2023 GCE A level FM P2 (Solution)

2023 GCE A level FM Paper 2 Solution Section A: Pure Mathematics [50 marks] 1 The diagram shows the curve with polar equation 𝑟 = 2 ln 𝜃 for 1 ≪ 𝜃 ≪ π. The area of the region bounded by the curve and the line 𝜃 = π is denoted by A. Determine the exact value of A. [6] 1 2 1 2 1 2 1 1 2 1 1 2 1 2 1 4(ln )2 12 (ln ) 2ln 12 (ln ) 2 ln 2 (ln ) 2 ln 2( 1) A r d d d d                         = = = −    = − −   = − + −    

2023 GCE A level FM Paper 2 Solution 2 Let f(𝑥) = 2𝑥 − 𝑥 − 3. (a) By sketching 𝑦 = 2𝑥 and 𝑦 = 𝑥 + 3 on a single diagram, show that the equation f(𝑥) = 0 has exactly two roots. [2] The two roots of f(𝑥) = 0 are denoted by  and , where 𝛼 < 𝛽 . (b) Show that 3 2.−  − [2] (c) (i) Find, in an exact form, the value of x for which f ( )x is a minimum. [2] (ii) Use the Newton-Raphson method, with initial approximation 0 0.5,x = to calculate the value of 1x correct to 4 significant figures. [2] (iii) With reference to your answer to (c)(i), explain why 0 0.5x = is an unsuitable starting-value for using this iterative method to find an approximation to . [2] (iv) Use the Newton-Raphson method to calculate the value of  correct to 4 decimal places. [2] (a) f ( ) 0 2 3. xxx=  = + From the graph, we see that the graphs of 2xy= and 3yx=+ intersect exactly twice. Hence the equation has exactly two real roots. (b) x 3− 2− 2x 1 8 1 4 3x+ 0 1 2 3, 3 2 3 for a certain ( 3, 2). 2 3, 2 x x x xx xx xx  + =−  = +  − − + =−  (c)(i) Consider f '( ) 0.x = 2 ln 2 1 0 12 ln 2 1 1 1ln ln(ln 2)ln 2 ln 2 ln 2 x x x −= = = =−  2f''( ) 2 (ln 2) 0 for all .xxx= 1f ( ) is a minimum at ln(ln 2).ln 2xx =− x y

2023 GCE A level FM Paper 2 Solution (ii) By Newton-Raphson’s formula, 0 0 0 10 23 105.22 ln 2 1 x x xxx −−= − =− − (4sf) (iii) The minimum point occurs at 1 ln(ln 2) 0.5288ln 2x=− − . Hence the tangent to the curve at 0.5x=− being near the minimum point has a very gentle slope. As a result, it intersects the x-axis at a faraway point (-105.2,0).  0 0.5x = is not a suitable starting value. (iv) 1 0 1 2 3 4 23 2 ln 2 1 2 2.907207 2.862572 2.862500 2.862500 n n x n nn x xxx x x x x x + −−=− − =− =− =− =− =− 5 5 f ( 2.86245) 4.56 10 f ( 2.86255) 4.49 10 2.86255 2.86245 2.8625 (4dp)   − − − =−  − =  −  −  =− 3 The transformation T is the linear mapping from 4 to 5 given by the matrix 1 2 1 3 2 3 1 5 1 5 1 3 3 2 1 13 1 1 2 0    = −  −− − M . (a) Showing all necessary working, use row operations to find the row rank of .M [5] (b) (i) State the column rank of ,M giving a reason for