mixed 1 [142 marks]

1. [Maximum mark: 5] 22M.1.SL.TZ2.2

The nth term of an arithmetic sequence is given by u n = 15 − 3n.

(a) State the value of the first term, u 1 . [1]

Markscheme

u 1 = 12 A1

[1 mark]

(b) Given that the nth term of this sequence is −33, find the value of
n. [2]

Markscheme

15 − 3n = −33 (A1)

n = 16 A1

[2 marks]

(c) Find the common difference, d. [2]

Markscheme

valid approach to find d (M1)

u 2 − u 1 = 9 − 12 OR recognize gradient is −3 OR attempts to solve

−33 = 12 + 15d

d = −3 A1
 [2 marks]

2. [Maximum mark: 5] 21M.1.SL.TZ1.3

Consider an arithmetic sequence where u 8 = S 8 = 8. Find the value
of the first term, u 1 , and the value of the common difference, d. [5]

Markscheme

METHOD 1 (finding u 1 first, from S 8 )

4(u 1 + 8) = 8 (A1)

u 1 = −6 A1

u 1 + 7d = 8 OR 4(2u 1 + 7d) = 8 (may be seen with their value of

u1 ) (A1)

attempt to substitute their u 1 (M1)

d = 2 A1

METHOD 2 (solving simultaneously)

u 1 + 7d = 8 (A1)

4(u 1 + 8) = 8 OR 4(2u 1 + 7d) = 8 OR u 1 = −3d (A1)

attempt to solve linear or simultaneous equations (M1)

u 1 = −6, d = 2 A1A1

[5 marks]
3. [Maximum mark: 5] 21M.1.SL.TZ2.4
In the expansion of (x + k) 7 , where k ∈ R, the coefficient of the term
in x 5 is 63.

Find the possible values of k. [5]

Markscheme

EITHER

7
attempt to use the binomial expansion of (x + k) (M1)

(or
7 7 0 7 6 1 7 5 2
C x k + C x k + C x k + …
0 1 2

+ …)
7 7 0 7 5 1 7 5 2
C k x + C k x + C k x
0 1 2

identifying the correct term 7C 2x 5 k 2 (or 7C 5k 2 x 5 ) (A1)

OR

attempt to use the general term 7C rx r k 7−r (or 7C rk r x 7−r ) (M1)

r = 2 (or r = 5) (A1)

THEN

7
C
2
= 21 (or 7C 5 = 21 (seen anywhere) (A1)

21x k
5 2
= 63x
5
(21k
2
= 63 , k
2
= 3) A1

k = ±√ 3 A1

Note: If working shown, award M1A1A1A1A0 for k = √ 3.

[5 marks]
4. [Maximum mark: 6] 19N.1.SL.TZ0.S_1
In an arithmetic sequence, u 2 = 5 and u 3 = 11.

(a) Find the common difference. [2]

Markscheme

valid approach (M1)

eg 11 − 5, 11 = 5 + d

d = 6 A1 N2

[2 marks]

(b) Find the first term. [2]

Markscheme

valid approach (M1)

eg u 2 − d, 5 − 6, u 1 + (3 − 1) (6) = 11

u 1 = −1 A1 N2

[2 marks]

(c) Find the sum of the first 20 terms. [2]

Markscheme

correct substitution into sum formula

eg 20

2
(2 (−1) + 19 (6)),
20

2
(−1 + 113) (A1)

S 20 = 1120 A1 N2
[2 marks]
5. [Maximum mark: 7] 23M.1.SL.TZ1.2
The function f is defined by f (x) for x ∈ R, x ≠ 2.
7x+7
=
2x−4

(a) Find the zero of f (x). [2]

Markscheme

recognizing f (x) = 0 (M1)

x = −1 A1

[2 marks]

(b) For the graph of y = f (x), write down the equation of

(b.i) the vertical asymptote; [1]

Markscheme

x = 2 (must be an equation with x) A1

[1 mark]

(b.ii) the horizontal asymptote. [1]

Markscheme

y =
7

2
(must be an equation with y) A1

[1 mark]
(c) Find f −1 (x), the inverse function of f (x). [3]

