HCI 2024 Term 3 Class Test Revision Paper 1 Solutions

Term 3 A.Math Class Test (Revision Paper 1) Solution Answer all the questions. Question 1: 1 (a) Given that ( )3log b x y p = and 2log b y q x  = , express log b xy in terms of p and q. [6] (b) Solve the equation 2log 8 2 log 7x x+= . [4] Solution: 1 (a) ( )3log b x y p = 3log log bb x y p+= 3 log log bb x y p+= ---------- (1) 2log b y q x  = 2log log bb y x q−= log 2 log bb y x q−= ---------- (2) (1) – (2), 5 log b x p q=− log 5b pqx −= ---------- (3) Substitute (3) into (2). log 2 5b pqyq −−=  log 2 5b pqyq −=+  22log 55b y q p q= + − 23log 55b y p q=+ log b xy ( ) 1 2 log b xy= ( )1 log log 2 bb xy=+

1 2 3 2 5 5 5 pq pq−= + + 1 2 3 2 5 5 5 pq pq−= + + 1 3 2 25 pq+=  ( )1 3 210 pq=+ 1(b) 2log 8 2 log 7x x+= 3 2log 2 2 log 7x x+= 2 2 3 2 log 7log xx += Let 2 log yx= . 3 2 7yy += 23 2 7yy+= 22 7 3 0yy − + = ( )( )2 1 3 0yy− − = 2 1 0y −= or 3 0y −= 1 2y = 3y = When 1 2y = , 2 1log 2x = 1 2 2x = 2x = 1.414x = 1.41x  When 3y = , 2log 3x = 3 2x = 8x =

 2x = [Accept 1.41x = ] or 8x = Question 2: 2 (i) Use the substitution 2 xy = to express the equation ( )3 12 7 2 4 4x x x+ + = − as a cubic equation in y. [1] (ii) Show that 1 2y = is the only real solution of this equation. [4] (iii) Hence solve the equation ( )3 12 7 2 4 4x x x+ + = − . [1] Solution: 2(i) ( )3 12 7 2 4 4x x x+ + = − ( ) ( )322 2 7 2 4 2x x x+ = − ( ) ( )322 2 2 7 2 4 0x x x+ + − = ( ) ( ) ( ) 32 2 2 2 7 2 4 0x x x + + − = 322 7 4 0y y y+ + − = 2(ii) Let ( ) 32f 2 7 4y y y y= + + − . 32 1 1 1 1f 2 7 42 2 2 2        = + + −               0= ( ) 2 1y− is a factor of ( )f y .

2 32 32 2 2 4 2 1 2 7 4 2 2 7 4 2 8 4 8 4 0 yy y y y y yy yy yy y y ++ − + + − − +− − − − ( )f 0y = 322 7 4 0y y y+ + − = ( )( )22 1 4 0y y y− + + = 2 1 0y −= or 2 4 0yy + + = 2 1y = ( )( ) ( ) 21 1 4 1 4 21y −  −= 1 2y = 1 15 2 −  −= (no real solution) Hence 1 2y = is the only real solution of this equation. 2(iii) 2 xy = 12 2 x = 12 2x −= 1x = −

Question 3: 3 The two variables x and y are related by the equation 2 5 1y xy−= . The diagram shows part of a straight line graph obtained by plotting 1 y against x. Find (i) the coordinates of P and Q, [3] (ii) the value of  in degrees. [2] Solution: 3(i) 2 5 1 (2 5 ) 1 1 25 1When 0, 2 (0,2) 2When 0, 5 2( ,0)5 y xy yx xy x y P yx Q −= −= =− ==  == 

3(ii) tan 5 78.7 180 78.7 101.3 (to 1 d.p.)    =− = = −  = Question 4: The expression 32ax x bx++ has remainder 3 when it is divided by 1x+ . When the expression is divided by 2x− , the remainder is 6. Find the remainder when the expression is divided by 3x− . [4] 4 Let 32ax x bx++ be f(x) f(x) = 32ax x bx++ f( 1) 3−= 2 (1)ab− − = −−− f(2) 6 8 4 2 6 8 2 2 4 1 (2) ab ab ab = + + = += + = −−− (1) + (2) : 33 1 3 a a b = = =− 2f ( ) ( 3)x x x x = + − f (3) 27= Therefore, remainder = 27 Question 5: 5 (i) Factorise the expression 3x3 – 8x2 – 5x + 6 completely. [3] (ii) Hence, sketch the graph of 323 – 8 – 5 6y x x x=+ , indicate clearly the coordinates of the x-intercepts and y- intercepts. [3]

Solution: Let f(x) = 3x3 − 8x2 − 5x + 6. f(−1) = 3(−1)3 − 8(−1)2 − 5(−1) + 6 = −3 − 8 + 5 + 6 = 0 (x + 1) is a factor of f(x) 3x3 − 8x2 − 5x + 6 = (x + 1)(ax2 + bx + c) Equating coefficients of x3: a = 3 Equating coefficients of x2: −8 = a + b b = −11 Equating constant term: 6 = c 3x3 − 8x2 − 5x + 6 = (x + 1)(3x2 − 11x + 6) = (x + 1)(3x − 2)(x − 3) (ii)

Question 6: The figure shows part of a straight line graph obtained by plotting values of the variables indicated. Express y as a function of x. [3] Solution: ( ) 2 62 31 2 ln 2 int : 1, 2 2 2( 1) 0 ln 2 x gradient gradient y x C Po C C yx ye − −= −+ =− =− + − =− − + = =− = Question 7: Using long division, find the quotient and remainder when 5 4 3 2 – 3 2 – 2 3 1x x x x x+ + + is divided by 223x + . [3] Solution: (–3, 6) (–1, 2)