2022 A Level 9649 FM P2 (Solutions)

Paper 2 Remarks 1 2d 32d y ax bxx =+ At the turning points, d 0.d y x = So, 23 2 0 (3 ) 0 20 or . 3 ax bx x ax b bxx a += += = =− When 0,x= .yc= When 2 ,3 bx a=− 32 32 32 3 2 3 2 22 33 84 27 9 84 27 9 4 27 bby a b c aa bba b c aa b ca b ca    = − + − +          = − + +       = − + + =+ For the equation 32 0ax bx c+ + = to have 3 distinct real roots, the y- coordinates of the two turning points must have different signs. Thus ( ) 3 2 33 3 2 3 4 027 4 27 0 27 4 0 (shown) bcc a c b a c a c b c  + + + 2 (a) ( )21 1 2 n n nu u u++ =+ 212 0n n nu u u++− − = The auxiliary equation of the RR is 2 10 (2 1)( 1) 0 2 1 or 12 mm m mm m −= + − = =− − = 1 (1)2 1 2 n n n n n u u cd cd = − + = − + 1 2 1 (1)2 1 (2)4 u u a c d a b c d b = − + = − = + = −

(2) – (1): 4 ()3c b a=− (1) + 2(2): 1 ( 2 )3d a b=+ 4 1 1( ) ( 2 ), 13 2 3 n nu b a a b n  = − − + +  (b)(i) 4 1 1( ) ( 2 ), 13 2 3 n nu b a a b n = − − + +  As , 3 1since 0 as , therefore2 1 ( 2 ) 33 29 n n nu n ab ab → → − → → += += Since a and b are distinct positive integers, hence possible pairs of values of a and b are: 1and 4, 5 and 2, 7 and 1a b a b a b= = = = = = (b)(ii) If ab= , then 4 1 1(0) ( 2 )3 2 3 n nu a a = − + + ,1nu a n= Then the sequence is a constant sequence.

3 (a) 00 ( ln )Let f ( , ) , 1, 1.y x yx y x y x += = =     1 0 0 0 1 0 0 0 1 1 2 1 1 1 2 1 1 1 2 2 1 ln10.5f ( , ) 1 0.5 1.5 1 0.5 1.5(1.5 ln1.5)f ( , ) f ( , ) 1 0.25 12 1.5 1.72637... 0.5f ( , ) 2.90376... 0.5 f ( , ) f ( , ) 3.42793... 3.43 (3sf)2 u y x y y y x y x u u y x y y y x y x u += + = + =  += + + = + +  = = + = = + + = = (b)(i) ,0 πsin xxyx x=   Taking logarithms, we have ln ln ln sin .y x x x=− 1 d 1 cos lnd sin yx xxy x x x = + − At turning point, d 0.d y x = 1 cot 1 ln 1tan 1 ln 1tan (shown)1 ln xx x x x x − =+ = + =  + (b)(ii) 10 g( ), with 0.9nnx x x+ == From GC, 12 0.84095, 0.87994, , converges to 0.864 (3s f).nx x x==  x-coordinate of the turning point is 0.864, y-coordinate is 1.16 (3sf) 4 (a)(i) sin cosy A x B x=+ ( ) ( ) 11 22d 1 1 cos sind 2 2 d2 cos sind y A x x B x xx yx A x B xx −−   =−         =− ( ) 2 2 2 2 dd4 2 sin cosdd dd4 2 0 (shown)dd yyx A x B xxx yyxy xx + =− + + + = ( ) ( ) 1 1 12 2 2 2 2 d d 1 12 sin cosd d 2 2 yyx x A x x B x xxx − − −     + = − −            (a)(ii) Let nP be the statement 21 21 d d d4 (4 2) 0d d d n n n n n n y y yxn x x x ++ ++ + + + = for all integers 0n . LHS of 22 0 22 d d d d4 [4(0) 2] 4