Markscheme

EITHER

interchanging x and y (M1)

2xy − 4x = 7y + 7

correct working with y terms on the same side: 2xy − 7y = 4x + 7

(A1)

OR

2yx − 4y = 7x + 7

correct working with x terms on the same side: 2yx − 7x = 4y + 7

(A1)

4y+7
interchanging x and y OR making x the subject x =
2y−7
(M1)

THEN

(or equivalent) A1
−1 4x+7 7
f (x) = (x ≠ )
2x−7 2

[3 marks]
6. [Maximum mark: 5] 23M.1.SL.TZ1.1
Point P has coordinates (−3, 2), and point Q has coordinates (15, − 8).

Point M is the midpoint of [PQ] .

(a) Find the coordinates of M. [2]

Markscheme

M (6, − 3) A1A1

[2 marks]

Line L is perpendicular to [PQ] and passes through M.

(b) Find the gradient of L. [2]

Markscheme

gradient of [PQ] = −
5

9
(A1)

gradient of L =
9

5
A1

[2 marks]

(c) Hence, write down the equation of L. [1]

Markscheme

9 9 69
y + 3 = (x − 6) OR y = x − (or equivalent) A1
5 5 5
Note: Do not accept L =
9

5
x −
69

5
.

[1 mark]
7. [Maximum mark: 5] 23M.1.SL.TZ2.3
A function f is defined by f (x) = 1 −
1

x−2
, where x ∈ R, x ≠ 2.

(a) The graph of y = f (x) has a vertical asymptote and a

horizontal asymptote.

Write down the equation of

(a.i) the vertical asymptote; [1]

Markscheme

x = 2 A1

[1 mark]

(a.ii) the horizontal asymptote. [1]

Markscheme

y = 1 A1

[1 mark]

(b) Find the coordinates of the point where the graph of y = f (x)

intersects

(b.i) the y-axis; [1]

Markscheme

(0,
3

2
) A1
 [1 mark]

(b.ii) the x-axis. [1]

Markscheme

(3, 0) A1

[1 mark]

(c) On the following set of axes, sketch the graph of y = f (x),

showing all the features found in parts (a) and (b).

[1]

Markscheme
two correct branches with correct asymptotic behaviour and intercepts clearly
shown A1

[1 mark]
8. [Maximum mark: 16] 23M.1.SL.TZ2.7
The following diagram shows part of the graph of a quadratic function f .

The vertex of the parabola is (−2, − 5) and the y-intercept is at point P.

(a) Write down the equation of the axis of symmetry. [1]

Markscheme

x = −2 (must be an equation) A1

[1 mark]

1 2
The function can be written in the form f (x) =
4
(x − h) + k, where h,

k ∈ Z.

(b) Write down the values of h and k. [2]

Markscheme

h = −2, k = −5 A1A1
 [2 marks]

(c) Find the y-coordinate of P. [2]

Markscheme

substituting x = 0 into f (x) (M1)

1 2
y = (0 + 2) − 5
4

y = −4 (accept P(0, − 4)) A1

[2 marks]

In the following diagram, the line L is normal to the graph of f at point P.

(d) Find the equation of the line L, in the form y = ax + b. [3]

Markscheme
 f ′(x) =
1

2
(x + 2) (=
1

2
x + 1) (A1)

substituting x = 0 into their derivative (M1)

f ′(0) = 1

gradient of normal is −1 (may be seen in their equation) A1

y = −x − 4 (accept a = −1, b = −4) A1

Note: Award A0 for L = −x − 4 (without the y =).

[4 marks]

The line L intersects the graph of f at a second point, Q, as shown above.

(e) Calculate the distance between P and Q. [8]

Markscheme

equating theirf (x) to their L (M1)

1 2
(x + 2) − 5 = −x − 4
4

1
+ 2x = 0 (or equivalent) (A1)
2
x
4

valid attempt to solve their quadratic (M1)

4
x(x + 8) = 0 OR x(x + 8) = 0

x = −8 A1
Note: Accept both solutions x = −8 and x = 0 here, x = −8 may be
seen in working to find coordinates of Q or distance.

substituting their value of x (not x = 0) into their f (x) or their L (M1)

2
OR
1
y = −(−8) − 4 y = (−8 + 2) − 5
4

Q(−8, 4) A1

correct substitution into distance formula (A1)

2 2
√ (−8 − 0) + (4 − (−4))

distance = √ 128 (= 8√ 2) A1

[8 marks]
9. [Maximum mark: 7] 22N.1.SL.TZ0.1
Let f (x) = −2x + 3, for x ∈ R.

