Special Maths Special Exam Paper September 20 Solutions Question One 1.1 Number line: A 2 1 0 1 2 7 8 9 B Using the number line, we have (a) A B = (, ); (b) A B = [ 1, 9]; (c) B A = [, 9]; (d) AΔB =(A B) (B A) =[ 1, ] [, 9]. () 1.2 dom(f) ={x : x 2 0} = {x : x 2 0} =[, ]. The range is R(f) = {f(x) : x dom(f)} = { x 2 : x dom(f)} =[, 0]. () 1. (a) f(x) =x 2 +x 2. Thus f 1 () = {x : f(x) =} = {x : x 2 +x 12 = 0} = {x :(x )(x +)=0} = { ; /}. () (b) Since f(x) =x 2 +x 2, the f( 2) = 12 2 = 0, implying f(f( 2)) = f(0) = 2. (2) 1. x afactoroff(x) =x +ax+27 implies f() = 0. Thus 27+9a+27 = 0, implying 9a =. Thus a =. Therefore f(x) =x x + 27. Dividing f(x) byx, we have f(x) =(x )(x 2 +x 9). () 1. First we graph the line 2x y + 8 = 0. In std form the line is given by y = 1 x + 2, then the required region lies above the line since we have 2 1
y 1 x + 2. The required region is shaded. 2 y 2 1 2 1 1 2 1 2 x () [2] Question Two 2.1 < x <, implying < x <9. Thus / <x<9/. () 2.2 L : y =x + 2. 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 =(x +2)=x +, giving y =x +9. (b) Since L 2 L, it follows that the slope of L 1 equals 1. Therefore the equation of L 2 is y +2= 1(x 1) = 1x +1/, giving y = 1x /. Graph: 2
y 9 L 1 2 1 2 1 1 2 1 L 2 2 L 7 x (8) { { 2x if 2x 0 2x if x /2 2. y = 2x + if 2x < 0 = 2x + if x < /2 So the graph is as follows: y x =/2 2 1 2 1 1 2 x 1 2 ()
2. y = x 2 + x 12 = (x +)(x ). Thus the x-intercepts of the graph are (, 0) and (, 0). The axis of symmetry is x = b/2a = 1/2. Therefore the turning point is ( 1/2, f( 1/2)) = ( 1/2, 9/ 12.). The y-intercept is (0, 12). The graph is concave up since the coefficient of x 2 is positive. See graph below: x = 1/2 y 2 1 2 1 1 2 1 x 2 () 12 ( 1/2, 12.)
2. x+2 x+2 1 implies ( 2x 2x )2 1. Therefore x 2 +x + x 2 12x +9, implying 0 x 2 1x +=(x 1)(x ). Using a sign table or the parabola method, we obtain 1/ x. 1/ x 1 0 +++++ + +++++ x - 0 +++++ (x 1)(x ) +++++ 0 0 +++++ We want all the real numbers x such that (x 1)(x ) has a negative sign. Thus the solution set is [1/, ]. () [2] Question Three.1 Given x+2 1: Suppose 2x > 0. Then x>/2 andx +2 2x, 2x i.e. x. Suppose 2x < 0. Then x</2 andx +2 2x. So x</2 and x, which is impossible. Hence this case does not arise, implying the solution is /2 <x and so the solution set is the interval (/2, ]. ().2 x+2 x 2 = 2 implies x (2 1/2) = 2, so that x = 2 2/2 = 2. Therefore x = 2, implying x =1/2. (2). x+1 x+2 2 x 1 x = x ( 2 ) x (2 1/ 1) = 1 0 (2/ 1/) = 2 1. (). (a) log 2 (x 2) + log 2 (x ) = 1 implies log 2 (x 2)(x ) = 1. Therefore (x 2)(x ) = 2, implying x 2 x +=0=(x 1)(x ). Therefore the only solution is x =, since 1 gives an undefined log. () (b) log x 2 = implies 2 = x. Thus =(1/x). This implies 1/x =, Hence x =1/. (2). 2x x =0. Lety = x,thenx = y 2. Therefore 2y 2 y = 0=(y + 1)(2y ). Thus y = 1 ory =/2. However, y 0, hence the only correct solution is y =/2, implying x = y 2 =9/. ()
