. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 Use the iterative formula = sin +, 4 n + 0 = n.5, to calculate the values of, and 3, giving your answers to 4 decimal places. (d) By considering the change of sign of f() in a suitable interval, verify that α =.9 correct to 3 decimal places. (Total 9 marks). f() = 3 + 3 (a) Show that f() = 0 can be rearranged as 3 + =,. + The equation f() = 0 has one positive root α.
The iterative formula 3n + = n+ is used to find an approimation to α. n + Taking = 0, find, to 3 decimal places, the values of, 3 and 4. Show that α =.057 correct to 3 decimal places. (Total 8 marks) 3. The diagram above shows part of the curve with equation y = 3 + +, which intersects the -ais at the point A where = α. To find an approimation to α, the iterative formula is used. = ( ) n + + n (a) Taking 0 =.5, find the values of,, 3 and 4. Give your answers to 3 decimal places where appropriate.
Show that α =.359 correct to 3 decimal places. (Total 6 marks) 4. f() = 3e The curve with equation y = f () has a turning point P. (a) Find the eact coordinates of P. (5) The equation f () = 0 has a root between = 0.5 and = 0.3 Use the iterative formula + = e 3 n n with 0 = 0.5 to find, to 4 decimal places, the values of, and 3. By choosing a suitable interval, show that a root of f() = 0 is = 0.576 correct to 4 decimal places. (Total marks) 5. f() = 3 3 6 (a) Show that f() = 0 has a root, α, between =.4 and =.45 Show that the equation f () = 0 can be written as = +, 0. 3
Starting with 0 =.43, use the iteration n+ = n + 3 to calculate the values of, and 3, giving your answers to 4 decimal places. (d) By choosing a suitable interval, show that α =.435 is correct to 3 decimal places. (Total marks) 6. f() = ln( + ) +, >,. (a) Show that there is a root of f() = 0 in the interval < < 3. Use the iterative formula n+ = n( n + ) +, 0 =.5 to calculate the values of, and 3 giving your answers to 5 decimal places. Show that =.505 is a root of f() = 0 correct to 3 decimal places. (Total 7 marks) 7. f() = 3 + 3. (a) Show that the equation f() = 0 can be rewritten as =. 3
Starting with = 0.6, use the iteration n+ = 3 n to calculate the values of, 3 and 4, giving all your answers to 4 decimal places. Show that = 0.653 is a root of f() = 0 correct to 3 decimal places. (Total 7 marks) 8. y O P π 4 The figure above shows part of the curve with equation y = ( ) tan, 0 < 4 π The curve has a minimum at the point P. The -coordinate of P is k. (a) Show that k satisfies the equation 4k + sin 4k = 0. (6)
The iterative formula is used to find an approimate value for k. = ( sin 4n ), 0 = 0.3, 4 n + Calculate the values of,, 3 and 4, giving your answers to 4 decimal places. Show that k = 0.77, correct to 3 significant figures. (Total marks) 9. f() = 3 4. (a) Show that the equation f() = 0 can be written as = + The equation 3 4 = 0 has a root between.35 and.4. Use the iteration formula n + = +, with 0 =.35, to find, to decimal places, the values of, and 3. The only real root of f() = 0 is α. By choosing a suitable interval, prove that α =.39, to 3 decimal places. (Total 9 marks)
0. f() = 3e ln, > 0. (a) Differentiate to find f (). The curve with equation y = f() has a turning point at P. The -coordinate of P is α. Show that α = 6 e α. The iterative formula n + = n e, 0 =, 6 is used to find an approimate value for α. Calculate the values of,, 3 and 4, giving your answers to 4 decimal places. (d) By considering the change of sign of f () in a suitable interval, prove that α = 0.443 correct to 4 decimal places. (Total 9 marks)
. y O A B C f() = + ln, > 0. The diagram above shows part of the curve with equation y = f(). The curve crosses the -ais at the points A and B, and has a minimum at the point C. (a) Show that the -coordinate of C is. (5) Find the y-coordinate of C in the form k ln, where k is a constant. Verify that the -coordinate of B lies between 4.905 and 4.95. (d) Show that the equation ( ) e -. + ln = 0 can be rearranged into the form = The -coordinate of B is to be found using the iterative formula ( - ) n + = e n, with 0 = 5.
(e) Calculate, to 4 decimal places, the values of, and 3. (Total 3 marks). f() = 3, 0. (a) Show that the equation f() = 0 has a root between and. An approimation for this root is found using the iteration formula n + = + n 3, with 0 =.5. By calculating the values of,, 3 and 4, find an approimation to this root, giving your answer to 3 decimal places. By considering the change of sign of f() in a suitable interval, verify that your answer to part is correct to 3 decimal places. (Total 8 marks) 3. The curve C with equation y = k + ln, where k is a constant, crosses the -ais at the point A,0. e (a) Show that k =. Show that an equation of the tangent to C at A is y = e.
