paper 1 most likely questions may 2018 [332 marks]...paper 1 most likely questions may 2018 [332...

Paper 1 Most likely questions May 2018 [332 marks]

1a. [2 marks]

In an arithmetic sequence, the first term is 8 and the second term is 5.

Find the common difference.

Markschemesubtracting terms (M1)

eg

A1 N2

[2 marks]

5 − 8, u2 − u1

d = −3

1b. [2 marks]Find the tenth term.

Markschemecorrect substitution into formula (A1)

eg

A1 N2

[2 marks]

u10 = 8 + (10 − 1)(−3), 8 − 27, − 3(10) + 11

u10 = −19

1c. [2 marks]Find the sum of the first ten terms.

Markschemecorrect substitution into formula for sum (A1)

eg

A1 N2

[2 marks]

S10 = (8 − 19), 5 (2(8) + (10 − 1)(−3))102

S10 = −55

[1 mark]2a.

The following diagram shows the graph of a function , with domain .

The points and lie on the graph of .

Write down the range of .

Markschemecorrect range (do not accept ) A1 N1

eg

[1 mark]

f −2 ⩽ x ⩽ 4

(−2, 0) (4, 7) f

f

0 ⩽ x ⩽ 7

[0, 7], 0 ⩽ y ⩽ 7

2b. [1 mark]Write down ;f(2)

Markscheme A1 N1

[1 mark]

f(2) = 3

2c. [1 mark]Write down .

Markscheme A1 N1

[1 mark]

f −1(2)

f −1(2) = 0

[3 marks]2d. On the grid, sketch the graph of .f −1

Markscheme

A1A1A1 N3

Notes: Award A1 for both end points within circles,

A1 for images of and within circles,

A1 for approximately correct reflection in , concave up then concave down shape (do not accept line segments).

[3 marks]

(2, 3) (0, 2)

y = x

^

3a. [4 marks]

The following diagram shows triangle ABC, with , , and .

Show that .

Markschemeevidence of choosing the cosine rule (M1)

eg

correct substitution into RHS of cosine rule (A1)

eg

evidence of correct value for (may be seen anywhere, including in cosine rule) A1

eg

correct working clearly leading to answer A1

eg

AG N0

Note: Award no marks if the only working seen is or (or similar).

[4 marks]

AB = 3 cm BC = 8 cm AB̂C =π

3

AC = 7 cm

c2 = a2 + b2 − ab cosC

32 + 82 − 2 × 3 × 8 × cos π

3

cos π

3

cos = , AC2 = 9 + 64 − (48 × ) , 9 + 64 − 24π

312

12

AC2 = 49, b = √49

AC = 7 (cm)

AC2 = 49 AC = √49

3b. [3 marks]The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

Find the exact perimeter of this shape.

Markschemecorrect substitution for semicircle (A1)

eg

valid approach (seen anywhere) (M1)

eg

A1 N2

[3 marks]

semicircle = (2π × 3.5), × π × 7, 3.5π12

12

perimeter = AB + BC + semicircle, 3 + 8 + ( × 2 × π × ) , 8 + 3 + 3.5π12

72

11 + π (= 3.5π + 11) (cm)72

4a. [2 marks]

Let and , for , where is a constant.

Find .

f(x) = 1 + e−x g(x) = 2x + b x ∈ R b

(g ∘ f)(x)

Markschemeattempt to form composite (M1)

eg

correct function A1 N2

eg

[2 marks]

g(1 + e−x)

(g ∘ f)(x) = 2 + b + 2e−x, 2(1 + e−x) + b

4b. [4 marks]Given that , find the value of .

Markschemeevidence of (M1)

eg , graph with horizontal asymptote when

Note: Award M0 if candidate clearly has incorrect limit, such as .

evidence that (seen anywhere) (A1)

eg , graph of or

with asymptote , graph of composite function with asymptote

correct working (A1)

eg

A1 N2

[4 marks]

limx→+∞(g ∘ f)(x) = −3 b

limx→∞(2 + b + 2e−x) = 2 + b + lim

x→∞(2e−x)

2 + b + 2e−∞ x → ∞

x → 0, e∞, 2e0

e−x → 0

limx→∞(e−x) = 0, 1 + e−x → 1, 2(1) + b = −3, elarge negative number → 0 y = e−x

y = 2e−x y = 0 y = −3

2 + b = −3

b = −5

R

5. [7 marks]Let , for . The following diagram shows part of the graph of and the rectangle OABC, where A is on thenegative -axis, B is on the graph of , and C is on the -axis.

Find the -coordinate of A that gives the maximum area of OABC.

f(x) = 15 − x2 x ∈ R f

x f y

x

Markschemeattempt to find the area of OABC (M1)

eg

correct expression for area in one variable (A1)

eg

valid approach to find maximum area (seen anywhere) (M1)

eg

correct derivative A1

eg


eg

A2 N3

[7 marks]

OA × OC, x × f(x), f(x) × (−x)

area = x(15 − x2), 15x − x3, x3 − 15x

A′(x) = 0

15 − 3x2, (15 − x2) + x(−2x) = 0, − 15 + 3x2

15 = 3x2, x2 = 5, x = √5

x = −√5 (accept A(−√5, 0))

6. [7 marks]Consider , for , where .

The equation has exactly one solution. Find the value of .

f(x) = logk(6x − 3x2) 0 < x < 2 k > 0

f(x) = 2 k

MarkschemeMETHOD 1 – using discriminant

correct equation without logs (A1)

eg

valid approach (M1)

eg

recognizing discriminant must be zero (seen anywhere) M1

eg

correct discriminant (A1)

eg


eg

A2 N2

METHOD 2 – completing the square

correct equation without logs (A1)

eg

valid approach to complete the square (M1)

eg


eg

recognizing conditions for one solution M1

eg


eg

A2 N2

[7 marks]

6x − 3x2 = k2

−3x2 + 6x − k2 = 0, 3x2 − 6x + k2 = 0

Δ = 0

62 − 4(−3)(−k2), 36 − 12k2 = 0

12k2 = 36, k2 = 3

k = √3

6x − 3x2 = k2

3(x2 − 2x + 1) = −k2 + 3, x2 − 2x + 1 − 1 + = 0k2

3

3(x − 1)2 = −k2 + 3, (x − 1)2 − 1 + = 0k2

3

(x − 1)2 = 0, − 1 + = 0k2

3

= 1, k2 = 3k2

3

k = √3

7a. [2 marks]

Let and , for .

