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1 The root of the equation f(x) = 0, where f(x) = x +ln2x - 4 is to be estimated using the iterative formula xn1 4 ln2xn , with x0 2.4. (a) Showing your values of x1, x2, x3,…, obtain the value, to 3 decimal places, of the root. (4) (b) By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a). (2) 1 The root of the equation f(x) = 0, where f(x) = x + ln2x - 4 is to be estimated using the iterative formula xn1 4 ln2xn , with x0 2.4. (a) Showing your values of x1, x2, x3,…, obtain the value, to 3 decimal places, of the root. (4) x1 4 ln2x0 4 ln2(2.4) 2.4313... x2 4 ln2(2.431...) 2.4183... x3 2.4237... x4 2.4215... x 2.422 x5 2.4224... 1 The root of the equation f(x) = 0, where f(x) = x + ln2x - 4 is to be estimated using the iterative formula xn1 4 ln2xn , with x0 2.4. (b) By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a) (2) x 2.422 f(2.4215) = -0.000965... f(2.4225) = 0.000447... The function is continuous and there is a change of sign, therefore x is a root 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (a) Show that f(x) = 0 can be rearranged as (2) 4x 1 x , x 1 x 1 The iterative formula xn + 1 = find an approximation to . 4xn 1 xn 1 is used to (b) Taking x1 = 1, find, to 2 decimal places, the values of (3) x2, x3 and x4. 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (a) Show that f(x) = 0 can be rearranged as 4x 1 x , x 1 x 1 x x 4x 1 0 3 2 x3 x2 4 x 1 x2 (x 1) 4 x 1 4x 1 2 x (x 1) (2) 4x 1 x x 1 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (b) Taking x1 = 1, find, to 2 decimal places, the values of (3) x2, x3 and x4. 4 xn 1 xn1 xn 1 4 1 1 x2 1 1 1.58 4 1.58... 1 x3 1.58... 1 1.68 x4 1.70 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (c) By choosing values of x in a suitable interval, prove that = 1.70, correct to 2 decimal places. (3) (d)Write down a value of x1 for which the iteration formula xn + 1 = 4 x n 1 x 1 n does not produce a valid value for x2. Justify your answer. (2) 2 4. 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (c) By choosing values of x in a suitable interval, prove that = 1.70, correct to 2 decimal places. (3) f(1.695) = -0.037… f(1.705) = 0.0435… The function is continuous and there is a change of sign, therefore is a root 2 f(x) = x3 + x2 4x 1. The equation f(x) = 0 has only one positive root, (d)Write down a value of x1 for which the iteration formula xn + 1 = 4 x n 1 x 1 n does not produce a valid value for x2. Justify your answer. (2) x = -1; Division by zero is not possible -1 < x < -¼; Cannot find the square root of a negative number 3 (a) Sketch, on the same set of axes, the graphs of y = 2 – ex and y = x. (3) [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = ex + x – 2, x 0, (b) explain how your graphs show that the equation f(x) = 0 has only one solution, (1) 3 (a) Sketch, on the same set of axes, the graphs of y = 2 – ex and y = x. (3) [It is not necessary to find the coordinates of any points of intersection with the axes.] 3 Given that f(x) = ex + x – 2, x 0, (b) explain how your graphs show that the equation f(x) = 0 has only one solution, (1) Where curves meet is solution to f(x) = 0; hence there is only one intersection 3 (c) show that the solution of f(x) = 0 lies between x = 3 and x = 4. (2) The iterative formula xn + 1 = (2 –e solve the equation f(x) = 0. (d) xn 2 ) is used to Taking x0 = 4, write down the values of x1, x2, x3 and x4, and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. (4) 3 (c) show that the solution of f(x) = 0 lies between x = 3 and x = 4. (2) f(x) = ex + x – 2 f(3) = -0.218… f(4) = 0.018… This is a continuous function and there is a change of sign over this interval, therefore there is a root within this interval. The iterative formula xn + 1 = (2 –e to solve the equation f(x) = 0. 3 (d) xn Taking x0 = 4, write down the values of x1, x2, x3 and x4, and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. (4) x1 =(2 – e–4)2 = 3.92707… x2 = 3.92158… x3 = 3.92115… x4 = 3.92111(9)… Approx. solution = 3.921 (3 dp) x0 = 4 )2 is used 4 (a) The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). Sketch the graph of y = ln 3x, showing the exact value of p (2) The normal to the curve at the point Q, with x-coordinate q, passes through the origin. (b) Show that x = q is a solution of the equation x2 + ln 3x = 0 5 8. (4) 4 (a) The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). Sketch the graph of y = ln 3x, showing the exact value of p (2) ln 1 = 0 ln3x = ln1 3x = 1 x = 1/3 1 3 5 8. 4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). The normal to the curve at the point Q, with x-coordinate q, passes through the origin. (b) Show that x = q is a solution of the equation x2 + ln 3x = 0 y = ln 3x, y’ = 1/x (4) Q At Q gradient is 1/q Gradient of normal is -q Equation of OQ is y – 0 = -q(x- 0) y = -qx ln 3x = -x2 x2 + ln 3x = 0 5 8. 4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). (c) Show that the equation2 in part (b) can be rearranged in the form x = 13 e xn (2) (d) x2 Use the iteration formula xn + 1 = e , with x0 = 13 , to find x1, x2, x3 and x4. Hence write down, to 3 decimal places, an approximation for q (3) 1 3 5 8. 4 The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). (c) Show that the equation2 in part (b) can be rearranged in the form x = 13 e xn (2) x2 + ln 3x = 0 ln 3x = - x2 xn2 3x = e x = 1/3e-x2 5 8. 4 (d) The curve with equation y = ln 3x crosses the x-axis at the point P (p, 0). x2 Use the iteration formula xn + 1 = e , with x0 = 13 , to find x1, x2, x3 and x4. Hence write down, to 3 decimal places, an approximation for q (3) 1 3 x1 = 0.298280; x2 = 0.304957, x3 = 0.303731, x4 = 0.303958 Root = 0.304 (3 decimal places) 5 8.