Solving equations numerically: OCR MEI A‑Level Further Maths questions

OCR MEI A-Level Further Maths past-paper questions on 6.1 Solving equations numerically, 2018–2025, with mark schemes and examiner’s reports where available. Add MadAsMaths questions to your practice, or browse them separately by topic and difficulty.

All 33 questions on Solving equations numerically, as a list

June 2025, Numerical Methods, Q4 (7 marks): The Newton-Raphson method is to be used to find the positive root of the equation 2 tanh x x 4 0 . - + = 2 The diagram shows part of the graph of y tanh x x + .
June 2025, Numerical Methods, Q5 (10 marks): The diagram shows part of the graph of y sinh x 3 x + . 1 = - y 10 5 x 0 –1 1 2 3 4 –5 The largest positive root, a, of the equation sinh x 3 x 1 0 is to be fou
June 2025, Numerical Methods, Q7 (11 marks): The equation 8 ln x ( 2 ) 3 x 1 0 has a positive root, c, and a negative root, d. + - + = It is proposed that the iterative formula x g ( x ) , where ( ) g x 8
June 2025, AS Numerical Methods, Q1 (5 marks): The method of interval bisection is used to find a sequence of approximations to one of the roots x 2 of the equation e x 3 x 0 . - - = The table shows the asso
June 2025, AS Numerical Methods, Q6 (6 marks): 6 The equation . ln x 0 5 x x 2 0 has a root a, such that 1 2 . + - - = 1 1 a A student attempts to find a using the iterative formula 2 x g ( x ) with x 0 = ,
June 2025, AS Numerical Methods, Q9 (8 marks): In this question you must show detailed reasoning 4 + The equation . 0 2 x 0 8 . x 0 2 . 0 has a root a such that 0 1 . - = 1 1 a (a) Use the Newton-Raphson met
June 2024, Numerical Methods, Q3 (6 marks): - 3 The equation x cosh ( x 2 ) 0 has two roots, a and b, such that 1 b . - = a (a) Use the iterative formula x g ( x ) where ( g x ) cosh ( x 2 ) , 1 = = - n n
June 2024, Numerical Methods, Q5 (10 marks): The root of the equation ( )x f 0 is being found using the method of interval bisection. Some of = the associated spreadsheet output is shown in the table below
June 2024, AS Numerical Methods, Q4 (11 marks): The graph shows part of the graph of y f ( ) x . = y 1.5 1 0.5 0 x 2 2.5 3 3.5 4 4.5 -0.5 The equation ( )x f 0 has two positive roots, a and b, where 2 b. = a
June 2024, AS Numerical Methods, Q6 (11 marks): The equation log x x 5 0 has two roots, a and b. 3 - = You are given that 1 1 1 2 . a x n (a) Use the iterative formula x x x and x g ( ) , where ( g ) 3 2 , to
June 2023, Numerical Methods, Q5 (6 marks): The equation 3 2 ln x x 0 has a root near x 1 8 . . - - = = A student proposes to use the iterative formula x g ( x ) 3 2 ln x to find this root. 1 = = - n n n
June 2023, Numerical Methods, Q7 (6 marks): d , when x y 7 y x , is given in Table 7.1. The value of a function, f ( ) , and its gradient function, x 2 = = d Table 7.1 d y x f ( )x d x 2 6 .2 8 - (a) Dete
June 2023, Numerical Methods, Q8 (8 marks): The graph of y 0 2 . cosh x 0 4 . x for values of x from 0 to 3.32 is shown on the graph below. = - y 1.2 1 0.8 0.6 0.4 0.2 x 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
June 2023, AS Numerical Methods, Q4 (9 marks): It is given that the function ( )x g is continuous and differentiable and that g ( . ) 1 8 0 576 . , correct l = to 3 decimal places. (a) Use this result to det
June 2023, AS Numerical Methods, Q7 (9 marks): 2 7 The equation . 5 39 x 2 17 . x 1 11 . x 0 45 . 0 has a root, a, close to .0 5 . - - + = - (a) Use the Newton-Raphson method with x 0 5 . to determine the va
June 2022, Numerical Methods, Q3 (7 marks): 3 The equation ( ) f x sin - ( ) x x 0 1 . 0 has a root a such that 1 1 1 0 . = - + = - a Alex uses an iterative method to find a sequence of approximations to
