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Pure Mathematics 1 Practice Paper J8 MARK SCHEME Q1.

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Q2.

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Q3.

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Q4.

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Q5.

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Q6.

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Q7.

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Q8.

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Q9.

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Q10.

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Q11.

Q Scheme Marks

States or implies the formula for differentiation from first principles.

3

0

f ( ) 5

f ( ) f ( )f ( ) lim

h

x x

x h xx

h

Correctly applies the formula to the specific formula and expands and

simplifies the formula.

3 3

0

3 2 2 3 3

0

2 2 3

0

5 5f ( ) lim

5 3 3 5f ( ) lim

15 15 5f ( ) lim

h

h

h

x h xx

h

x x h xh h xx

h

x h xh hx

h

M1

Factorises the ‘h’ out of the numerator and then divides

by h to simplify.

2 2

0

2 2

0

15 15 5f ( ) lim

f ( ) lim 15 15 5

h

h

h x xh hx

h

x x xh h

A1

States that as h → 0, 9x2 + 9xh + 3h2 → 9x2 o.e.

so derivative = 9x2 * A1*

3

3 3

3 3

9 9 3

9 9 3

3 9 9

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Q12.

Q Scheme Marks

(a)

Gradient = 4.55−2.3

12−.0= 0.1875

Creating log P = log a+ t log b Sub 2.3 and 0.1875 in above equation Equation of line: log P = 0.1875 t + 2.3 (accept 0.19 as gradient)

M1 A1

(b)

log 𝑎 = 2.3 𝑎 = 102.3 = 199.526… = 200 (3 S.F)

log 𝑏 = 0.1875

𝑏 = 100.1875 = 1.5399… = 1.54 (3 S.F)

M1 A1 M1 A1

(c)

The initial population of bacteria

A1

(d)

Sub 𝑡 = 20, 𝑎 = 200 , 𝑏 = 1.54 in 𝑃 = 𝑎𝑏𝑡

= 200(1.54)20 = 1125756.368 = 1126000 (nearest 1000)

M1 A1