. The seq 1; 11; 7; ; is arithmetic with common difference d = 7 +11 = = 11 + 1 and a = T 1 = 1. The nth term is T n = a +(n 1)d = implies 1 + (n 1)() =. Therefore n 1 = [ + 1]/ =80/ = 20, whence n =21isthenumberofthe term whose value is. () [21] Question Four.1 Graph of y = 2 x+ Soln. Asymptotes are the lines x = andy = 0. Consider the following table of points on the curve: x - - -2-1 0 1 2 As x approaches, y -1-2 2 1 2/ 1/2 2/ 1/ 2/7 y approaches ±. Asx gets larger in the negative and positive directions, y approaches 0. Thus the graph is as shown below: y 2/ 2 1 0 1 2 x x = ()
.2 (a) An arithmetic series is of form n=1 [a+(n 1)d]. Thus the sum of the first n terms of the series is S n = a+(a+d)+(a+2d)+ +(a+(n 1)d). Reversing the order of the terms, we have S n =(a +(n 1)d) +(a + (n 2)d) + +(a +2d) +(a + d) +a. Adding the two equations, we have 2S n =[2a +(n 1)d] +[2a +(n 1)d] + +[2a +(n 1)d] = n[2a +(n 1)d]. Thus S n = n [2a +(n 1)d]. We may further write 2 S n = n[a + a +(n 1)d] = n[t 2 2 1 + T 2 ], where T i is the i-th term of the series. (2) (b) S 12 = 12 [2a +11d] = /2, implying 12a +d = /2 (1) 2 Now T + T =7. implies a +2d + a +d =7., thus 2a +d =1/2. Multiplying by, we have 12a +d = (2). Therefore (1) - (2) yields 0d =1/2, implying d =1/ =0, 2. Substitute into (2), then a =. Hence T 2 = a +2d =+2 (1/) = 9. (). T = a +d = andt 8 = a +7d = 17. Solving the two equations simultaneously, we have d = 12. Thus d =. Therefore a+( ) =, whence a = 11. Sum is S n = n 2 [2a+(n 1)d] = 19 implies S n = n 2 [2(11)+(n 1)( )] = n[11 + (n 1)( 2)] = 19. Thus n[11 + 2 2n] = 19, implying 2n 2 +1n + 19 = 0. So 2n 2 1n 19 = 0 = (2n + 1)(n 1), whence n = 1. So the series must have 1 terms. (). The constant ratio is r = x 2 = x+2,thus(x x x 2 2)2 = x(x +2)=x 2 +2x. Therefore x 2 x+ = x 2 +2x, whence x =2/ andr =(2/ 2)/(2/) = 2. T n = ar n 1 and a = x =2/ implies T =(2/)( 2) 17 = 2 /. () [2] Question Five.1 The seq ;;9/; has constant ratio r = / anda =. T n = ar n 1 < 1/99 implies (/) n 1 < 1/99. Therefore (/) n 1 < 1/9, implying (/) n 1 > 9. Thus (n 1) log (/) > log 9, implying log 9 n>1+ =1.11. Hence n = 1, implying T log(/) 1 =(/) 1 is the required term. () 7