June 2022, Numerical Methods, Q4 (8 marks): x 2 4 Fig. 4.1 shows part of the graph of y e x x 1 1 . . = - - - y 3 2 1 x 1 2 3 4 0 –1 Fig. 4.1 x 2 The equation e x x 1 1 . 0 has a root a such that 1 1 1 2
June 2022, Numerical Methods, Q6 (11 marks): Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation 1 2 x x sin - ( ) 1 0 . - + = a k. 2 x x x x Charlie uses th
June 2022, AS Numerical Methods, Q4 (12 marks): 2 4 The equation x 3 13 x 11 0 has three roots, a, b and c such that a 1 b 1 c. - + = 5 2 Fig. 4.1 shows part of the graph of y 3 x 13 x 11 . = - + y 10 8 6 4 2
June 2022, AS Numerical Methods, Q5 (7 marks): - 5 The equation x 3 x 1 0 has two positive roots, a and b, where a 1 b. + = 5 x 1 + n Charlie is using the iterative formula x g ( x ) where ( g x ) to try to
November 2021, Numerical Methods, Q3 (7 marks): The method of False Position is used to find a sequence of approximations to the root of an equation. The spreadsheet output showing these approximations, toget
November 2021, Numerical Methods, Q6 (12 marks): + + 6 The equation . ln x 0 5 x x 1 0 has two roots a and b, such that 0 1 1 1 and 1 1 1 2 . - = a b (a) Use the Newton-Raphson method with x 0 = to obtain b co
November 2021, AS Numerical Methods, Q4 (13 marks): Fig. 4.1 shows the graph of y ln x 3 x + . 4 = - y 2 1 x –1 0 1 2 3 4 –1 -1 Fig. 4.1 The equation ( ) f x 0 where f ( ) x ln x 3 x 4 has two roots, a and b, whe
November 2021, AS Numerical Methods, Q7 (12 marks): x 2 7 The equation ( ) f x 0 where ( ) f x 0 9 . x - - has a root a, such that x 1 .1 87 . = = . a # x 1 + n Sundip is considering using the iterative formula x
November 2020, Numerical Methods, Q5 (34 marks): You are given that 3 x x 25 + g ( )x . = 2 Fig. 5.1 shows two values of x and the associated values of ( )x g . x 1.45 1.55 g ( )x 1.49468 1.49949 Fig. 5.1 (a)
November 2020, AS Numerical Methods, Q6 (11 marks): 6 Fig. 6.1 shows the graph of y x 3 x + for 1 2 x 2 . = - - G G y 5 4 3 2 1 x 0 –3 –2 –1 1 2 3 –1 –2 Fig. 6.1 3 - The equation x 3 x 1 0 has three roots, a, b a
November 2020, AS Numerical Methods, Q7 (6 marks): - 7 The equation x 2 x 1 0 has one root, which is close to x = . 1 - = (a) Use the secant method with x 0 5 . and x 0 8 . to obtain x2, a first approximation to
June 2019, Numerical Methods, Q2 (8 marks): 2 e x 2 2 Fig. 2.1 shows the graph of y x 5 x 0 5 . . = - + y 2 1 x –1 0 1 –1 Fig. 2.1 2 2 e x 2 x x There are three roots of the equation 5 0 5 . 0 . The roots
June 2019, Numerical Methods, Q4 (13 marks): 4 Fig. 4 shows the graph of y x 6 x + . 4 = - y 6 4 2 x 0 2 4 Fig. 4 5 - There are two roots of the equation x 6 x 4 0 . The roots are a and b, such that b. 1 +
June 2019, AS Numerical Methods, Q1 (8 marks): 2 1 The graph of y 3 15 . x 0 75 . x 5 2 . x 2 2 . is shown in Fig. 1.1. = + - + y 6 5 4 3 2 1 x –2 –1 0 1 2 3 –1 –2 Fig. 1.1 3 2 There are three roots to the e
June 2019, AS Numerical Methods, Q6 (6 marks): Fig. 6 shows some spreadsheet output concerning the solution of an equation, together with some further analysis. The iterates are shown in column C, the difference
June 2018, AS Numerical Methods, Q4 (11 marks): x 2 4 The equation e x x 2 0 has a root a, where 0 1 1 1 and a root b, where 2 1 1 3 . - - = a b (i) Show how to obtain the iterative formula x 2 e x - r r x .
June 2018, AS Numerical Methods, Q6 (8 marks): - 6 (i) Show that the equation . x x has a root a, where 3 [1] 0 1 2 3 0 1 1 4 . + = a The method of false position is used to find a. The spreadsheet output in

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