Special Maths Exam Paper 1 November 2014 Solutions

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1 Special Maths Exam Paper 1 November 2014 Solutions Question One 1.1 L : y = x 4. Therefore the slope of L is. (a) Since L 1 L, it follows that the slope of L 1 equals. Therefore the equation of L 1 is y + 1 = (x + 2) = x + 6, giving y = x + 5. (b) Since L 2 L, it follows that the slope of L 1 equals 1. Therefore the equation of L 2 is y = 1 (x +2) = 1x 2/, giving y = 1x +7/. Graph: y 4 L L L x (8) { x 2 if x 2 0 i.e. x 2/ 1.2 y = x 2 = x + 2 if x 2 < 0 i.e. x < 2/ 1

2 { x + 2 if x 2/ Thus y = x 2 if x < 2/ The line x = 2/ is the axis of symmetry. y x = 2/ (2, 4) x (4) 1. See graph below: y 4 2 (1, 1) x x = 1 5 By symmetry, x = is another x-intercept. The general eqn of this parabola is y = ax 2 + bx + c. From this, we construct eqns using the intercepts and the turning point (1, 1). The axis of symmetry is given by x = 1. We solve simultaneously 0 = a b + c and 0 = 9a + b + c. Subtracting, we have 4b = 8a, thus b = 2a. Using the turning point we have 1 = a + b + c. Subtracting the first eqn from this one, we have 1 = 2b, implying b = 1/2. Substitute into 2

3 b = 2a. Then 1/2 = 2a, implying a = 1/4. Substitute into the first eqn: thus 0 = 1/4 1/2 + c. Therefore c = /4. Hence the eqn of the parabola is y = 1 4 x x + 4. (7) 1.4 (i) f(x) = y = 1 + 2, thus dom(f) = {x : } = {x : x x x+2 0}. Case x > 0: Then x + 2 0, implying x 2. So for this x case the solution is 0 < x 2. Case x < 0: Then x > 0 implies x < 0 and x + 2 < 0, implying x > 2 > 0, a contradiction! Thus this case does not arise. Therefore dom(f) = {x : 0 < x 2} = (0, 2]. (ii) The range is R(f) = {f(x) : 0 < x 2} = (, 0]. (4) [2] Question Two 2.1 (a) f(x) = 2x 2 + 7x 5. Thus f 1 () = {x : f(x) = } = {x : 2x 2 + 7x 5 = } = {x : 2x 2 + 7x 15 = 0} = {x : (2x )(x + 5) = 0} = { 5; /2}. (2) (b) f(x) = 2x 2 +7x 5. Thus f( 2) = 2(4)+7( 2) 5 = 11, implying f(f( 2)) = f( 11) = 2(121) 77 5 = 160. (2) 2.2 x a factor of f(x) = x + 7ax 6 implies Remainder = f() = 0 when f(x) is divided by x. So 0 = f() = a 6 = a. Therefore a = 1. So f(x) = x 7x 6. Dividing f(x) by x, we get x 2 + x +2. Thus f(x) = (x )(x 2 + x + 2) = (x )(x + 1)(x + 2). (5) 2. x +4 x+2 = x (1+4 2 ) = 7 = 7 = 111. () 5 x+1 2 x 1 x (5 2 1 ) 15 2/ (a) (0, 0016) x = 0, 008 implies (16/000) x = 8/00. Therefore ( 2 ( 2 ), implying 4x =. Hence x = /4. )4x =

4 (b) ( x ) 2 28( x ) + 9 = 0. Let y = x, then y 2 28y + 9 = 0 = (y 1)(y 9). Therefore y = 1/ or y = 9. Thus x = 1 or x = 2. Hence x = 1 or x = 2. (5) 2.5 (a) Consider (8/00) 1/ = 5 x, then (1/125) 1/ = 5 x, i.e. 5 / = 5 x. Thus x = 1 (2) (b) log 1/ 24 = log (1/24) = log ( 5 ) = 5. (2) (5+x) (5+x) 2.6 log ) = 1, implying = (x 1) (x 1) 1 =. Thus 5 + x = x, implying 9x = 15. Therefore x = 5. () Question Three 1.1 log 1/ (2x + ) = log ( ) < 8 = log 2x+ 8 1, implying < 2x+ 8 = Therefore 2x + > 1/6561, implying 2x > + 1/6561, whence x > () [24].2 y = log 2 ( x). Thus (i) the asymptote is the line x =. (ii) When y = 0 then x = 1, implying the x-intercept is (2, 0). (iii) y = log 2 ( x) iff x = 2 y, i.e. x = 2 y. A table of points may be useful. x , 5 2, y Thus the graph is as follows: y x x =