Find .

f(x) = 5x g(x) = x2 + 1 x ∈ R

f −1(x)

Markschemeinterchanging and (M1)

eg

A1 N2

[2 marks]

x x

x = 5y

f −1(x) = x

5

7b. [3 marks]Find .

MarkschemeMETHOD 1

attempt to substitute 7 into or (M1)

eg

(A1)

A1 N2

METHOD 2

attempt to form composite function (in any order) (M1)

eg

correct substitution (A1)

eg

A1 N2

[3 marks]

(f ∘ g)(7)

g(x) f(x)

72 + 1, 5 × 7

g(7) = 50

f (50) = 250

5(x2 + 1), (5x)2 + 1

5 × (72 + 1)

(f ∘ g)(7) = 250

2

8a. [4 marks]Find .

Markschemevalid approach to set up integration by substitution/inspection (M1)

eg

correct expression (A1)

eg

A2 N4

Notes: Award A1 if missing “ ”.

[4 marks]

∫ xex2−1dx

u = x2 − 1, du = 2x, ∫ 2xex2−1dx

∫ 2xex2−1dx, ∫ eudu12

12

ex2−1 + c12

+c

8b. [3 marks]Find , given that and .f(x) f ′(x) = xex2−1 f(−1) = 3

Markschemesubstituting into their answer from (a) (M1)

eg


eg

A1 N2

[3 marks]

x = −1

e0, e1−1 = 312

12

+ c = 3, c = 2.512

f(x) = ex2−1 + 2.512

9a. [3 marks]

The first three terms of a geometric sequence are , , , for .

Find the common ratio.

Markschemecorrect use A1

eg

valid approach to find (M1)

eg

A1 N2

[3 marks]

lnx16 lnx8 lnx4 x > 0

logxn = n logx

16 lnx

r

, , , lnx4 = lnx16 × r2un+1

un

lnx8

lnx16

4lnx

8lnx

r = 12

∞

9b. [5 marks]Solve .

Markschemerecognizing a sum (finite or infinite) (M1)

eg

valid approach (seen anywhere) (M1)

eg recognizing GP is the same as part (a), using their value from part (a),

correct substitution into infinite sum (only if is a constant and less than 1) A1

eg


eg

A1 N3

[5 marks]

∞∑k=125−k lnx = 64

24 lnx + 23 lnx, , S∞, 16 lnx + …a

1−r

r r = 12

|r|

, , 32 lnx24 lnx

1− 12

lnx16

12

lnx = 2

x = e2

10a. [4 marks]

The vectors a = and b = are perpendicular to each other.

Find the value of .

Markschemeevidence of scalar product M1

eg a b,

recognizing scalar product must be zero (M1)

eg a b

correct working (must involve combining terms) (A1)

eg

A1 N2

[4 marks]

( 42

) ( k + 3k

)

k

∙ 4(k + 3) + 2k

∙ = 0, 4k + 12 + 2k = 0

6k + 12, 6k = −12

k = −2

10b. [3 marks]Given that c = a + 2b, find c.

Markschemeattempt to substitute their value of (seen anywhere) (M1)

eg b = , 2b =


eg

c = A1 N2

[3 marks]

k

( −2 + 3−2

) ( 2−4

)

( 42

) + ( 2−4

) , ( 4 + 2k + 62 + 2k

)

( 6−2

)

11a. [1 mark]

The random variable is normally distributed with a mean of 100. The following diagram shows the normal curve for .

Let be the shaded region under the curve, to the right of 107. The area of is 0.24.

Write down .

Markscheme A1 N1

[1 mark]

X X

R R

P(X > 107)

P(X > 107) = 0.24 (= , 24%)625

11b. [3 marks]Find .P(100 < X < 107)

Markschemevalid approach (M1)

eg


eg

A1 N2

[3 marks]

P(X > 100) = 0.5, P(X > 100) − P(X > 107)

0.5 − 0.24, 0.76 − 0.5

P(100 < X < 107) = 0.26 (= , 26%)1350

11c. [2 marks]Find .


eg

A1 N2

[2 marks]

P(93 < X < 107)

2 × 0.26, 1 − 2(0.24), P(93 < X < 100) = P(100 < X < 107)

P(93 < X < 107) = 0.52 (= , 52%)1325

12. [6 marks]Let . Given that , find .f ′(x) = 3x2

(x3+1)5f(0) = 1 f(x)


eg

correct integration by substitution/inspection A2

eg

correct substitution into their integrated function (must include ) M1

eg

Note: Award M0 if candidates substitute into or .

(A1)

A1 N4

[6 marks]

∫ f ′dx, ∫ dx3x2

(x3+1)5

f(x) = − (x3 + 1)−4 + c, 14

−1

4(x3+1)4

c

1 = + c, − + c = 1−1

4(03+1)4

14

f ′ f ′′

c = 54

f(x) = − (x3 + 1)−4 + (= + , )14

54

−1

4(x3+1)4

54

5(x3+1)4−1

4(x3+1)4

13a. [2 marks]

The values of the functions and and their derivatives for and are shown in the following table.

Let .

Find .

Markschemeexpressing as a product of and (A1)

eg

A1 N2

[2 marks]

f g x = 1 x = 8

h(x) = f(x)g(x)

h(1)

h(1) f(1) g(1)

f(1) × g(1), 2(9)

h(1) = 18

13b. [3 marks]Find .

Markschemeattempt to use product rule (do not accept ) (M1)

eg

correct substitution of values into product rule (A1)

eg

A1 N2

[3 marks]

h′(8)

h′ = f ′ × g′

h′ = fg′ + gf ′, h′(8) = f ′(8)g(8) + g′(8)f(8)

h′(8) = 4(5) + 2(−3), − 6 + 20

h′(8) = 14

5

14. [7 marks]Solve , for .