(a) The graph of a linear function g is parallel to the graph of f and

passes through the origin. Find an expression for g(x). [2]

Markscheme

gradient of g is −2 (may be seen in function, do not accept −2x + 3)

(A1)

g(x) = −2x A1

[2 marks]

(b) The graph of a linear function h is perpendicular to the graph of f

and passes through the point (−1, 2) . Find an expression for
h(x). [3]

Markscheme

gradient is 1

2
(may be seen in function) (A1)

attempt to substitute their gradient and (−1, 2) into any form of equation
for straight line (M1)

y − 2 =
1

2
(x + 1) OR 2 =
1

2
⋅ (−1) + c

h(x) =
1

2
(x + 1) + 2 (=
1

2
x +
5

2
) A1

[3 marks]

(c) Find (g ∘ h)(0). [2]

Markscheme

(g ∘ h)(x) = −2(
1

2
x +
5

2
) OR h(0) =
5

2
OR g(
5

2
) (A1)

(g ∘ h)(0) = −5 A1

[2 marks]
10. [Maximum mark: 16] 22N.1.SL.TZ0.7
(a) The graph of a quadratic function f has its vertex at the point
(3, 2) and it intersects the x-axis at x = 5. Find f in the form
2
f (x) = a(x − h) + k.
[3]

Markscheme

correct substitution of h = 3 and k = 2 into f (x) (A1)

2
f (x) = a(x − 3) + 2

correct substitution of (5, 0) (A1)

2 1
0 = a(5 − 3) + 2 (a = − )
2

Note: The first two A marks are independent.

2
A1
1
f (x) = − (x − 3) + 2
2

[3 marks]

The quadratic function g is defined by g(x) = px

2
+ (t − 1)x − p where

x ∈ R and p, t ∈ R, p ≠ 0.

In the case where g(−3) = g(1) = 4,

(b.i) find the value of p and the value of t. [4]

Markscheme

METHOD 1
 correct substitution of (1, 4) (A1)

p + (t − 1) − p = 4

t = 5 A1

substituting their value of t into 9p − 3(t − 1) − p = 4 (M1)

8p − 12 = 4

p = 2 A1

METHOD 2

correct substitution of ONE of the coordinates (−3, 4) or (1, 4) (A1)

9p − 3(t − 1) − p = 4 OR p + (t − 1) − p = 4

valid attempt to solve their two equations (M1)

p = 2, t = 5 A1A1

2
(g(x) = 2x + 4x − 2)

[4 marks]

(b.ii) find the range of g. [3]

Markscheme

attempt to find the x-coordinate of the vertex (M1)

2
OR OR OR
−3+1 −4
x = (= −1) 4x + 4 = 0 2(x + 1) − 4
2 2×2

y-coordinate of the vertex = −4 (A1)

correct range A1
 [−4, + ∞[ OR y ≥ −4 OR g ≥ −4 OR [−4, ∞)

[3 marks]

(c) The linear function j is defined by j(x) = −x + 3p where

x ∈ R and p ∈ R, p ≠ 0.

Show that the graphs of j(x) = −x + 3p and

+ (t − 1)x − p have two distinct points of

2
g(x) = px

intersection for every possible value of p and t. [6]

Markscheme

equating the two functions or equations (M1)

g(x) = j(x) OR px
2
+ (t − 1)x − p = −x + 3p

px
2
+ tx − 4p = 0 (A1)

attempt to find discriminant (do not accept only in quadratic formula)

(M1)

Δ = t
2
+ 16p
2
A1

> 0, because t ≥ 0 and p > 0, therefore the sum

2 2 2 2
Δ = t + 16p

will be positive R1R1

Note: Award R1 for recognising that Δ is positive and R1 for the reason.