5 . Let x 1/2 = y, then x = y 2. So 2y 2 + 7y 15 = 0 = (2y )(y + 5), implying y = /2 or y = 5. Since x 1/2 is defined only for x 0, it follows that y 0. So the correct solution is y = /2, thus x = y 2 = 9/4. (4) (4).4 Number line: A B Using the number line, we have (a) A B = (4, 6); (b) A B = (0, 9]; (c) B A = [6, 9]. (4).5 T 4 = a + d = 1 and T 7 = a + 6d = 1. Solving the two equations simultaneously, we have d = 12. Thus d = 4. Therefore a + ( 4) = 1, whence a = 11. Sum is S n = n 2 [2a+(n 1)d] = 24 implies S n = n 2 [2(11)+(n 1)( 4)] = n[11 + (n 1)( 2)] = 24. Thus n[ n] = 24, implying 2n 2 + 1n + 24 = 0. So 2n 2 1n 24 = 0 = (2n + )(n 8), whence n = 8. So the series must have 8 terms. (6).6 The constant ratio is r = x 1 = x, thus (x x+1 x 1 1)2 = x(x + 1). Therefore x 2 2x + 1 = x 2 + x, implying x = 1, whence x = 1/ and r = (1/)/( 2/) = 1/2. T n = ar n 1 and a = x + 1 = 4/ implies T 20 = (4/)( 1/2) 19 = (1/)(1/2) 19. (4) Question Four 4.1 A geometric series is of form n=0 arn. Thus the sum of the first n terms of the series is S n = a+ar+ar 2 + +ar n 1, implying S n r = ar+ar [25]

6 +ar n 1 +ar n. Therefore S n S n r = a ar n, i.e. S n () = a( n ), whence S n = a(n ). Now S n = a(n ) becomes a(1 0) for r < 1 as n becomes large since r n a becomes 0. For r = 1, is undefined. For r > 1, S n = a(n ) becomes arbitrarily large as n gets larger and larger, thus S does not exist. Thus we conclude that S = a is true only for r < 1, i.e. the geometric series n=0 arn converges to a for r < 1, otherwise it diverges (for a 0). (4) 4.2 T 2 = ar = 6 and T 4 = ar = 54 implies arr 2 = 54. Thus 6r 2 = 54, implying r 2 = 9, whence r = ±. Hence ar = 6 implies a = 6/r = ±2. Now S n = a(n ) implies S 25 = a(25 ). Case r = : Then a = 2. Therefore S 25 = 2(1 ( )25 ) = (25 1) Case r = : Then a = 2. Therefore S 25 = 2(1 25 ) = 2(1 25 ) = (5) = 0.1 = = 1/ + / 2 + / + / 4 +, a geometric series from the 2nd term onward with a a = /0 and r = 1/. The sum of the series is = /0 = /0 = 1 1/ 9/ 1/0. Hence 0.1 = 1/ + 1/0 = 4/0 = 2/15. (4) 4.4 Amount after n years is A n = P(1 + r 0 )n where r = rate and P is the principal amount initially invested. Thus log(a n /P) = nlog(1 + r ), 0 implying n = log(an/p) = log(4450 2/4450) = half years = 6.47 log(1+ r 0 ) log( years. ) (4) 4.5 Value after n years is V n = P(1 r 0 )n where r = rate and P is the original value. Thus log(v n /P) = nlog(1 r log(vn/p) ), implying n = = 0 log(1 r 0 ) log( 1 4 P/P) log(1 9 0 ) = years. (4) [21] 6

7 Question Five 5.1 Let g = green; b = blue and w = white. Then P(g) = 1/, P(b) = 1/5 and P(w) = 1 1/ 1/5 = 7/15. (a) P(w or b) = P(w) + P(b) = 7/15 + 1/5 = /15 = 2/. (b) P(w or g) = P(w) + P(g) = 7/15 + 1/ = 12/15 = 4/5. (4) 5.2 Let = not; Tue = Tuesday, Wed = Wednesday, Thu = Thursday. 7 Tue Wed Thu 9 Tue Wed Thu P( before Friday) = P(Tue or Wed or Thu) = P(Tue) + P(Wed) + P(Thu) = = 994. (4) 00 sum of items 5. (a) Mean = = 97 = 46,. number of items 21 (b) Median = (n+1) -th item = (21+1) -th item = 11th item = (c) Mode = 50 (d) Q 1 = (n+1) -th item = (21+1) -th item = 5,5th item = 0+ = 1, 5, and Q = (n+1) -th item = (21+1) -th item = 16,5th item = 59+6 = Therefore IQR = Q Q 1 = = n (e) Standard Deviation is given by S D = n i=1 (a i x) 2 where x = mean of the data a 1, a 2,, a n. Now x = 46, implies n i=1 (a i x) 2 = (5 46.) 2 +(12 46.) 2 +(19 46.) 2 +2(0 46.) 2 + ( 46.) 2 + (5 46.) 2 + (9 46.) 2 +(41 46.) 2 +(49 46.) 2 +4(50 46.) 2 +(54 46.) 2 +(59 46.) 2 +2(6 46.) 2 +(70 46.) 2 +(81 46.) 2 +(90 46.) 2 = 906. Therefore variance = 906 = 41, 57, implying Standard deviation is S 21 D = = () 7

8 5.4 Consider the Venn Diagram below: P = pop; K = Kwaito, R = Rap. x + y = 6; x + z = 16; x + y + z + 16 = 6, implying x + y + z = 20. Thus 6 + z = 20, so z = 14. Therefore x + 14 = 16, implying x = 2, thus y = 6 2 = 4. Now 0 = 8 + x + z + a = a, implying a = 6. Therefore the number who like Kwaito only is 41 (a + x + y) = 41 (6 + 6) = 29. (a) (i) Number who like Pop and Kwaito only is a = 6, (ii) Number who like Kwaito only is 29, (iii) Number who like all three is x = P 8 a K z x 29 y R 16 (b) (i) P(Rap only) = 16/79 (ii) P(Pop + Kwaito) = 8/79 (iii) P(Kwaito + Rap - Pop) = P(Kwaito + Rap only) = y/79 = 4/79. (6) ********************END******************** [24] 8