Markschemecorrect application of (A1)

eg

correct equation without logs A1

eg

recognizing double-angle identity (seen anywhere) A1

eg

evaluating (A1)

correct working A1

eg and , one correct final answer

(do not accept additional values) A2 N0

[7 marks]

log2(2 sin x) + log2(cosx) = −1 2π < x < 5π

2

loga + logb = logab

log2(2 sin xcosx), log2 + log(sin x) + log(cosx)

2 sin xcosx = 2−1, sin xcosx = , sin 2x =14

12

log(sin 2x), 2 sin xcosx = sin 2x, sin 2x = 12

sin−1( ) = (30∘)12

π

6

x = + 2π, 2x = , , 750∘, 870∘, x =π

1225π

629π

6π

12x = 5π

12

x = , 25π

1229π

12

15a. [2 marks]

Let .

Find the equation of the axis of symmetry of the graph of .

Markschemecorrect approach (A1)

eg

(must be an equation) A1 N2

[2 marks]

f(x) = x2 − 4x + 5

f

, f ′(x) = 2x − 4 = 0, (x2 − 4x + 4) + 5 − 4−(−4)

2

x = 2

15b. [4 marks]

The function can also be expressed in the form .

(i) Write down the value of .

(ii) Find the value of .

f(x) = (x − h)2 + k

h

k

Markscheme(i)

A1 N1

(ii) METHOD 1

valid attempt to find (M1)

eg

correct substitution into their function (A1)

eg

A1 N2

METHOD 2

valid attempt to complete the square (M1)

eg


eg

A1 N2

[4 marks]

h = 2

k

f(2)

(2)2 − 4(2) + 5

k = 1

x2 − 4x + 4

(x2 − 4x + 4) − 4 + 5, (x − 2)2 + 1

k = 1

16a. [3 marks]

Let , where is acute.

Find .

sin θ =√53

θ

cosθ

Markschemeevidence of valid approach (M1)

egright triangle,


egmissing side is 2,

A1 N2

[3 marks]

cos2θ = 1 − sin2θ

√1 − ( )2√53

cosθ = 23


Markschemecorrect substitution into formula for (A1)

eg

A1 N2

[2 marks]

cos2θ

cos2θ

2 × ( )2− 1, 1 − 2( )2

, ( )2− ( )2

23

√53

23

√53

cos2θ = − 19

17a. [2 marks]

The values in the fourth row of Pascal’s triangle are shown in the following table.

Write down the values in the fifth row of Pascal’s triangle.

Markscheme1, 5, 10, 10, 5, 1 A2 N2

[2 marks]

17b. [5 marks]Hence or otherwise, find the term in in the expansion of .

Markschemeevidence of binomial expansion with binomial coefficient (M1)

eg

, selecting correct term,

correct substitution into correct term (A1)(A1)(A1)

eg

Note: Award A1 for each factor.

A1 N2

Notes: Do not award any marks if there is clear evidence of adding instead of multiplying.

Do not award final A1 for a final answer of 720, even if is seen previously.

[5 marks]

x3 (2x + 3)5

( n

r) an−rbr (2x)5(3)0 + 5(2x)4(3)1 + 10(2x)3(3)2 + …

10(2)3(3)2, ( 53

) (2x)3(3)2

720x3

720x3

18a. [2 marks]

Events and are independent with and .

Find .

Markschemevalid interpretation (may be seen on a Venn diagram) (M1)

eg

A1 N2

[2 marks]

A B P(A ∩ B) = 0.2 P(A′ ∩ B) = 0.6

P(B)

P(A ∩ B) + P(A′ ∩ B), 0.2 + 0.6

P(B) = 0.8


Markschemevalid attempt to find (M1)

eg

correct working for (A1)

eg

correct working for (A1)

eg

A1 N3

[4 marks]

P(A ∪ B)

P(A)

P(A ∩ B) = P(A) × P(B), 0.8 × A = 0.2

P(A)

0.25, 0.20.8

P(A ∪ B)

0.25 + 0.8 − 0.2, 0.6 + 0.2 + 0.05

P(A ∪ B) = 0.85

( )

19. [7 marks]Let . Find , given that .

Markschemeevidence of integration (M1)

eg

correct integration (accept missing ) (A2)

eg

substituting initial condition into their integrated expression (must have ) M1

eg

Note: Award M0 if they substitute into the original or differentiated function.

recognizing (A1)

eg

(A1)

A1 N5

[7 marks]

f ′(x) = sin3(2x)cos(2x) f(x) f ( ) = 1π

4

∫ f ′(x)dx

C

× , sin4(2x) + C12

sin4(2x)

418

+C

1 = sin4( ) + C18

π

2

sin( ) = 1π

2

1 = (1)4 + C18

C = 78

f(x) = sin4(2x) +18

78

R

20a. [1 mark]

Let and , for .

Write down .

Markscheme A1 N1

[1 mark]

f(x) = 8x + 3 g(x) = 4x x ∈ R

g(2)

g(2) = 8


Markschemeattempt to form composite (in any order) (M1)

eg

A1 N2

[2 marks]

(f ∘ g)(x)

f(4x), 4 × (8x + 3)

(f ∘ g)(x) = 32x + 3

20c. [2 marks]Find .f −1(x)

Markschemeinterchanging and (may be seen at any time) (M1)

eg

A1 N2

[2 marks]

x y

x = 8y + 3

f −1(x) = (accept , y = )x−38

x−38

x−38

21a. [3 marks]

The following Venn diagram shows the events and , where and

. The values and are probabilities.

(i) Write down the value of .

(ii) Find the value of .

Markscheme(i)

A1 N1

(ii) appropriate approach (M1)

eg

A1 N2

[3 marks]

A B

P(A) = 0.4, P(A ∪ B) = 0.8P(A ∩ B) = 0.1 p q

q

p

q = 0.1

P(A) − q, 0.4 − 0.1

p = 0.3



eg

correct values (A1)

eg

A1 N2

[3 marks]

P(B)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B), P(A ∩ B) + P(B ∩ A′)

0.8 = 0.4 + P(B) − 0.1, 0.1 + 0.4

P(B) = 0.5

22a. [2 marks]

Consider . The graph of has a minimum value when .