.2 0.1 =0.1 =0.1+0.0 + 0.00 + 0.000 + =1/ + / 2 + / +/ +, a geometric series from the 2nd term onward with a a =/0 and r =1/. The sum of the series is = /0 = /0 = 1 r 1 1/ 9/ 1/0. Hence 0.1 = 1/ + 1/0 = /0 = 2/1. (). Let r = red; g = green; y = yellow. (a) r g r y r g g 8 8 y y r 8 8 g y (i) P(rr or gg or yy) = P(rr) + P(gg) + P(yy) = P(r)P(r) + P(g)P(g) + P(y)P(y) = ( )2 +( )2 +( 8 )2 = 1++ = 11 =29/81 2 2 (ii) P(rg or ry or gr or yr or yg or gy) = 2P(rg) + 2P(ry) + 2P(gy)= 2P(r)P(g) + 2P(r)P(y) + 2P(g)P(y) = 2 +2 8 +2 8 = 2[ 2+2+8 ]= =2/81 2 2 (iii) P(ry or yr or yy or rr) = 2P(ry) + P(yy) + P(rr) = 2( )( 8 )+ ( 8 )( 8 )+( )( )=2++1 = 112 = 28 2 2 81 8
(b) For the 2nd pick, the total number of balls is 17. 17 r g 17 r 8 y 17 r 17 g g 17 8 8 y 17 y 17 r 7 17 17 g y. Amount after n years is A n = P (1 + r 0 )n where r =rateandp is the principal amount initially invested. Thus log(a n /P )=nlog(1 + r ), 0 implying n = log(an/p ) = log(70 2/70) = log(2) =11.891 half years log(1+ r 0 ) log(1+ 12 200 ) log( 212 ) 200 =.978 years years. (). Value after n years is V n = P (1 r 0 )n where r =rateandp is the original value. Thus log(v n /P )=nlog(1 r log(vn/p ) ), implying n = = 0 log(1 r 0 ) log( 1 P/P) log(1 11 = log( 1 ) 0 ) log( 89 0 =1.89088 years 1 years. () ) [2] (8) 9
Question Six.1 Let R = rainy; F = fine. R R 7 F R R 7 F F 7 R F R F F R 7 F (a) P(Sat R and Sun F) = P(R)P(F) = 7 = 21. 0 (b) P(Sat F and Sun F) = P(F)P(F) = = 1 =/20. 0 (c) P(Mon R) = P(FFR or FRR or RFR or RRR) = 7 + 7 7 7 = 7+++ = 28 = 17 00 00 + 7 + 20. (8)
.2 Let = not; Tue = Tuesday, Wed = Wednesday, Thu = Thursday. Tue Wed Thu 8 Tue Wed Thu 7 2 P( before Friday) = P(Tue or Wed or Thu) = P(Tue) + P(Wed) + P(Thu) = + 7 8 =0.97. () + = 00+280+9 = 97 = 122 00 00 12 sum of items. (a) Mean = = 82 =.78. number of items (b) Median = (n+1) -th item = (+1) -th item = 9.th item = 0. 2 2 (c) Mode = 0 (d) Q 1 = (n+1) -th item = (+1) -th item =,7th item = 2+2 =28., 2 and Q = (n+1) -th item = (+1) -th item = 1,2th item = 0+2 = 1. 2 Therefore IQR = Q Q 1 =1 28. =2. 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 = implies n i=1 (a i x) 2 =( ) 2 +( ) 2 +2(2 ) 2 + (2 ) 2 +( ) 2 +(0 ) 2 +(2 ) 2 +(0 ) 2 +( ) 2 + (8 ) 2 +(0 ) 2 +(2 ) 2 +2( ) 2 +(90 ) 2 = 70. Therefore variance = 70 = 22., implying Standard deviation is S D = 0. = 20.. (8) 11
. (a) Consider the Venn Diagram below: M = Maths; P = Physics, A = Accounting. x + y = 1; x + z = ; x + y + z + 22 = 0, implying + y + 22 = 0. Thus (i) y =0 2 = 28 is the number who take Maths and Physics only. (ii) Therefore x =1 y =1 28 = is the number who take all three subjects. This implies z = x = =7. Sox + a + z + = 1, implies +a +7+=1,thusa =2. (iii) Now = a + x + y + b =2++28+b, therefore b =2isthe number taking Physics only. 7 M 22 y P z x b a A (b) (i) P(Maths only) = 22/7 (ii) P(P and A) = = (x + a)/7 = /7 (iii) P(A + M - P) = P(A + M only) = z/7 = 7/7. () [2] END 12