There are two distinct points of intersection between the graphs of g and j.
AG

[6 marks]
11. [Maximum mark: 6] 23M.1.SL.TZ1.6
The function f is defined by f (x) = sin qx, where q > 0. The following
diagram shows part of the graph of f for 0 ≤ x ≤ 4m, where x is in radians.
There are x-intercepts at x = 0, 2m and 4m.

(a) Find an expression for m in terms of q. [2]

Markscheme

recognition that period is 4m OR substitution of a point on f (except the

origin) (M1)

4m =
2π

q
OR 1 = sin qm

m =
2q
π
A1

[2 marks]

2qx
The function g is defined by g(x) = 3 sin
3
, for 0 ≤ x ≤ 6m.

(b) On the axes above, sketch the graph of g. [4]

Markscheme
horizontal scale factor is 3

2
(seen anywhere) (A1)

Note: This (A1) may be earned by seeing a period of 6m, half period of 3m or
the correct x-coordinate of the maximum/minimum point.

A1A1A1

Note:
Curve must be an approximate sinusoidal shape (sine or cosine).
Only in this case, award the following:
A1 for correct amplitude.
A1 for correct domain.
A1 for correct max and min points and correct x-intercepts.

[4 marks]
12. [Maximum mark: 7] 22M.1.SL.TZ1.4

Consider the functions f (x) = √ 3 sin x + cos x where 0 ≤ x ≤ π and

g(x) = 2x where x ∈ R.

(a) Find (f ∘ g)(x). [2]

Markscheme

(f ∘ g)(x) = f (2x) (A1)

f (2x) = √ 3 sin 2x + cos 2x A1

[2 marks]

(b) Solve the equation (f ∘ g)(x) = 2 cos 2x where

0 ≤ x ≤ π. [5]

Markscheme

√ 3 sin 2x + cos 2x = 2 cos 2x

√ 3 sin 2x = cos 2x

recognising to use tan or cot M1

OR cot 2x = √ 3 (values may be seen in right triangle)

1
tan 2x =
√3

(A1)

1 π
(arctan( ) =)
6
(seen anywhere) (accept degrees) (A1)
√3

π 7π
2x = ,
6 6

x =
π

12
,
7π

12
A1A1
Note: Do not award the final A1 if any additional solutions are seen.
Award A1A0 for correct answers in degrees.
Award A0A0 for correct answers in degrees with additional values.

[5 marks]
13. [Maximum mark: 7] 21N.1.SL.TZ0.1
Consider the function f (x) = −2(x − 1)(x + 3), for x ∈ R. The following

diagram shows part of the graph of f .

For the graph of f

(a.i) find the x-coordinates of the x-intercepts. [2]

Markscheme

setting f (x) = 0 (M1)

x = 1, x = −3 (accept (1, 0), (−3, 0)) A1

[2 marks]

(a.ii) find the coordinates of the vertex. [3]

Markscheme

METHOD 1

x = −1 A1
 substituting their x-coordinate into f (M1)

y = 8 A1

(−1, 8)

METHOD 2

attempt to complete the square (M1)

2
−2((x + 1) − 4) (M1)

x = −1, y = 8 A1A1

(−1, 8)

[3 marks]

(b) The function f can be written in the form

2
f (x) = −2(x − h) + k.

Write down the value of h and the value of k. [2]

Markscheme

h = −1 A1

k = 8 A1

[2 marks]
14. [Maximum mark: 7] 21N.1.SL.TZ0.5
The function f is defined for all x ∈ R. The line with equation y = 6x − 1 is

the tangent to the graph of f at x = 4.