The distance between the two zeros of is 9.

Show that the two zeros are 3 and .

f(x) = x2 + qx + r f x = −1.5

f

−6

Markschemerecognition that the -coordinate of the vertex is

(seen anywhere) (M1)

eg axis of symmetry is , sketch,

correct working to find the zeroes A1

eg

and AG N0

[2 marks]

x

−1.5

−1.5 f ′(−1.5) = 0

−1.5 ± 4.5

x = −6 x = 3

22b. [4 marks]Find the value of and of .q r

MarkschemeMETHOD 1 (using factors)

attempt to write factors (M1)

eg

correct factors A1

eg

A1A1 N3

METHOD 2 (using derivative or vertex)

valid approach to find (M1)

eg

A1

correct substitution A1

eg

A1

N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations (M1)

eg

one correct value

eg A1

correct substitution A1

eg

second correct value A1

eg

N3

[4 marks]

(x − 6)(x + 3)

(x − 3)(x + 6)

q = 3, r = −18

q

f ′(−1.5) = 0, − = −1.5q

2a

q = 3

32 + 3(3) + r = 0, (−6)2 + 3(−6) + r = 0

r = −18

q = 3, r = −18

9 + 3q + r = 0, 36 − 6q + r = 0

q = 3, r = −18

32 + 3(3) + r = 0, (−6)2 + 3(−6) + r = 0, 32 + 3q − 18 = 0, 36 − 6q − 18 = 0

q = 3, r = −18

q = 3, r = −18

23. [8 marks]Let and , for, where. The graphs of and intersect at exactly one point. Find the value of .

Markschemediscriminant

(seen anywhere) M1

valid approach (M1)

eg

rearranging their equation (to equal zero) (M1)

eg

recognizing LHS is quadratic (M1)

eg

correct substitution into discriminant A1

eg

correct working to find discriminant or solve discriminant (A1)

eg

correct simplification (A1)

egx

A1 N2

[8 marks]

f(x) = 3tan4x + 2k g(x) = −tan4x + 8ktan2x + k

0 ⩽ x ⩽ 10 < k < 1 f g k

= 0

f = g, 3tan4x + 2k = −tan4x + 8ktan2x + k

4tan4x − 8ktan2x + k = 0, 4tan4x − 8ktan2x + k

4(tan2x)2 − 8ktan2x + k = 0, 4m2 − 8km + k

(−8k)2 − 4(4)(k)

= 0

64k2 − 16k, −(−16)±√162

2×64

16k(4k − 1), 322×64

k = 14

24a. [2 marks]

There are 10 items in a data set. The sum of the items is 60.

Find the mean.


eg

A1 N2

6010

mean = 6

24b. [3 marks]

The variance of this data set is 3. Each value in the set is multiplied by 4.

(i) Write down the value of the new mean.

(ii) Find the value of the new variance.

Markscheme(i) new mean A1 N1

(ii) valid approach (M1)

eg , new standard deviation

new variance A1 N2

[3 marks]

= 24

variance × (4)2, 3 × 16 = 4√3

= 48

25a. [2 marks]

Let and . Write the following expressions in terms of and .

.


eg

A1 N2

[2 marks]

x = ln3 y = ln5 x y

ln( )53

ln5 − ln3

ln( ) = y − x53

25b. [4 marks].

Markschemerecognizing factors of 45 (may be seen in log expansion) (M1)

eg

correct application of (A1)

eg


eg

A1 N3

[4 marks]

ln45

ln(9 × 5), 3 × 3 × 5, log32 × log5

log(ab) = loga + logb

ln9 + ln5, ln3 + ln3 + ln5, ln32 + ln5

2 ln3 + ln5, x + x + y

ln45 = 2x + y

26a. [5 marks]

Let, for , and

, for .

Let .

Write in the form , where .

Markschemeattempt to form composite in any order (M1)

eg


eg

correct application of Pythagorean identity (do not accept ) (A1)

eg

valid approach (do not accept ) (M1)

eg

A1 N3

[5 marks]

f(x) = 6x√1 − x2 −1 ⩽ x ⩽ 1g(x) = cos(x) 0 ⩽ x ⩽ π

h(x) = (f ∘ g)(x)

h(x) asin(bx) a, b ∈ Z

f (g(x)) , cos(6x√1 − x2)

6 cosx√1 − cos2x

sin2x + cos2x = 1

sin2x = 1 − cos2x, 6 cosx sin x, 6 cosx |sin x|

2 sin xcosx = sin 2x

3(2 cosx sin x)

h(x) = 3 sin 2x

26b. [2 marks]Hence find the range of .h


eg amplitude , sketch with max and min -values labelled,

correct range A1 N2

eg

, from to 3

Note: Do not award A1 for or for “between and 3”.

[2 marks]

= 3 y −3 < y < 3

−3 ⩽ y ⩽ 3 [−3, 3] −3

−3 < y < 3−3

27. [7 marks]Let ui

j k and v

jk , where . Given that v is a unit vector perpendicular to u, find the possible values of and of .

= −3++= m

+ n m, n ∈ R m n

Markschemecorrect scalar product (A1)

eg

setting up their scalar product equal to 0 (seen anywhere) (M1)

eg u v

correct interpretation of unit vector (A1)

eg

valid attempt to solve their equations (must be in one variable) M1

eg

correct working A1

eg

both correct pairs A2 N3

eg and and ,

and and

Note: Award A0 for , or any other answer that does not clearly indicate the correct pairs.

[7 marks]

m + n

∙ = 0, − 3(0) + 1(m) + 1(n) = 0, m = −n

√02 + m2 + n2 = 1, m2 + n2 = 1

(−n)2 + n2 = 1, √1 − n2 + n = 0, m2 + (−m)2 = 1, m − √1 − m2 = 0

2n2 = 1, 2m2 = 1, √2 = , m = ±1n

1√2

m = 1√2

n = − , m = −1√2

1√2

n = 1√2

m = (0.5)12 n = −(0.5) , m = −√1

212

n = √ 12

m = ± , n = ±1√2

1√2

28. [6 marks]Let . Given that , find .

