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 Since the question do not require exact answer, students can solve the equation using GC directly Normal // y-axis → Tangent // x-axis → d d y x = 0 → 25 3 0yx−= Note making y the subject would be easier than x

4 Coordinates of the point is 335 2 5 4,33   . (c) For stationary point, d 0d y x = From part (a), 22 dd3 3 5 0 dd yyx y x y xx + − + =  . Differentiate wrt x : 222 2 22 d d d d d6 3 6 5 0d d d d d y y y y yx y y x x x x x x + + − + + =   222 2 22 d d d d6 3 6 5 10 0d d d d y y y yx y y x x x x x + + − − =  At the stationary point 352 3x= , 35 4 d, 03d yy x== . 2 22 d6 1.2d 3 5 yx x y x −= =−− < 0 Hence, maximum point. Note that finding 2nd derivative is much more efficient than 1st derivative method Note ( ) 2dd 36dd yyyxx = Ensure clear working is shown (substitute values of x, y and d d y x OR calculate correct value of 2nd derivative) to determine the sign of 2nd derivative and hence nature of stationary point Total marks: 7 Apply product rule

5 Qn Suggested Solution Comments 4(a) Let AD k AB= . ( ) ( ) ( )1 1 , 1 kkOD k k kkk +−= = + − +− ba ba ( ): ( 1 ), ODl k k = + − r b a • Qn is asking for vector eqn of line OD. Line OD  OD Eqn of line OD is of the form: =+r a b where a is the position vector of a fixed point on the line (choose vector O or OD ) and b is the direction vector of the line (choose OD ). (b) ( ) ( ) ( ) 1 2 1 2 1 2 3 32 OE OC OB=+ = + + =+ a b b ab Since O, D and E are collinear, point E lies on ODl . Method 1 ( ) ( ) ( ) ( ) 1 2 3 2 ( 1 ) 3 2 ( 1 ) 1 OE k k kk kk    = + − + = + − +− ba a b b a a b = a + b Since a and b are non-zero and non-parallel vectors, ( ) 33 22 1 ... (1) 1 ... (2) k kk     = − =  − = Subst (1) into (2): 35 221− =  = From (1): 2 51kk =  = Method 2 ( ) ( ) 3 22 Let 1 3 2 OD OE kk    = + − = + = +b a a b a b • 1 2OE BC as both OE and BC are not even parallel. • ( )1 2OE OC OB=+ using Ratio Thm O A E C a b B D 1 1 k

6 Since a and b are non-zero and non-parallel vectors, 3 2 ... (1) 1 ... (2) k k  = −= Subst (1) into (2): 3 2 521  − =  = From (1): 2 5k= (c) 1, 2 and 3 2 31.= = − =a b a b 2 22 22 6 1 12 2 3 2 31 3 2 31 (3 2 ).(3 2 ) 31 9 . 12 . 4 . 31 9 12 . 4 31 9(1) 12 . 4(2) 31 . −= −= − − = − + = − + = − + = =− =− ab ab a b a b a a a b b b a a b b ab ab Let  the angle between a and b. 1 2 o .1cos 1 2 4 104.5 (1.82 rad)   −= = =−  = ab ab • Concept (derived from scalar product formula): ( ) ( ) 2 2 2 (3 2 ).(3 2 ) 32 3 2 or 3 2 a b a b ab ab ab −− =− − − (d) ˆ=a b b a is the length of projection of b onto a. (note that a is the unit vector) • Note that a is the unit vector. So, it should be length of projection of b onto a and not the other way round. Total marks: 11 Qn Suggested Solution Comments 5(a) ( )d 30d 1 dd30 ln 30 30 e e e ( e ) 30 e c kt kt c kt T kTt T k tT T kt c T A A TA −− − =− − =−− − =− + − = = = =+  Subst 0, 160, 160 30 130 30 130e kt tT AA T − == = +  =  = + You will need to “remove” the modulus before applying the initial conditions k is given as a positive constant. Since T is decreasing, a negative sign must be applied.

7 (b) As the actual cooling starts 5 minutes later, we need to shift the cooling function 5 minutes to the right. Replace t with 55ta−  =− . (c) When 5,t ( 5)( ) ( 5) 30 130e kt pT t T t −−= − = + 10 10 Subst 15, 100 100 30 130e 70e 130 17ln 0.06190410 13 p k k tT k − − == =+ =  =− = 0.061904( 5) 0.061904 0.0619 ( 5) 30 130e 30 177.16e 30 177e (shown) t t t Tt −− − − − = + =+ =+ Alternatively, you can think of using part (a), can think of it as 10 mins has lapsed to find k. 10 10 Subst 10, 100 100 30 130e 70e 130 17ln 0.06190410 13 k k tT k − − == =+ =  =− = (d) 0.061904 0.061904 60 30 177.16e 30e 177.16 28.687 28.7 min t t t − − =+ = == Please be aware that you can do part (c) and (d) without attempting the earlier parts. (e) Total marks: 10 t Tp = 30 O 5 160 If there is a horizontal asymptote, you will need draw and label it.

8 Section B: Probability and Statistics [60 marks] Qn Suggested solutions Comments 6(ai) Note: P( ) P( ) 0.1B C A B C =   = since BA . Max x occurs when the part in C not in B (0.5) is entirely in A. Hence, max x = 0.5. Min x occurs when as much of the part in C not in B is outside of A as possible. ( ) P( ) 1 ( ) ( ) ( ) 1 0.8 0.6 0.1 1 0.3 AC P A P C P A C x x  + −   + − +   Hence, min x = 0.3. From the diagram, min x = 0.3, max x = 0.5 Alternative Taking Intersection 0.6 0 0.6 0.3 0 0.3 0.3 0.5 0.5 0 0.5 xx x x x xx −     −    ⎯⎯⎯⎯⎯⎯ →   −     Hence, min x = 0.