(a) Write down the value of f ′(4). [1]

Markscheme

f ′(4) = 6 A1

[1 mark]

(b) Find f (4). [1]

Markscheme

f (4) = 6 × 4 − 1 = 23 A1

[1 mark]

The function g is defined for all x ∈ R where g(x) = x

2
− 3x and

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

(c) Find h(4). [2]

Markscheme

h(4) = f (g(4)) (M1)

2
h(4) = f (4 − 3 × 4) = f (4)

h(4) = 23 A1
 [2 marks]

(d) Hence find the equation of the tangent to the graph of h at

x = 4. [3]

Markscheme

attempt to use chain rule to find h′ (M1)

f ′(g(x)) × g′(x) OR (x
2
− 3x)′×f ′(x
2
− 3x)

h′(4) = (2 × 4 − 3)f ′(4

2
− 3 × 4) A1

= 30

y − 23 = 30(x − 4) OR y = 30x − 97 A1

[3 marks]
15. [Maximum mark: 5] 21M.1.SL.TZ1.1
The graph of y = f (x) for −4 ≤ x ≤ 6 is shown in the following diagram.

(a.i) Write down the value of f (2). [1]

Markscheme

f (2) = 6 A1

[1 mark]

(a.ii) Write down the value of (f ∘ f )(2). [1]

Markscheme

(f ∘ f )(2) = −2 A1
 [1 mark]

(b) Let g(x) =

1
f (x) + 1 for −4 ≤ x ≤ 6. On the axes above,
2

sketch the graph of g.

[3]

Markscheme

M1A1A1

Note: Award M1 for an attempt to apply any vertical stretch or vertical

translation, A1 for a correct horizontal line segment between −4 and 0
(located roughly at y = 3),
A1 for a correct concave down parabola including max point at (2, 4) and
for correct end points at (0, 3) and (6, 0) (within circles). Points do not need
to be labelled.

[3 marks]
16. [Maximum mark: 14] 21M.1.SL.TZ1.7
Let f (x) = mx
2
− 2mx, where x ∈ R and m ∈ R. The line y = mx − 9

meets the graph of f at exactly one point.

(a) Show that m = 4. [6]

Markscheme

METHOD 1 (discriminant)

mx
2
− 2mx = mx − 9 (M1)

2
mx − 3mx + 9 = 0

recognizing Δ = 0 (seen anywhere) M1

2
Δ = (−3m) − 4(m)(9) (do not accept only in quadratic formula for
x) A1

valid approach to solve quadratic for m (M1)

√ 2
36± 36 −4×9×0
9m(m − 4) = 0 OR m =
2×9

both solutions m = 0, 4 A1

m ≠ 0 with a valid reason R1

the two graphs would not intersect OR 0 ≠ −9

m = 4 AG

METHOD 2 (equating slopes)

mx
2
− 2mx = mx − 9 (seen anywhere) (M1)

f ′(x) = 2mx − 2m A1

equating slopes, f ′(x) = m (seen anywhere) M1

2mx − 2m = m

x =
3

2
A1

substituting their x value (M1)

2
3 3 3
( ) m − 2m × = m × − 9
2 2 2

4
m −
12

4
m =
6

4
m − 9 A1

−9m
= −9
4

m = 4 AG

METHOD 3 (using −b

2a
)

mx
2
− 2mx = mx − 9 (M1)

2
mx − 3mx + 9 = 0

attempt to find x-coord of vertex using

−b

2a
(M1)

−(−3m)

2m
A1

x =
3

2
A1

substituting their x value (M1)

2
3 3
( ) m − 3m × + 9 = 0
2 2

9 9

4
m −
2
m + 9 = 0 A1

−9m = −36

m = 4 AG

[6 marks]
The function f can be expressed in the form f (x) = 4(x − p)(x − q), where

p, q ∈ R.

(b) Find the value of p and the value of q. [2]

Markscheme

4x(x − 2) (A1)

p = 0 and q = 2 OR p = 2 and q = 0 A1

[2 marks]

2
The function f can also be expressed in the form f (x) = 4(x − h) + k,

where h, k ∈ R.

(c) Find the value of h and the value of k. [3]

Markscheme

attempt to use valid approach (M1)

0+2 −(−8)

2
,
2×4
, f (1), 8x − 8 = 0 OR
2 2
4(x − 2x + 1 − 1)(= 4(x − 1) − 4)

h = 1, k = −4 A1A1

[3 marks]

(d) Hence find the values of x where the graph of f is both negative
and increasing.
 [3]

Markscheme

EITHER

recognition x = h to 2 (may be seen on sketch) (M1)

OR

recognition that f (x) < 0 and f ′(x) > 0 (M1)

THEN

1 < x < 2 A1A1

Note: Award A1 for two correct values, A1 for correct inequality signs.

