2025 A Levels H3 Math Paper 1 Soln updated

2025 GCE ‘A’ Level H3 Mathematics (9820/01) Solutions 1 (a) Solve the equation 6 4 23 13 13 3 0x x x− + − = . Give your answers in exact form. [3] (b) The points ( ),xy which satisfy 6 4 2 2 4 63 13 13 3 0x x y x y y− + − = lie on six straight lines through the origin. Find the equations of these lines. [3] (c) Define regions A and B in the xy-plane as follows: ( )  ( )  6 4 2 2 4 6 6 4 2 2 4 6 , : 3, 1 and 3 13 13 3 0 , , : 3, 1 and 3 13 13 3 0 . A x y x y x x y x y y B x y x y x x y x y y =   − + −  =   − + −  Draw a diagram to illustrate these two regions. [3] (d) Calculate the ratio Area of Area of A B , giving your answer to 3 decimal places. [3] Solution: 1 (a) Since the polynomial 6 4 23 13 13 3x x x− + − is an even function in x, ( )( )( ) 6 4 2 2 2 23 13 13 3 3x x x x a x b x c− + − = − − − . By inspection, 1x= are roots. Without loss of generality, assume 1a= . ( )( )( ) ( ) ( ) ( ) ( ) 6 4 2 2 2 2 2 4 2 6 4 2 3 13 13 3 3 1 3 1 3 1 x x x x x b x c x x b c x bc x b c x bc b c x bc − + − = − − − = − − + + = − + + + + + − Equating corresponding terms in the above identity, ( ) 103 1 13 and 3 3 1 3b c b c bc bc− + + =−  + = − =−  = Substitute 1c b= into 10 3bc+= gives 31 10 3 10 3 03b b bb+ =  − + = ( )( )3 1 3 0 11 or 3 3 or respectively33 bb bc  − − =  =  = Hence ( )( ) 6 4 2 2 2 2 13 13 13 3 3 1 3 3x x x x x x − + − = − − −  . The roots are therefore 11, 3, 3    .

2 (b) The point ( )0,0 satisfies 6 4 2 2 4 63 13 13 3 0x x y x y y− + − = . ------ (*) [Note: 00xy=  = ] Consider the points ( ), where , 0x y x y  which satisfy (*). Since 0x , dividing above equation by 6x− gives 6 4 2 By (a) 13 13 13 3 0 1, 3, 3 1 , 3 , 3 y y y y x x x x y x y x y x      − + − =  =              = = = (c) y 1 R B A B B A B A A A A A A A A - 3 A A O A A 3 x A A B A B B A B -1 Let R denote the rectangle  3, 3 1,1−  − . Blue: yx= ; Green: 3yx= ; Red: 1 3 yx= . 6 4 2 6 4 2 2 4 6 6 2 2 2 6 3 13 13 3 3 13 13 3 1 3 1 3 3 y y yx x y x y y x x x x y y yx x x x      − + − =− − + −                    =− − − −                    2 1 0 1yy yxxx −      , 2 3 0 3 3yy yxxx −      , 2 1 1 103 33yy yxxx −      . (d) By symmetry, we need only consider the first quadrant. 1 1 131Area of Area of 2 3 22 3 Area of Area of Area of 1 1 1 433 3 122 3 AA B R A +−  += = = −  −− + −  .

3 2025 GCE ‘A’ Level H3 Mathematics 9820/01 2 (a) Use the substitution y ux= , where u is a function of x, and 0x , to solve the differential equation 2 2 2d d yx x xy yx = + + subject to the condition that 0 when 1yx== . [5] (b) Solve the differential equation 4 4 3 2d 2dyx x x y yx = + + subject to the condition that 0 when 1yx== . [5] Solution: 2 (a) dd dd yuy ux u x xx=  = + . Substitute into DE gives ( ) ( ) 22 2 2 dd 1dduux u x x x ux ux x uxx + = + +  = + 2 01 11 d d 1 ux ux xu= + ( 1, 0 0x y u= =  = )   ( ) ( ) 1 10 1 tan ln tan ln tan ln tan ln u x ux ux ux y x x − − = = = = 22 dd 2ddyuy ux ux x xx=  = + . Substitute into DE gives 4 2 4 5 2 4 2 2 dd2 2 1 dduux ux x x ux u x x uxx + = + +  = + 22 01 11 d d 1 ux ux ux = + ( 1, 0 0x y u= =  = ) 22 01 1 2 11 d d 1 1tan 1 1tan 1tan ux ux ux u x xu x xyx x − = +  = − −=  −=  

4 3 Throughout this question, 0 and 0xy . (a) By considering the three terms 22 2,, 22yyx , use the AM -GM inequality to prove that if 22 1xy+= , then 2 2 33 xy  . State the values of x and y for which equality is achieved. [3] (b) Use the AM-GM inequality to prove that if 22 1xy+= , then 23 63 25 5 xy  . State the values of x and y for which equality is achieved. [4] (c) Let m and n be fixed positive integers. Given that 22 1xy+= , find an expression in terms of m and n for the maximum possible value of mnxy . State, in terms of m and n, the values of x and y for which this maximum is achieved. [5] Solution: 3 (a) By AM-GM inequality, 1 2 2 2 2 322 1 2 2 3 2 2 y y y yxx      + +        ( ) 22 1 24 13 22 24 2 11 4 3 3 4 27 2 since , 0 33 xyxy xy xy xy x y +=  + =      Equality is achieved when 222 1 2 2 2 2 2 122 1 22 33xyyx y x x x x y +==  =  − =  =  = . (b) Apply AM-GM inequality to the 5 numbers 2 2 2 2 2 , , , ,2 2 3 3 3x x y y y , we have ( ) 1 2 2 2 2 2 2 2 2 2 2 5 22 46 2 3 5 1 1 1 2 2 3 3 3 5 2 2 3 3 3 5 5 1 2 3 5 x x y y y x x y y y xy xy         + + + + = + =         

5 2025 GCE ‘A’ Level H3 Mathematics 9820/01 23 23 5 2 2 3 2 3 3 6 3 5 5 5 25 5 xy   = = . Equality is achieved when 2222 1 2 2 2 2 233 2 3 2 22 3 5 5 xyxy x y x x x y +==  =  = −  =  = . (c) Applying AM-GM inequality to the mn+ numbers 2 2 2 2 identical terms identical terms , , , , , mn x x y y m m n n   , ( ) ( ) 22 1 2 2 2 2 2 2 identical terms identical terms 1 22 11 1 mn mn mn xy mn mn mn mn mn x y x x y y m n m n m m n n m n xy mn mn mnxy mn + = + = + +       + + + + + =    ++             +     +    Maximum possible value of mnxy is ( ) mn mn mn mn + + . This maximum is achieved when 2222 1 2 2 2 2 xyx y m n nx my nx m mx x ym n m n m n +==  =  = −  =  = ++ . 4 Throughout this question, whenever a ball is selected at random from a bag, all of the balls in the bag are equally likely to be selected. (a) A bag contains 100 balls, of which w are white and b are black (so 100wb+= ). One ball is selected at random from the bag and is not replaced. Then a second ball is selected. (i) Find, in terms of w, the probability that the selected balls are the same colour. [3] (ii) Find the value of w for which the probability in part (a)(i) is minimised and state this minimum probability. [3] (b) Starting with the bag containing 45 white balls and 55 black balls, the following process is repeated.

6 While the bag contains at least two balls, two balls are selected at random. If they are both white, then they are discarded, and a further black ball is added to the bag. If one or both of the balls are black, then a black ball is discarded and the other