Markschemeevidence of antidifferentiation (M1)

eg

correct integration (accept absence of ) (A1)(A1)

attempt to substitute into their integrated expression (must have ) M1

eg

Note: Award M0 if substituted into original or differentiated function.

correct working to find (A1)

eg

A1 N4

[6 marks]

f ′(x) = 6x2 − 5 f(2) = −3 f(x)

f = ∫ f ′

C

f(x) = − 5x + C, 2x3 − 5x6x3

3

(2, − 3) C

2(2)3 − 5(2) + C = −3, 16 − 10 + C = −3

C

16 − 10 + C = −3, 6 + C = −3, C = −9

f(x) = 2x3 − 5x − 9

R

29a. [3 marks]

Let , for .

Find .

Markschemeinterchanging and (seen anywhere) (M1)

eg

evidence of correct manipulation (A1)

eg

A1 N2

Notes: If working shown, and they do not interchange and , award A1A1M0 for .

If no working shown, award N1 for .

f(x) = (x − 5)3 x ∈ R

f −1(x)

x y

x = (y − 5)3

y − 5 = 3√x

f −1(x) = 3√x + 5 (accept 5 + x , y = 5 + 3√x )13

x y 3√y + 5

3√y + 5

29b. [3 marks]Let be a function so that . Find

.g (f ∘ g)(x) = 8x6

g(x)

MarkschemeMETHOD 1

attempt to form composite (in any order) (M1)

eg


eg

A1 N2

METHOD 2

recognising inverse relationship (M1)

eg

correct working

eg (A1)

A1 N2

g ((x − 5)3) , (g(x) − 5)3 = 8x6, f(2x2 + 5)

g − 5 = 2x2, ((2x2 + 5) − 5)3

g(x) = 2x2 + 5

f −1(8x6) = g(x), f −1(f ∘ g)(x) = f −1(8x6)

g(x) = 3√(8x6) + 5

g(x) = 2x2 + 5

30. [7 marks]In the expansion of, the coefficient of the term in is

, where . Find .(3x + 1)n x2

135n n ∈ Z+ n

MarkschemeNote: Accept sloppy notation (such as missing brackets, or binomial coefficient which includes ).

evidence of valid binomial expansion with binomial coefficients (M1)

eg

attempt to identify correct term (M1)

eg

setting correct coefficient or term equal to (may be seen later) A1

eg

correct working for binomial coefficient (using formula) (A1)

eg

EITHER

evidence of correct working (with linear equation in ) (A1)

eg

correct simplification (A1)

eg

A1 N2

OR

evidence of correct working (with quadratic equation in ) (A1)

eg

evidence of solving (A1)

eg

A1 N2

Note: Award A0 for additional answers.

[7 marks]

x2

( n

r) (3x)r(1)n−r, (3x)n + n(3x)n−1 + ( n

2) (3x)n−2 + … , ( n

r) (1)n−r(3x)r

( n

n − 2) , (3x)2, n − r = 2

135n

9 ( n

2) = 135n, ( n

n − 2) (3x)2 = 135n, = 135nx29n(n−1)

2

nCr

, n(n−1)(n−2)(n−3)…

2×1×(n−2)(n−3)(n−4)…

n(n−1)

2

n

= 135, x2 = 135x29(n−1)

2

9(n−1)

2

n − 1 = , = 15135×29

(n−1)

2

n = 31

n

9n2 − 279n = 0, n2 − n = 30n, (9n2 − 9n)x2 = 270nx2

9n(n − 31) = 0, 9n2 = 279n

n = 31

31. [6 marks]An arithmetic sequence has the first term and a common difference .

The 13th term in the sequence is . Find the value of.

lna ln3

8 ln9a

MarkschemeNote: There are many approaches to this question, and the steps may be done in any order. There are 3 relationships they may needto apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is

log rules: ,

exponent rule: .

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3 relationships, they may be awardedthe A marks, even if they have made a previous error. However all applications of a relationship need to be correct. Once an error hasbeen made, do not award A1FT for their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into formula (A1)

eg

set up equation for in any form (seen anywhere) (M1)

eg

correct application of relationships (A1)(A1)(A1)

A1 N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs (A1)

eg

correct application of addition rule for logs (A1)

eg

substituting for 9 or 3 in ln expression in equation (A1)

eg

Example 2

recognising (A1)

eg

one correct application of exponent rule for logs relating to (A1)

eg

another correct application of exponent rule for logs (A1)

eg

32

lnb + lnc = ln(bc), lnb − lnc = ln( )b

c

lnbn = n lnb

u13

lna + (13 − 1) ln3

u13

lna + 12 ln3 = 8 ln9

a = 81

lna + ln312 = ln98

ln(a312) = ln98

ln(a312) = ln316, ln(a96) = ln98

9 = 32

lna + 12 ln3 = 8 ln32, lna + 12 ln9 = 8 ln912

ln9 ln3

lna + 12 ln3 = 16 ln3, lna + 6 ln9 = 8 ln9

lna = ln34, lna = ln92

32a. [2 marks]Given that and , write down the value of and of .2m = 8 2n = 16 m n

Markscheme A1A1 N2

[2 marks]

m = 3, n = 4

32b. [4 marks]Hence or otherwise solve .

Markschemeattempt to apply (M1)

eg

equating their powers of (seen anywhere) M1

eg

correct working A1

eg

A1 N2

[4 marks]

Total [6 marks]

82x+1 = 162x−3

(2a)b = 2ab

6x + 3, 4(2x − 3)

2

3(2x + 1) = 8x − 12

8x − 12 = 6x + 3, 2x = 15

x = (7.5)152

33a. [4 marks]

Given that, where is an obtuse angle,

find the value of

sin x = 34

x

cosx;


eg ,


eg

correct calculation (A1)

eg

A1 N3

[4 marks]

sin2x + cos2x = 1

42 − 32, cos2x = 1 − ( )234

, cos2x =√7

47

16

cosx = −√7

4

33b. [3 marks]find the value of

Markschemecorrect substitution (accept missing minus with cos) (A1)

eg

correct working A1

eg

A1 N2

[3 marks]

Total [7 marks]

cos2x.