[3 marks]
17. [Maximum mark: 6] 20N.1.SL.TZ0.S_5
Let f (x) = −x
2
+ 4x + 5 and g(x) = −f (x) + k.

Find the values of k so that g(x) = 0 has no real roots. [6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.

METHOD 1 – (discriminant)

correct expression for g (A1)

eg −(−x
2
+ 4x + 5) + k , x
2
− 4x − 5 + k = 0

evidence of discriminant (M1)

eg b
2
− 4ac, Δ

correct substitution into discriminant of g (A1)

2
eg (−4) − 4(1)(−5 + k) , 16 − 4(k − 5)

recognizing discriminant is negative (M1)

eg
2
Δ < 0 , (−4) − 4(1)(−5 + k) < 0 , 16 < 4(k − 5) , 16 − 4(−1)(5) < 0

correct working (must be correct inequality) (A1)

eg −4k < −36 , k − 5 > 4 , 16 + 20 − 4k < 0

k > 9 A1 N3

METHOD 2 – (transformation of vertex of f )

valid approach for finding f (x) vertex (M1)

eg
b
− = 2 , f ′(x) = 0
2a

correct vertex of f (x) (A1)

eg (2, 9)

correct vertex of −f (x) (A1)

eg (2, − 9)

correct vertex of g(x) (A1)

2 0
eg ( ) + ( ) , (2, − 9 + k)
−9 k

recognizing when vertex is above x-axis (M1)

eg −9 + k > 0, sketch

k > 9 A1 N3

METHOD 3 – (transformation of f )

recognizing vertical reflection of f (x) (M1)

eg −f (x) , x
2
− 4x − 5 , sketch

correct expression for g(x) (A1)

eg x
2
− 4x − 5 + k

valid approach for finding vertex of g(x) (M1)

eg
b
− = 2 , g′(x) = 0
2a

correct y coordinate of vertex of g(x) (A1)

eg y = −9 + k , (2, − 9 + k)

recognizing when vertex is above x-axis (M1)

eg −9 + k > 0 , sketch

k > 9 A1 N3

[6 marks]
18. [Maximum mark: 7] 19N.1.SL.TZ0.S_3
Let g (x) = x
2
+ bx + 11. The point (−1, 8) lies on the graph of g.

(a) Find the value of b. [3]

Markscheme

valid attempt to substitute coordinates (M1)

eg g (−1) = 8

correct substitution (A1)

2
eg (−1) + b (−1) + 11 = 8, 1 − b + 11 = 8

b = 4 A1 N2

[3 marks]

(b) The graph of f (x) = x

2
is transformed to obtain the graph of g
.

Describe this transformation. [4]

Markscheme

valid attempt to solve (M1)

eg + 4x + 4) + 7, h = ,
2 −4
(x k = g (−2)
2

correct working A1

2
eg (x + 2) + 7, h = −2, k = 7

−2
translation or shift (do not accept move) of vector ( ) (accept left by 2
7

and up by 7) A1A1 N2
[4 marks]
19. [Maximum mark: 6] 17N.1.SL.TZ0.S_8
Let f (x) = x
2
− x, for x ∈ R. The following diagram shows part of the graph
of f .

The graph of f crosses the x-axis at the origin and at the point P(1, 0).

The line L intersects the graph of f at another point Q, as shown in the following
diagram.

(d) Find the area of the region enclosed by the graph of f and the
line L. [6]

Markscheme

valid approach (M1)

 1
eg ∫ L − f, ∫
−1
(1 − x )dx, splitting
2
area into triangles and integrals

correct integration (A1)(A1)

1
3 3 2 2
x x x x
eg [x −
3
] , −
3
−
2
+
2
+ x
−1

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

eg 1 −
1

3
− (−1 −
−1

3
)

Note: Award M0 for substituting into original or differentiated function.

area = 4

3
A2 N3

[6 marks]

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