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Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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13 questions

SA-9709-P3-NM4-G013 marksL1 FoundationCalculatorNewton–Raphson
Let f(x)=x2−2f(x)=x^2-2. Use the Newton–Raphson method with x1=1x_1=1 to find x2x_2 and x3x_3, giving x3x_3 to 4 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-Q011 markL1 FoundationCalculatorNewton–Raphson
The Newton–Raphson formula is
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-Q021 markL1 FoundationCalculatorNewton–Raphson
On the graph of y=f(x)y=f(x), the Newton–Raphson estimate x2x_2 is
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0013 marksL2 StandardCalculatorNewton–Raphson
Use the Newton–Raphson method with x1=2x_1=2 to find x3x_3 for the equation x3−2x−5=0x^3-2x-5=0. Give your answer to 4 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0023 marksL2 StandardCalculatorNewton–Raphson
Use one step of the Newton–Raphson method with x1=1x_1=1 to find an approximation x2x_2 to the root of x=cos⁡xx=\cos x (xx in radians). Give x2x_2 to 4 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0031 markL2 StandardCalculatorFailures
Why does the Newton–Raphson method fail for f(x)=x3−3x+1f(x)=x^3-3x+1 with x1=1x_1=1?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0042 marksL2 StandardCalculatorBisection and interpolation
f(x)=x3−2x−5f(x)=x^3-2x-5 has f(2)=−1f(2)=-1 and f(3)=16f(3)=16. Use linear interpolation once on the interval [2,3][2,3] to find an approximation to the root. Give your answer to 3 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0053 marksL2 StandardCalculatorBisection and interpolation
Starting from the interval [2,3][2,3], apply interval bisection three times to f(x)=x3−2x−5f(x)=x^3-2x-5. The final interval is [2,c][2,c]. Find cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0083 marksL2 StandardCalculatorNewton–Raphson
Use one step of the Newton–Raphson method with x1=1.5x_1=1.5 on f(x)=ex−3xf(x)=e^x-3x to find x2x_2, giving your answer to 4 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0092 marksL2 StandardCalculatorBisection and interpolation
f(x)=x3+x−3f(x)=x^3+x-3 has f(1)=−1f(1)=-1 and f(1.5)=1.875f(1.5)=1.875. Use linear interpolation once on [1,1.5][1,1.5] to estimate the root, giving your answer to 3 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0071 markL3 ExamCalculatorFailures
For f(x)=x3−2x−5f(x)=x^3-2x-5, the Newton–Raphson method with x1=0.85x_1=0.85 gives x2=37.2x_2=37.2, although the only root is 2.092.09. Why?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0103 marksL3 ExamCalculatorNewton–Raphson
Applying the Newton–Raphson method to f(x)=1x−7f(x)=\dfrac1x-7 gives xn+1=xn(2−7xn)x_{n+1}=x_n(2-7x_n), which finds 17\tfrac17 without dividing. With x1=0.1x_1=0.1, find x3x_3 to 4 decimal places.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-NM4-0064 marksL4 ChallengingCalculatorNewton–Raphson
Show that the Newton–Raphson method for f(x)=x2−7f(x)=x^2-7 gives xn+1=12(xn+7xn)x_{n+1}=\tfrac12\left(x_n+\dfrac7{x_n}\right). Starting with x1=3x_1=3, find x3x_3 to 6 decimal places and x4x_4 to 8 decimal places.

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Salam Academy Original Practice · Proposed mark scheme