Complete the table below, giving your answers to 3 significant figures..5.5 3 + ln.0.6.79 (d) Use the trapezium rule, with four equal intervals, to estimate the value of 3 ( + ln ) d. (Total marks) 4. f() = 3 + 4. The equation f() = 0 has only one positive root, α. (a) Show that f() = 0 can be rearranged as = 4 +,. + The iterative formula n + = 4n + n + is used to find an approimation to α. Taking =, find, to decimal places, the values of, 3 and 4. By choosing values of in a suitable interval, prove that α =.70, correct to decimal places.
(d) Write down a value of for which the iteration formula n + = produce a valid value for. Justify your answer. 4n + n + does not (Total 0 marks) 5. f() = + e,. 5 (a) Find f (). The curve C, with equation y = f(), crosses the y-ais at the point A. Find an equation for the tangent to C at A. e Complete the table, giving the values of + to decimal places. 5 0 0.5.5 e 0.45 0.9 + 5
(d) Use the trapezium rule, with all the values from your table, to find an approimation for the value of 0 e + 5 d. (Total marks) 6. y C R O The curve C has equation y = f(), 0.. The diagram above shows the part of C for which Given that dy d = e, and that C has a single maimum, at = k, (a) show that.48 < k <.49.
Given also that the point (0, 5) lies on C, find f(). The finite region R is bounded by C, the coordinate aes and the line =. Use integration to find the eact area of R. (Total marks) 7. A student tests the accuracy of the trapezium rule by evaluating I, where I =.5 0.5 3 4 + d. (a) Complete the student s table, giving values to decimal places where appropriate. 0.5 0.75.5.5 3 4 + 6.06 4.3 Use the trapezium rule, with all the values from your table, to calculate an estimate for the value of I. Use integration to calculate the eact value of I. (d) Verify that the answer obtained by the trapezium rule is within 3% of the eact value. (Total marks)
8. The curve C has equation y = f(), where f() = 3 ln +, > 0. The point P is a stationary point on C. (a) Calculate the -coordinate of P. Show that the y-coordinate of P may be epressed in the form k k ln k, where k is a constant to be found. The point Q on C has -coordinate. Find an equation for the normal to C at Q. The normal to C at Q meets C again at the point R. (d) Show that the -coordinate of R (i) satisfies the equation 6 ln + + 3 = 0, (ii) lies between 0.3 and 0.4. (Total 4 marks) 9. (a) Sketch, on the same set of aes, the graphs of y = e and y =. [It is not necessary to find the coordinates of any points of intersection with the aes.]
Given that f() = e +, 0, eplain how your graphs show that the equation f() = 0 has only one solution, () show that the solution of f() = 0 lies between = 3 and = 4. The iterative formula n + = ( n e ) is used to solve the equation f() = 0. (d) Taking 0 = 4, write down the values of,, 3 and 4, and hence find an approimation to the solution of f() = 0, giving your answer to 3 decimal places. (Total 0 marks) 0. The curve with equation y = ln 3 crosses the -ais at the point P (p, 0). (a) Sketch the graph of y = ln 3, showing the eact value of p. The normal to the curve at the point Q, with -coordinate q, passes through the origin. Show that = q is a solution of the equation + ln 3 = 0. Show that the equation in part can be rearranged in the form = e 3. e 3 n (d) Use the iteration formula n + =, with 0 = 3, to find,, 3 and 4. Hence write down, to 3 decimal places, an approimation for q. (Total marks)
. y O A B The diagram above shows a sketch of the curve with equation y = f() where the function f is given by f: e,. The curve meets the -ais at the point A and the y-ais at the point B. (a) Write down the coordinates of A and B. Find, in the form f ():..., the inverse function of f and state its domain. (5) Prove that the equation f() = has a root α in the interval [3, 4]. (d) Use the iterative formula n + = f ( n ), with = 3.5, to find α to 3 decimal places. Prove that your answer is correct to 3 decimal places. (5) (Total 4 marks)
. (a) Sketch the curve with equation y = ln. Show that the tangent to the curve with equation y = ln at the point (e, ) passes through the origin. Use your sketch to eplain why the line y = m cuts the curve y = ln between = and = e if 0 < m < e. Taking 0 =.86 and using the iteration n + = 3 n e, (d) calculate,, 3, 4 and 5, giving your answer to 5 to 3 decimal places. The root of ln 3 = 0 is α. (e) By considering the change of sign of ln 3 over a suitable interval, show that your answer for 5 is an accurate estimate of α, correct to 3 decimal places. (Total 3 marks)