HCI 2024 Term 3 Class Test Revision Paper 2 Solution

Term 3 A.Math Class Test (Revision Paper 2) Solution Answer all the questions. Question 1: It is given that h(x) = (3x + 1)(x + 5)(2x – 1). (a) Express h(x) in the form Ax3 + Bx2 + Cx + D, giving the values of the constants A, B, C and D. [2] (b) Find the value of the constant a, given that (x + 3) is a factor of h(x) + ax. [2] Solution: (a) Expanding h(x), (3x2 + 16x + 5)(2x – 1) = 6x3 + 29x2 – 6x – 5  A = 6, B = 29, C = –6 and D = –5 (b) h(x) + ax = 6x3 + 29x2 – 6x – 5 + ax Let g(x) = 6x3 + 29x2 – 6x – 5 + ax. Since x + 3 is a factor of g(x), by Factor Theorem, g(–3) = 0 6(–3)3 + 29(–3)2 – 6(–3) – 5 + (–3)a = 0 a = 37 1 3 Question 2: Given that P(x) = 3x3 – 7x2 – x – 55 and Q(x) = 6x3 – 5x2 – 3x + 2, find the value(s) of q if P(q) = Q(q). [4] Solution: P(q) = 3q3 – 7q2 – q – 55 and Q(q) = 6q3 – 5q2 – 3q + 2. P(q) = Q(q) 3q3 – 7q2 – q – 55 = 6q3 – 5q2 – 3q + 2 3q3 + 2q2 – 2q + 57 = 0 ⎯⎯ → to be solved Let f(q) = 3q3 + 2q2 – 2q + 57 f(–3) = 3(–3)3 + 2(–3)2 – 2(–3) + 57 = 0

By Factor Theorem, q + 3 is a factor of 3q3 + 2q2 – 2q + 57. So, 3q3 + 2q2 – 2q + 57 = (q + 3)(aq2 + bq + c). Equating coefficient of q3: a = 3 Equating coefficient of q2: 3a + b = 2 3(3) + b = 2 b = –7 Equating constant term: 3c = 57 c = 19 3q3 + 2q2 – 2q + 57 = 0 (q + 3)(3q2 – 7q + 19) = 0 q = –3 and 3q2 – 7q + 19 = 0 Discriminant, b2 – 4ac = (–7)2 – 4(3)(19) < 0  3q2 – 7q + 19 has no solutions. Question 3: The figure shows part of a straight line graph obtained by plotting 1y x + against x2. Express y in terms of x. [3] Solution:

Gradient of line 31 23m −= − = –2 Equation of the line  Y = –2X + c (2, 3) lies on the line  3 = –2(2) + c c = 7 Hence, the equation of the line is 1y x + = –2x2 + 7 y = –2x3 + 7x – 1. Question 4: In the figure, x and y are related by 1 2 32e x x y =− . Find (a) the value of , [3] (b) the coordinates of P and Q. [2] Solution: a) 1 2 3e2 x x y =− e 2 3x y x y=− Y = mX + c Gradient = tan 

= 2  = tan–1 2 = 1.11 (b) Equation of line  Y = 2X – 3 P lies on y-axis  coordinates of P are (0, Y) Y = –3 Thus, the coordinates of P are (0, –3). Q lies on x-axis  coordinates of Q are (X, 0) X = 1.5 Thus, the coordinates of Q are (1.5, 0). Question 5: Given that x and y are known to be related by the equation 3 xy st= where s and t are constants. Express the equation in a form suitable for drawing a straight line graph. [3] Question 6: Solve log3 (x + 6) = 2 – log3 (x – 4) + log9 x2, giving your answers in exact values. [5] Solution: log3 (x + 6) = log3 32 – log3 (x – 4) + 2 3 3 log log 9 x log3 (x + 6) = log3 32 – log3 (x – 4) + 3 3 2log 2log 3 x log3 (x + 6) = log3 32 – log3 (x – 4) + log3 x log3 (x + 6) = log3 23 ( ) 4 x x− x + 6 = 9 4 x x− (x + 6)(x – 4) = 9x x2 + 2x – 24 = 9x x2 – 7x – 24 = 0 x = 7 49 4(24) 2 + x = 7 145 2 + or 7 145 2x −=

Question 7 : Given that (log3 7)(log7 k) = 2, find the value of k. [2] Solution : ( ) 3 3 3 loglog 7 2 log 7 k =  3 2 log 2 3 9 k k = = = Thus, k = 9. Question 8: (a) (i) Solve 323 7 7 3 0x x x− + + − = , showing your working clearly. [4] (ii) Sketch the graph of 323 7 7 3y x x x=− + + − . Indicate clearly, the coordinates of the x- intercept(s) and y-intercept. [3] (b) When a polynomial ()Px is divided by ( 1)x− and ( 2)x− , the remainders are 2 and 3 respectively. Find the remainder when ()Px is divided by ( 1)( 2)xx−− . [4] Solution: (a) (i)

32 32 2 2 2 3 7 7 3 0 Let ( ) 3 7 7 3 ( 1) 3 7 7 3 ( 1) 0 ( 1) is a factor of ( ). ( ) ( 1)( 3 3) By comparing the coefficient of : 37 10 ( ) ( 1)( 3 10 3) ( ) ( 1)( 3 1)( 3) ( ) 0 ( x x x f x x x x f f x f x f x x x kx x k k f x x x x f x x x x fx x − + + − = =− + + − − = + − − −= + = + − + − −= = = + − + − = + − + − = 1)( 3 1)( 3) 0 11 or or 3 3 xx x x x + − + − = =− =− = (b) (ii) (b)

Remainder ( ) ( ) ( ) ( ) ( ) ( ) ( 1)( 2) (1) 2 (2) 3 2 (1) 2 3 (2) (2) (1) : 1 1 Re 1 ax b P x Q x D x R x P x Q x x x ax b P P ab ab a b mainder x =+ =  + =  − − + + = = + = −−− + = −−− − = = =+