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12 questions
SA-9709-P3-NM-003 1 mark L1 Foundation Calculator Rearranging
Which of these is a rearrangement of x 3 − 2 x − 5 = 0 x^3-2x-5=0 x 3 − 2 x − 5 = 0 x = F ( x ) x=F(x) x = F ( x )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-004 1 mark L1 Foundation Calculator Convergence
Near a root α \alpha α − 1 < F ′ ( α ) < 0 -1<F'(\alpha)<0 − 1 < F ′ ( α ) < 0 x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-G01 2 marks L1 Foundation Calculator Rearranging
The equation x 3 + 4 x − 1 = 0 x^3+4x-1=0 x 3 + 4 x − 1 = 0 x = 1 − x 3 4 x=\dfrac{1-x^3}{4} x = 4 1 − x 3 x n + 1 = 1 − x n 3 4 x_{n+1}=\dfrac{1-x_n^3}{4} x n + 1 = 4 1 − x n 3 x 1 = 0 x_1=0 x 1 = 0 x 2 x_2 x 2 x 3 x_3 x 3 x 3 x_3 x 3
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-Q01 1 mark L1 Foundation Calculator Convergence
An iteration x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n ) α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-Q02 1 mark L1 Foundation Calculator Rearranging
If x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n ) α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-004 2 marks L2 Standard Calculator Rearranging
The sequence defined by x n + 1 = 1 2 ( x n + 7 x n ) x_{n+1}=\tfrac12\left(x_n+\dfrac7{x_n}\right) x n + 1 = 2 1 ( x n + x n 7 ) x 1 = 2 x_1=2 x 1 = 2 α \alpha α α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-008 2 marks L2 Standard Calculator Rearranging
The iteration x n + 1 = x n 3 + 3 5 x_{n+1}=\dfrac{x_n^3+3}{5} x n + 1 = 5 x n 3 + 3 α \alpha α α \alpha α x 3 + p x + q = 0 x^3+px+q=0 x 3 + p x + q = 0 p p p q q q
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-001 2 marks L2 Standard Calculator Rearranging
The sequence x n + 1 = 2 x n 3 + 4 3 x n 2 x_{n+1}=\dfrac{2x_n}{3}+\dfrac{4}{3x_n^2} x n + 1 = 3 2 x n + 3 x n 2 4 x 1 = 2 x_1=2 x 1 = 2 α \alpha α α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-002 1 mark L2 Standard Calculator Convergence
The equation x 3 − 2 x − 5 = 0 x^3-2x-5=0 x 3 − 2 x − 5 = 0 α = 2.09 \alpha=2.09 α = 2.09 F ( x ) = x 3 − 5 2 F(x)=\dfrac{x^3-5}{2} F ( x ) = 2 x 3 − 5 F ′ ( α ) ≈ 6.6 F'(\alpha)\approx6.6 F ′ ( α ) ≈ 6.6 x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n ) x 1 x_1 x 1 α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-005 3 marks L2 Standard Calculator Rearranging
The equation e x = 3 x e^x=3x e x = 3 x x n + 1 = ln ( 3 x n ) x_{n+1}=\ln(3x_n) x n + 1 = ln ( 3 x n ) x 1 = 1.5 x_1=1.5 x 1 = 1.5
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-003 2 marks L3 Exam Calculator Convergence
Let F ( x ) = 2 x + 5 3 F(x)=\sqrt[3]{2x+5} F ( x ) = 3 2 x + 5 x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n ) α = 2.0946 \alpha=2.0946 α = 2.0946 F ′ ( α ) F'(\alpha) F ′ ( α )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM3-006 4 marks L4 Challenging Calculator Convergence
The equation e x = 3 x e^x=3x e x = 3 x α < β \alpha<\beta α < β β = 1.512 \beta=1.512 β = 1.512 F ( x ) = 1 3 e x F(x)=\tfrac13e^x F ( x ) = 3 1 e x F ′ ( β ) F'(\beta) F ′ ( β ) x n + 1 = F ( x n ) x_{n+1}=F(x_n) x n + 1 = F ( x n ) β \beta β x 1 = 1 x_1=1 x 1 = 1 α \alpha α α \alpha α
Salam Academy Original Practice · Proposed mark scheme