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 your answer. [1] (ii) Write down a basis for the range of T. [1] (c) The kernel of T, ker(T), is defined as the set of vectors x y z w       in 4 that map to the zero vector in 5 . Determine ker(T). [4]

2023 GCE A level FM Paper 2 Solution (a) 2 2 1 3 3 1 4 4 1 5 5 1 22 33 44 55 2 3 1 3 1 41 3 1 2 1 3 1 2 1 3 2 3 1 5 0 1 1 1 1 5 1 3 0 3 0 6 3 2 1 13 0 8 4 4 1 1 2 0 0 3 3 3 1 2 1 3 0 1 1 1 01 R R R R R R R R R R R R RR RR RR RR →− →− →− →+ →− → →− →         − − −        ⎯⎯⎯⎯⎯ →−−    − − − −       −    ⎯⎯⎯⎯ → 1 1 2 3 3 2 4 4 2 5 5 2 33 4 4 3 2 2 ' 02 0 2 1 1 0 1 1 1 1 0 1 1 0 1 1 1 0 0 1 3 0 0 1 3 0 0 0 0 1 0 1 1 01 R R R R R R R R R R R R RR R R R →− →− →− →− →− →+     −  −  −   ⎯⎯⎯⎯⎯ → −−  −−  − ⎯⎯⎯⎯⎯ → 11 0 0 1 3 0 0 0 0 0 0 0 0        Row rank of M = number of non-zero rows in the echelon form = 3 (b)(i) Column rank = row rank = 3 (ii) Basis for the range of T= 1 2 1 2 3 1 ,1 5 1 3 2 1 1 1 2                        −−        −     (c) 01 0 1 1 00 1 1 1 00 0 1 3 00 0 0 0 0 0 0 0 0 0 0 43 0 3 22 34 1 x y z w x z w y z w xz w z w yy w w zxw w −                =               − + = + + = −   + =  =−         =  =      −=−         ker(T) = 4 4 2: 3 1 xx yy zz ww   −                =      −          

2023 GCE A level FM Paper 2 Solution 4 A sequence is given by 0 4u = and ( ) 2 1 22n n nu u u+ = − + for 0n . By considering 1nnvu=− , for all 0n , find an expression for nu in terms of n and prove the result by induction. [7] 4 ( ) ( ) 2 1 2 2 22 11 1 n n n n n u u u u v + = − + = − + =+ Considering 1nnvu=− , 11 1 1 nn nn uv uv++  = +  = + Comparing with 2 1 1nnuv+ =+ , we have ( ) ( ) ( ) ( ) ( ) 2 3 1 2 1 2 1 22 1 2 1 42 2 2 2 2 0 11 n nn nn n n n n vv vv v v v v v + + + − − − − + = + = = = = = = = ( ) 2 0 n nvv= ( ) 2 0Since 1, 1. n n n nu v u v= + = + Given 0 4u = and 1nnvu=− , 00 13vu= − = 231 n nu = + Let nP be the proposition that 231 n nu =+ for all 0n . For 0n= , LHS of 00 4 (given)Pu== RHS of 021 0 3 1 3 1 4P = + = + = 0P is true. Assume that kP is true for some 0k , 231 k ku =+ To show 1kP + is true also, i.e. 12 1 31 k ku + + =+ , LHS of 1kP +

2023 GCE A level FM Paper 2 Solution ( ) ( ) ( ) ( ) ( ) 1 1 1 2 222 2 2 2 2 1 22 3 1 2 3 1 2 3 2 3 1 2 3 2 2 31 RHS of kk k k k k k kk k u uu P + + + + = = − + = + − + + = + + − − + =+ = 1 is true is truekkPP + Since 0P is true and 1 is true is truekkPP + , by mathematical induction, nP is true for all 0n

2023 GCE A level FM Paper 2 Solution 5 The diagram shows a sector ABC of a circle with radius r and centre C on the negative y-axis. The angle ACB is 2, where 10 π.2 The shaded region, bounded by the arc AB and the x-axis, is rotated completely about the x-axis to form a solid of revolution, S. (a) (i) By considering the circle with equation 2 2 2 ,x y r+= or otherwise, show that sin 2 2 2 2 0 11d sin cos .22 r r x x r r    − = + [2] (ii) Hence, or otherwise, show that the volume of S is 332 π (3sin sin 3 cos ).3 r    −− [5] (b) Determine the corresponding expression, in terms of r and , for the surface area of S. [7] (a)(i) sin 22 0 d area of shaded region r r x x  −= ( ) 2 22 area of sector + area of triangle 11 sin ( cos )22 11 sin cos (shown)22 OAB OBC r r r rr       = =+ =+ (ii) Equation of arc AB: 2 2 2 ( cos )x y r r + + = V olume of S sin 2 0 2d r yx  =  ( ) ( ) 2sin 22 0 sin 2 2 2 2 2 2 0 sin 2 2 3 2 2 0 3 2 3 3 2 2 3 2 cos d 2 2 cos cos d 1 1 12 (1 cos ) 4 cos sin cos 3 2 2 1 1 12 sin (1 cos ) sin 4 cos sin cos 3 2 2 2 3 r r r r x r x r x r r x r x r x x r r r r r r r r r                         = − − = − − − +    = + − − +        = + − − +     =   2 3 2 33 3sin (1 cos ) sin 3 cos 3sin cos 2 3sin sin 3 cos (shown)3 r              + − − − = − − 

2023 GCE A level FM Paper 2 Solution (b) Surface area 2sin 0 d2 1 d d r yyx x   =+  2sin 0 2sin 0 sin 2 2 0 sin 0 sin 0 sin 220 si 1 0 d2 1 d d 2 1 d cos 2 d ( cos ) 2 d cos cos2 1 d cos cos2 1 d 2 cos sin r r r r r r r yyx x xyx yr ryx yr yrx yr rrx yr rrx rx xr x r r                  − =+   −=+  + = + = + =−  + =−  − =−          n 2 2 sin