1 − 2( )2, 2(− )2

− 1, ( )2− ( )23

4

√7

4

√7

434

2 ( ) − 1, 1 − , −716

1816

716

916

cos2x = − (= − )216

18

5

34a. [3 marks]

Let .

Show that the discriminant of is .

Markschemecorrect substitution into A1

eg

correct expansion of each term A1A1

eg

AG N0

[3 marks]

f(x) = px2 + (10 − p)x + p − 554

f(x) 100 − 4p2

b2 − 4ac

(10 − p)2 − 4(p)( p − 5)54

100 − 20p + p2 − 5p2 + 20p, 100 − 20p + p2 − (5p2 − 20p)

100 − 4p2

34b. [3 marks]Find the values of so that has two equal roots.

Markschemerecognizing discriminant is zero for equal roots (R1)

eg


eg , correct value of

both correct values A1 N2

[3 marks]

Total [6 marks]

p f(x) = 0

D = 0, 4p2 = 100

p2 = 25 1 p

p = ±5

35. [8 marks]Let , for . The following diagram shows the graph of .

There are-intercepts at .

The shaded region is enclosed by the graph of , the line , where , and the -axis. The area of is

. Find the value of .

Markschemeattempt to set up integral (accept missing or incorrect limits and missing ) M1

eg

correct integration (accept missing or incorrect limits) (A1)

eg

substituting correct limits into their integrated function and subtracting (in any order) (M1)

eg

(seen anywhere) (A1)

setting their result from an integrated function equal to M1

eg

evaluating or (A1)

eg

identifying correct value (A1)

eg

A1 N3

[8 marks]

f(x) = cosx 0 ≤ x ≤ 2π f

x x = , π

23π

2

R f x = b b > 3π

2x R

(1 − )√3

2b

dx

∫ b cosxdx, ∫ b

a cosxdx, ∫ bfdx, ∫ cosx3π

23π

2

[sin x]b , sin x3π

2

sin b − sin( ), sin( ) − sin b3π

23π

2

sin( ) = −13π

2

(1 − )√3

2

sin b = −√3

2

sin−1( ) =√3

2π

3sin−1(− ) = −

√3

2π

3

b = , − 60∘π

3

2π − , 360 − 60π

3

b = 5π

3

Z

36a. [3 marks]Write the expression in the form , where .

Markschemecorrect application of (seen anywhere) (A1)

eg


eg

A1 N2

[3 marks]

3 ln2 − ln4 lnk k ∈ Z

lnab = b lna

ln4 = 2 ln2, 3 ln2 = ln23, 3 log2 = log8

3 ln2 − 2 ln2, ln8 − ln4

ln2 (accept k = 2)

36b. [3 marks]Hence or otherwise, solve .

MarkschemeMETHOD 1

attempt to substitute their answer into the equation (M1)

eg

correct application of a log rule (A1)

eg

A1 N2

METHOD 2

attempt to rearrange equation, with written as or (M1)

eg

correct working applying (A1)

eg

A1 N2

[3 marks]

Total [6 marks]

3 ln2 − ln4 = −lnx

ln2 = −lnx

ln , ln = lnx, ln2 + lnx = ln2x (= 0)1x

12

x = 12

3 ln2 ln23 ln8

lnx = ln4 − ln23, ln8 + lnx = ln4, ln23 = ln4 − lnx

lna ± lnb

, 8x = 4, ln23 = ln48

4x

x = 12

37a. [2 marks]

Let. The vertex of the graph of

is at and the graph passes through.

Write down the value of

and of .

Markscheme A1A1 N2

[2 marks]

f(x) = a(x − h)2 + k

f(2, 3)(1, 7)

h

k

h = 2, k = 3

37b. [3 marks]Find the value of.

Markschemeattempt to substitute

in any order into their

(M1)

eg

correct equation (A1)

eg

a = 4 A1 N2

[3 marks]

a

(1, 7)

f(x)

7 = a(1 − 2)2 + 3, 7 = a(1 − 3)2 + 2, 1 = a(7 − 2)2 + 3

7 = a + 3

38a. [4 marks]

Let.

Find.

Markschemesubstituting for

(may be seen in integral) A1

eg

correct integration,

(A1)

substituting limits into their integrated function and subtracting (in any order) (M1)

eg

A1 N2

[4 marks]

f(x) = x2

∫ 21 (f(x))2dx

(f(x))2

(x2)2, x4

∫ x4dx = x515

− , (1 − 4)25

515

15

∫ 21 (f(x))2dx = (= 6.2)31

5

38b. [2 marks]The following diagram shows part of the graph of

.

The shaded region

is enclosed by the graph of

, the

-axis and the lines

and

.Find the volume of the solid formed when

is revolved about the

-axis.

Markschemeattempt to substitute limits or function into formula involving

(M1)

eg

A1 N2

[2 marks]

f

R

f

x

x = 1

x = 2

R360∘

x

f 2

∫ 21 (f(x))2dx, π ∫ x4dx

π (= 6.2π)315

39. [7 marks]Let . Find the value of

.∫ a

πcos 2xdx = , where π < a < 2π1

2a

Markschemecorrect integration (ignore absence of limits and “

”) (A1)

eg


eg

(A1)

setting their result from an integrated function equal to

M1

eg

recognizing

(A1)

eg

correct value (A1)

eg

A1 N3

[7 marks]

+C

, ∫ aπ

cos 2x = [ sin(2x)]a

π

sin(2x)2

12

sin(2a) − sin(2π), sin(2π) − sin(2a)12

12

sin(2π) = 0

12

sin 2a = , sin(2a) = 112

12

sin−11 = π2

2a = , a =π2

π4

+ 2π, 2a = , a = + ππ2

5π2

π4

a = 5π4

[2 marks]40a.

