaneous equations x − 4y = 2, x2 − 4xy = 8. [6]

5. The diagram shows the curve y = f(x), where a is a positive constant. The

maximum and minimum points on the curve are (−a, 3a) and (a, 0) respectively.

Sketch the following curves, on separate diagrams, in each case stating the

co−ordinates of the maximum and minimum points:

i) y = f(x − a)

ii) y = −2f(x).

[6]

(−a, 3a)

(a, 0) 0

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6. Factorise the polynomial 2x3 − x

2 − 15x. [3]

Hence, or otherwise, i) solve the equation 2x3 − x

2 − 15x = 0 [2]

ii) sketch the graph of y = 2x3 − x

2 − 15x. [3]

7. The diagram shows the graph of the curve with equation y = 1 − x

4.

x

y

Find the co-ordinates of the point, A, where the curve cuts the x-axis. [3]

The line given by y = x + a intersects the curve at one point only.

Find the possible values of a. [6]

State the range of values of a for which the line given by y = x + a does not

intersect the curve. [2]

8. i) The tenth term of an arithmetic progression is 36, and the sum of the first

ten terms is 180. Find the first term and the common difference. [5]

ii) Evaluate ( )200

1

2 5

n

n

=

−∑ . [4]

0

A

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9. A curve has equation y = f(x).

Given that the curve passes through the point A(−1, 1) and is such that

d

d

y

x = 2 − 1

2x

2, find the equation of the curve. [4]

Find the equation of the tangent to the curve at the point A, giving your answer in

the form ax + by = c where a, b and c are integers. [4]

10.

The diagram shows the curve

y = x2 − 3x + 3.

The curve crosses the y-axis at the point Q with coordinates (0, 3), and P is the

point with coordinates (3, 3).

The normal to the curve at P meets the y-axis at S.

Calculate the distance QS. [6]

x

y

P Q

O

S

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SOLUTIONS.

1.

( )5

6

3

1

x

=

( )6

1 53

1

x

= 61

3 5

1

x×

= 25

1

x =

25x

−.

This means that n = −2

5.

2. 21x2 + 4x − 1 = (7x − 1)(3x + 1).

The equation 0 1 4 21 3

1

3

2

=−+ yy is equivalent to ( )( )1 13 37 1 3 1y y− + = 0

⇒ 13y =

1

7 or

13y = −

1

3 and hence we have solutions y =

343

1 or y =

27

1− .

3. Gradient of the line given by x + 2y = 1 is − 12

.

Required equation is of the form y = mx + c.

m = the gradient and is such that − 12

× m = −1 ⇒ m = 2.

Therefore, y = 2x + c for some constant c.

Since the line passes through A(a, 2), put x = a, y = 2 to obtain c = 2 − 2a.

Therefore, the line p has equation y = 2x + 2 − 2a.

Put x = 0, y = 3 to obtain y = 2x + 2 − 2a

3 = 2 − 2a

⇒ a = − 12

.

4. We have: x − 4y = 2, x2 − 4xy = 8.

Substitute x = 4y + 2 to get (4y + 2)2 − 4(4y + 2)y = 8

⇒ 16y2 + 16y + 4 − 16y

2 − 8y = 8

8y = 4

and thus y = 12

.

Substitute in x = 4y + 2 to get x = 4.

5. i) (Translates by a units along the x-axis.)

max = (0, 3a), min = (2a, 0)

ii) (Stretch, scale-factor ×2 along the y-axis and then reflects in the x-axis.)

max = (a, 0), min = (−a, −6a).

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6. Taking out common factors gives 2x3 − x

2 − 15x = x(2x

2 − x − 15)

= x(2x + 5)(x − 3).

i) The equation 2x3 − x

2 − 15x = 0 can be written x(2x + 5)(x − 3) = 0

⇒ x = 0 or x = −2.5 or x = 3.

ii) The solutions in i) are the x coordinates of the points where the graph of

y = 2x3 − x

2 − 15x intersects the x axis.

This quickly leads to the correct graph.

7. A has y-coordinate 0 and so 0 = 1 − x

4

⇒ 0 = x − 4.

This means that A has co-ordinates (4, 0).

If the graphs of y = 1 − x

4 and y = x + a intersect at one point only, then the equation

x + a = 1 − x

4 has only one solution.

Rearrange to get x2 + ax = x − 4

⇒ the equation x2 + (a − 1)x + 4 = 0 has just one solution.

The only way this equation can have a single solution is if it’s discriminant is zero

and hence (a − 1)2 − 4 × 1 × 4 = 0, from which we obtain a = −3 or a = 5.

The range of values of a for which the line given by y = x + a does not

intersect the curve is now easily seen to be −3 < a < 5.

8. i) Denote the first term by a and the common difference by d.

We have u10 = 36 ⇒ a + 9d = 36

and S10 = 180 ⇒ ( )102

2 9a d+ = 180 or 2a + 9d = 36.

Solve simultaneously to get a = 0 and d = 4.

x

y y = 2x

3 − x

2 − 15

0 −2.5 3

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ii) ( )200

1

2 5

n

n

=

−∑ = −3 + (−1) + 1 + 3 + …… + 195.

This is the sum to two hundred terms of an arithmetic progression with first

term a = − 3 and common difference d = 2.

Thus ( )200

1

2 5

n

n

=

−∑ = ( )( )2 12

na d n+ − = 100 × (−6 + 199 × 2) = 39200.

9. Integrate both sides of d

d

y

x = 2 − 1

2x

2 to get y = 2x − 1

6x

3 + K for some constant K.

Now substitute x = −1, y = 1 to get 1 = 2 × (−1) − 16

× (−1)3 + K from which we

easily obtain that K = 176

.

This gives y = 2x − 16

x3 + 17

6.

For the tangent, note first that the gradient of the curve at the point A = d

d

y

x = 2 − 1

2x

2

= 2 − 12

× (−1)2 = 3

2.

Hence the tangent has an equation of the form y = 32

x + c for some constant c.

This is equivalent to 2y = 3x + K for some constant K.

Substitute x = −1, y = 1 to get K = 5.

Therefore, the required tangent has equation 2y = 3x + 5 or, equivalently, 3x − 2y = −5.

10. Find the equation of the normal at P: y = mx + c where m is the gradient.

Differentiate y = x2 − 3x + 3 to get

d

d

y

x = 2x − 3.

⇒ Gradient of tangent at P = 2 × 3 − 3 = 3.

⇒ Gradient of normal at P = M = − 13

(since the product of the gradients of

perpendicular lines = −1).

Thus, the normal at P has equation y = − 13

x + c for some constant c.

Since the normal passes through P(3, 3), substitute x = 3, y = 3 to obtain

3 = − 13

× 3 + c which leads to c = 4.

Therefore, the normal at P has equation y = − 13

x + 4.

Co-ordinates of S are now obtained by substituting x = 0 in the equation for the above

normal.

⇒ S = (0, 4).

Finally, the distance QS = (y co-ordinate of S) − (y co-ordinate of Q)

= 4 − 3

= 1.