MARK SCHEME – AS MATHEMATICS – 7356/1 – JUNE 2022

6

Q Marking instructions AO Marks Typical solution

1 Circles correct answer. 1.1b B1 log10(

2

????

)

Question 1 Total 1

Q Marking instructions AO Marks Typical solution

2 Ticks correct box. 1.1b B1

Question 2 Total 1

Q Marking instructions AO Marks Typical solution

3 Expands at least one term

correctly.

PI

Other terms: (3x)

4 + 4C2(3x)

2(

1

2

)

2

+ 4C3(3x)(

1

2

)

3 + (

1

2

)

4

Condone use of coefficients

only.

1.1a M1 4C1 (3x)

3 (

1

2

)

Selects their x

3 term

PI

1.1a M1 Coefficient = 54

States correct coefficient. 1.1b A1

Total 3

MARK SCHEME – AS MATHEMATICS – 7356/1 – JUNE 2022

7

Q Marking instructions AO Marks Typical solution

4 Uses identity cos2θ = 1– sin2θ 1.2 B1 1 – sin2θ = 10sinθ +4

0 = sin2θ + 10sinθ +3

sinθ = –0.3095 or –9.6904

sinθ = –9.6904 (not valid)

sin–1

(–0.3095) = –18.03°

θ = 198° or θ = 342°

Solves their quadratic in sinθ

PI by at least one correct value

of θ or - 18º

1.1a M1

Explains that the second

solution or both solutions is/are

inappropriate.

Accept N/A, out of range, no

solutions, math error, reject OE

Do not accept a ‘×’ or –9.69

(OE) crossed out alone.

2.4 E1F

Obtains one correct value for θ

AWRT 198° or 342°

1.1b A1

Obtains two correct values for θ

Condone - 18º

included, but no

other answers.

1.1b A1

Question 4 Total 5

Q Marking instructions AO Marks T