Find the value of each of the following, giving your answer as an integer.


eg

A1 N2

[2 marks]

log636

6x = 36, 62

2

[2 marks]40b.

Markschemecorrect simplification (A1)

eg

A1 N2

[2 marks]

log64 + log69

log636, log(4 × 9)

2

[3 marks]40c. log62 − log612

Markschemecorrect simplification (A1)

eg


eg

A1 N2

[3 marks]

log6 , log(2 ÷ 12)212

log6 , 6−1 = , 6x =16

16

16

−1

41a. [2 marks]

The line

is parallel to the vector

.

Find the gradient of the line

.

Markschemeattempt to find gradient (M1)

eg reference to change in

is

and/or

is

,

gradient

A1 N2

[2 marks]

L

( 32

)

L

x

3

y

232

= 23

41b. [3 marks]

The line passes through the point

.

Find the equation of the line

in the form

.

Markschemeattempt to substitute coordinates and/or gradient into Cartesian equationfor a line (M1)eg

correct substitution (A1)eg

A1 N2

[3 marks]

L(9, 4)

L

y = ax + b

y − 4 = m(x − 9), y = x + b, 9 = a(4) + c23

4 = (9) + c, y − 4 = (x − 9)23

23

y = x − 2 (accept a = , b = −2)23

23

41c. [2 marks]Write down a vector equation for the line

.

Markschemeany correct equation in the form r = a + tb (any parameter for t), where a indicates position eg

or

, and b is a scalar multiple of

eg r =

, r = 0i − 2 j + s(3i + 2 j) A2 N2

Note: Award A1 for a + tb, A1 for L = a + tb, A0 for r = b + ta.

[2 marks]

L

( 94

)( 0

−2)

( 32

)

( 94

) + t ( 32

) , ( x

y) = ( 3t + 9

2t + 4)

42. [6 marks]The graph of a function h passes through the point

.

Given that

, find

.

Markschemeevidence of anti-differentiation (M1)

eg

correct integration (A2)

eg

attempt to substitute

into their equation (M1)

eg


eg

A1 N5

[6 marks]

( , 5)π12

h′(x) = 4 cos 2x

h(x)

∫ h′(x), ∫ 4 cos 2xdx

h(x) = 2 sin 2x + c, 4 sin 2x2

( , 5)π12

2 sin(2 × ) + c = 5, 2 sin( ) = 5π12

π6

2 ( ) + c = 5, c = 412

h(x) = 2 sin 2x + 4

43a. [4 marks]

The sums of the terms of a sequence follow the pattern

Given that

, find

and

.

S1 = 1 + k, S2 = 5 + 3k, S3 = 12 + 7k, S4 = 22 + 15k, … , where k ∈ Z.

u1 = 1 + k

u2, u3

u4

Markschemevalid method (M1)

eg

A1A1A1 N4

[4 marks]

u2 = S2 − S1, 1 + k + u2 = 5 + 3k

u2 = 4 + 2k, u3 = 7 + 4k, u4 = 10 + 8k

43b. [4 marks]Find a general expression for

.

Markschemecorrect AP or GP (A1)

eg finding common difference is

, common ratio is

valid approach using arithmetic and geometric formulas (M1)

eg

and

A1A1 N4

Note: Award A1 for

, A1 for

.

[4 marks]

un

3

2

1 + 3(n − 1)

rn−1k

un = 3n − 2 + 2n−1k

3n − 2

2n−1k

44a. [2 marks]

Consider a function

such that

.

Find

.

Markschemeappropriate approach (M1)

eg

A1 N2

[2 marks]

f(x)

∫ 61 f(x)dx = 8

∫ 61 2f(x)dx

2 ∫ f(x), 2(8)

∫ 61 2f(x)dx = 16

44b. [4 marks]Find

.∫ 61 (f(x) + 2) dx

Markschemeappropriate approach (M1)

eg

(seen anywhere) (A1)


eg

A1 N3

[4 marks]

∫ f(x) + ∫ 2, 8 + ∫ 2

∫ 2dx = 2x

2(6) − 2(1), 8 + 12 − 2

∫ 61 (f(x) + 2) dx = 18

45. [6 marks]Let. The line

is the tangent to the curve of at

.Find the equation of

in the form.

Markschemerecognising need to differentiate (seen anywhere) R1

eg

attempt to find the gradient when

(M1)

eg

(A1)

attempt to substitute coordinates (in any order) into equation of a straight line (M1)

eg


eg

A1 N3

[6 marks]

f(x) = e2x

Lf(1, e2)

Ly = ax + b

f ′, 2e2x

x = 1

f ′(1)

f ′(1) = 2e2

y − e2 = 2e2(x − 1), e2 = 2e2(1) + b

y − e2 = 2e2x − 2e2, b = −e2

y = 2e2x − e2

46a. [4 marks]

Consider .

Find .

f(x) = x2 sin x

f ′(x)

Markschemeevidence of choosing product rule (M1)

eg

correct derivatives (must be seen in the product rule) ,

(A1)(A1)

A1 N4

[4 marks]

uv′ + vu′

cos x2x

f ′(x) = x2 cos x + 2x sin x

46b. [3 marks]Find the gradient of the curve of at

.

Markschemesubstituting

into their (M1)

eg

,

correct values for both and seen in (A1)

eg

A1 N2

[3 marks]

fx = π

2

π2f ′(x)

f ′ ( )π2

( )2cos( ) + 2 ( ) sin( )π

2π2

π2

π2

sin π2

cos π2

f ′(x)

0 + 2 ( ) × 1π2

f ′ ( ) = ππ2

47a. [3 marks]

Let , for

.

Find .

f(x) = √x − 5x ≥ 5

f −1(2)

MarkschemeMETHOD 1

attempt to set up equation (M1)

eg ,


eg ,

A1 N2

METHOD 2

interchanging and (seen anywhere) (M1)

eg


eg ,

A1 N2

[3 marks]

2 = √y − 52 = √x − 5

4 = y − 5x = 22 + 5

f −1(2) = 9

xy

x = √y − 5

x2 = y − 5y = x2 + 5

f −1(2) = 9

47b. [3 marks]Let be a function such that

exists for all real numbers. Given that , find

.

