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14 questions
SA-9709-P3-NM2-G01 2 marks L1 Foundation Calculator Iteration
The iteration x n + 1 = x n + 3 x_{n+1}=\sqrt{x_n+3} x n + 1 = x n + 3 has x 1 = 2 x_1=2 x 1 = 2 . Find x 2 x_2 x 2 and x 3 x_3 x 3 , each correct to 4 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-Q01 1 mark L1 Foundation Calculator Accuracy
An iteration gives x 5 = 2.0942 x_5=2.0942 x 5 = 2.0942 and x 6 = 2.0945 x_6=2.0945 x 6 = 2.0945 . The root correct to 2 decimal places is
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-Q02 1 mark L1 Foundation Calculator Iteration
The iteration x n + 1 = x n + 6 3 x_{n+1}=\sqrt[3]{x_n+6} x n + 1 = 3 x n + 6 starts at x 1 = 2 x_1=2 x 1 = 2 . Find x 2 x_2 x 2 .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-Q03 1 mark L1 Foundation Calculator Accuracy
To give a root correct to 2 decimal places from an iteration, you should
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-002 3 marks L2 Standard Calculator Iteration
Use the iteration x n + 1 = 2 x n + 5 3 x_{n+1}=\sqrt[3]{2x_n+5} x n + 1 = 3 2 x n + 5 with x 1 = 2 x_1=2 x 1 = 2 to find the root of x 3 − 2 x − 5 = 0 x^3-2x-5=0 x 3 − 2 x − 5 = 0 correct to 2 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-005 3 marks L2 Standard Calculator Iteration
Use the iteration x n + 1 = ln ( x n + 4 ) x_{n+1}=\ln(x_n+4) x n + 1 = ln ( x n + 4 ) with x 1 = 2 x_1=2 x 1 = 2 to find the root of x = ln ( x + 4 ) x=\ln(x+4) x = ln ( x + 4 ) correct to 3 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-001 3 marks L2 Standard Calculator Iteration
Use the iteration x n + 1 = 1 + 1 x n x_{n+1}=1+\dfrac1{x_n} x n + 1 = 1 + x n 1 with x 1 = 1 x_1=1 x 1 = 1 to find the positive root of x 2 − x − 1 = 0 x^2-x-1=0 x 2 − x − 1 = 0 correct to 2 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-003 3 marks L2 Standard Calculator Iteration
Use the iteration x n + 1 = x n + 3 x_{n+1}=\sqrt{x_n+3} x n + 1 = x n + 3 with x 1 = 2 x_1=2 x 1 = 2 to find the positive root of x 2 − x − 3 = 0 x^2-x-3=0 x 2 − x − 3 = 0 correct to 3 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-004 2 marks L2 Standard Calculator Accuracy
An iteration suggests that the root of x 3 − x − 4 = 0 x^3-x-4=0 x 3 − x − 4 = 0 is 1.796 1.796 1.796 to 3 decimal places. Let f ( x ) = x 3 − x − 4 f(x)=x^3-x-4 f ( x ) = x 3 − x − 4 . Find f ( 1.7955 ) f(1.7955) f ( 1.7955 ) and f ( 1.7965 ) f(1.7965) f ( 1.7965 ) , each to 2 significant figures.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-007 3 marks L3 Exam Calculator Iteration
Use the iteration x n + 1 = 1 2 ( x n + 2 cos x n ) x_{n+1}=\tfrac12\left(x_n+2\cos x_n\right) x n + 1 = 2 1 ( x n + 2 cos x n ) with x 1 = 1 x_1=1 x 1 = 1 to find the root of x = 2 cos x x=2\cos x x = 2 cos x correct to 3 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-010 3 marks L3 Exam Calculator Iteration
Use the iteration x n + 1 = 12 − x n 3 x_{n+1}=\sqrt[3]{12-x_n} x n + 1 = 3 12 − x n with x 1 = 2 x_1=2 x 1 = 2 to find the root of x 3 + x − 12 = 0 x^3+x-12=0 x 3 + x − 12 = 0 correct to 3 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-002 3 marks L3 Exam Calculator Iteration
Use the iteration x n + 1 = e − x n x_{n+1}=e^{-x_n} x n + 1 = e − x n with x 1 = 0.5 x_1=0.5 x 1 = 0.5 to find the root of x = e − x x=e^{-x} x = e − x correct to 2 decimal places. Give each iterate to 4 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM-009 5 marks L4 Challenging Calculator Iteration
It is given that ∫ 1 a ( 1 x + 1 ) d x = 3 \displaystyle\int_1^a\left(\frac1x+1\right)dx=3 ∫ 1 a ( x 1 + 1 ) d x = 3 , where a > 1 a>1 a > 1 . Show that a = 4 − ln a a=4-\ln a a = 4 − ln a , and use an iteration based on this, with a 1 = 3 a_1=3 a 1 = 3 , to find a a a correct to 2 decimal places.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM2-005 5 marks L4 Challenging Calculator Iteration
A chord of a circle with radius r r r subtends an angle θ \theta θ radians at the centre. The segment it cuts off has area equal to one quarter of the area of the circle. Show that θ = sin θ + π 2 \theta=\sin\theta+\tfrac\pi2 θ = sin θ + 2 π , and use the iteration θ n + 1 = sin θ n + π 2 \theta_{n+1}=\sin\theta_n+\tfrac\pi2 θ n + 1 = sin θ n + 2 π with θ 1 = 2.3 \theta_1=2.3 θ 1 = 2.3 to find θ \theta θ correct to 2 decimal places.
Salam Academy Original Practice · Proposed mark scheme