2025 DHS H2 Math Prelim P2 Soln

1 2025 Year 6 H2 Math Prelim Exam P2 Suggested Solutions Section A: Pure Mathematics [40 marks] Qn Suggested Solutions Comments 1(a) ( ) ( ) 2 2 32 9 1 d9 1d 1 d 18= d 1 yx x y x x y x x =+ + =− + + C concaves upwards  ( ) 2 32 d 18= d 1 y x x+ > 0 ( ) 3 1 0 1xx+   − When a function f( )yx= concaves upwards, the gradient of its gradient function is positive; i.e. 2 2 d 0d y x  . (b) 1x=− C yx= x A Stationary points: ( 4, 7), (2,5)AB−− y-intercept: ( )0,9C For ayx xb=+ + , where a & b are constants, there will be a vertical asymptote ( xb=− ) & an oblique asymptote ( yx= ). For any axial intercept(s), they can be found by letting x & y be zero respectively. For any stationary point(s), they can be found from d 0d y x = or using the GC. Total marks: 5

2 Qn Suggested Solution Comments 2(a) ( ) ( ) ( ) 2 2 1 1 1 e tan e d dtan tan d (shown) xx x ttt t t t t − − − = =    • Clear working needs to be shown to get full credit. (b) ( ) ( ) ( ) 22 1 2 22 1 22 2 1 2 2 11 1 2 1tan d 1tan d2 2 1 1 1 1tan d2 2 1 1 11tan 1 d2 2 1 1tan tan22 tan 122 t t t tt tt t tt tt tt t tt t t t t t C tt tC − − − −− − − =− +  += − −  ++ = − −  + = − − + = + − +     ( ) ( ) ( ) ( ) 1 2 0 1 2 1 21 1 2 1 1 e e tan e d tan 122 tan e e 1e 1 tan 12 2 2 tan e e 1e12 2 4 2 x x x tt t  − − − − − = + −  = + − − −  = + − − +  • Use LIATE to decide on the choice of term to differentiate. • 2 21 t t+ can be made proper by splitting the numerator or by long div • Note that 1 πtan 1 4 − = Total marks: 7 e d e d ddd e xx x t t x ttx t =  =  = = 1 2 2 dtan t d du 1 d 1 2 vut t tvtt −== ==+ 00, e 1 1, e xt xt = = = ==

3 Suggested Solution Comments 3(a) 33 50x y xy+ − = Differentiate wrt x : ( ) 22 22 2 2 dd3 3 5 0 dd d3 5 5 3 d d 5 3 d 3 5 yyx y x y xx yy x y x x y y x x y x + − + =  − = − −= − Note that product rule have to be applied dd( ) (1)dd yxy x yxx =+ (b) Gradient of normal = 2 2 35 53 yx yx −− − For normal to be parallel to the y-axis, gradient of normal is undefined. Hence, 25 3 0yx−= 23 5yx= ---- (1) Subst (1) into equation of curve, 3 3 2 2 63 33 5055 27 20125 x x x x xx    + − =       −= 2.0999 2.10 (3s.f ) or 0(rejected 0) 2.6457 2.65 (3s.f ) Using GC, x x x y = = =  == Coordinates of the point is ( )2.10, 2.65 . Alternative (if exact answers needed) ( ) 3 6 3 63 3 3 27 30125 27 20125 27 250 0125 x x x xx x x + − = −= −= Since x > 0, 3 3250 5 233x== . 2 3 23 33 250 250 5 45 3 15 3y = = = Some students had the misconception of 235yx− = 0 but notice gradient of normal would be 0 not undefined