Markschemerecognizing

(M1)

eg


eg ,

A1 N2

Note: Award A0 for multiple values, eg .

[3 marks]

gg−1

g(30) = 3(f ∘ g−1)(3)

g−1(3) = 30

f(30)

(f ∘ g−1)(3) = √30 − 5√25

(f ∘ g−1)(3) = 5

±5

48a. [2 marks]

Let and .

Find .

log3p = 6log3q = 7

log3p2

MarkschemeMETHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using (M1)

eg ,

,

A1 N2

[2 marks]

log un = n log u2log3p

log3(p2) = 12

p = 36

log3(36)2

log 312

12log33

log3(p2) = 12

48b. [2 marks]Find

.

MarkschemeMETHOD 1


eg

,

A1 N2

METHOD 2

valid method using and (M1)

eg

,

,

A1 N2

[2 marks]

log3 ( )p

q

log( ) = log p − log qp

q

6 − 7

log3 ( ) = −1p

q

p = 36

q = 37

log3 ( )36

37

log 3−1

−log33

log3 ( ) = −1p

q

48c. [3 marks]Find .log3(9p)

MarkschemeMETHOD 1


eg ,

(may be seen in expression) A1

eg

A1 N2

METHOD 2

valid method using (M1)

eg ,

correct working A1

eg ,

A1 N2

[3 marks]

Total [7 marks]

log3uv = log3u + log3vlog 9 + log p

log39 = 2

2 + log p

log3(9p) = 8

p = 36

log3(9 × 36)log3(32 × 36)

log39 + log336

log338

log3(9p) = 8

[3 marks]49a.

The following table shows the probability distribution of a discrete random variable X .

Find the value of k .

Markschemeevidence of summing to 1 (M1)

e.g.


e.g.

A1 N2

[3 marks]

∑p = 1, 0.3 + k + 2k + 0.1 = 1

0.4 + 3k, 3k = 0.6

k = 0.2

49b. [3 marks]Find .E(X)

Markschemecorrect substitution into formula

(A1)

e.g.

correct working

e.g. (A1)

= 3.3 A1 N2

[3 marks]

E(X)

0(0.3) + 2(k) + 5(2k) + 9(0.1), 12k + 0.9

0(0.3) + 2(0.2) + 5(0.4) + 9(0.1), 0.4 + 2.0 + 0.9

E(X)

50a. [3 marks]Let. Find an expression for

in terms of m.

MarkschemeNote: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer,do not award the final A1FT mark.

METHOD 1

valid approach involving Pythagoras (M1)

e.g.

, labelled diagram

correct working (may be on diagram) (A1)

e.g. ,

A1 N2

[3 marks]

METHOD 2

valid approach involving tan identity (M1)

e.g.


e.g.

A1 N2

[3 marks]

sin 100∘ = mcos 100∘

sin2x + cos2x = 1

m2 + (cos 100)2 = 1√1 − m2

cos 100 = −√1 − m2

tan = sincos

cos 100 = sin 100tan 100

cos 100 = mtan 100

50b. [1 mark]Let . Find an expression for

in terms of m.sin 100∘ = mtan 100∘

MarkschemeMETHOD 1

(accept

) A1 N1

[1 mark]

METHOD 2

A1 N1

[1 mark]

tan 100 = − m

√1−m2

m

−√1−m2

tan 100 = mcos 100

50c. [2 marks]Let. Find an expression for

in terms of m.

MarkschemeMETHOD 1

valid approach involving double angle formula (M1)

e.g.

(accept

) A1 N2

Note: If candidates find , award full FT in parts (b) and (c), even though the values may not have appropriate signs for the

angles.

[2 marks]

METHOD 2

valid approach involving double angle formula (M1)

e.g. ,

A1 N2

[2 marks]

sin 100∘ = msin 200∘

sin 2θ = 2 sin θcosθ

sin 200 = −2m√1 − m2

2m (−√1 − m2)

cos 100 = √1 − m2

sin 2θ = 2 sin θ cos θ2m × m

tan 100

sin 200 = (= 2m cos 100)2m2

tan 100

51a. [2 marks]

The first two terms of an infinite geometric sequence, in order, are

, where .

Find .

Markschemeevidence of dividing terms (in any order) (M1)

eg

A1 N2

[2 marks]

2log2x, log2x x > 0

r

, μ2

μ1

2log2x

log2x

r = 12

51b. [2 marks]Show that the sum of the infinite sequence is .

Markschemecorrect substitution (A1)

eg

correct working A1

eg

AG N0

[2 marks]

4log2x

2log2x

1− 12

2log2x

12

S∞ = 4log2x

51c. [4 marks]

The first three terms of an arithmetic sequence, in order, are

, where .

Find, giving your answer as an integer.

Markschemeevidence of subtracting two terms (in any order) (M1)

eg

correct application of the properties of logs (A1)

eg


eg

A1 N3

[4 marks]

log2x, log2( ) , log2( )x

2x

4x > 0

d

u3 − u2, log2x − log2x

2

log2( ) , log2( × ) , (log2x − log22) − log2xx

2x

x

21x

log2 , − log2212

d = −1

51d. [2 marks]

Let be the sum of the first 12 terms of the arithmetic sequence.

Show that .

S12

S12 = 12log2x − 66

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Markschemecorrect substitution into the formula for the sum of an arithmetic sequence (A1)

eg

correct working A1

eg

AG N0

[2 marks]

(2log2x + (12 − 1)(−1))122

6(2log2x − 11), (2log2x − 11)122

12log2x − 66

51e. [3 marks]Given that is equal to half the sum of the infinite geometric sequence, find , giving your answer in the form , where .

Markschemecorrect equation (A1)

eg


eg

(accept) A1 N2

[3 marks]

S12 x 2p p ∈ Q

12log2x − 66 = 2log2x

10log2x = 66, log2x = 6.6, 266 = x10, log2( ) = 66x12

x2

x = 26.6

p = 6610

paper 1 most likely questions may 2018 [332 marks]...paper 1 most likely questions may 2018 [332...

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