C3 MS Jun 2009

Page 1

June 2009 6665 Core Mathematics C3 Mark Scheme Question Number Q1

(a)

Scheme Iterative formula: xn +1 =

Marks

2 + 2 , x0 = 2.5 ( xn ) 2

An attempt to substitute x0 = 2.5 into the iterative formula. M1 Can be implied by x1 = 2.32 or 2.320 Both x1 = 2.32(0) A1 and x2 = awrt 2.372 Both x3 = awrt 2.356 A1 cso and x4 = awrt 2.360 or 2.36

2 x1 = +2 (2.5) 2

x1 = 2.32 x2 = 2.371581451... x3 = 2.355593575... x4 = 2.360436923...

(3) (b)

Let f ( x) = − x 3 + 2 x 2 + 2 = 0 f (2.3585) = 0.00583577... f (2.3595) = − 0.00142286... Sign change (and f ( x) is continuous) therefore a root

α is such that α ∈ ( 2.3585, 2.3595 ) ⇒ α = 2.359 (3 dp)

Choose suitable interval for x, M1 e.g. [2.3585, 2.3595] or tighter any one value awrt 1 sf dM1 or truncated 1 sf both values correct, sign change A1 and conclusion At a minimum, both values must be correct to 1sf or truncated 1sf, candidate states “change of sign, hence root”.

(3)

[6]

8371_9374 GCE Mathematics January 2009

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