SECTION A(). x + is the longest side of the triangle. ( x + ) = x + ( x 7) (Pyth. theroem) x x + x + = x 6x + 8 ( x )( x ) + x x + 9 x = (rejected) or x = +. AP and PB are in the golden ratio and AP > PB. AB AP = AP PB k + = k k k + = k k k = 0 Using the quadratic formula, ( ) ± ( ) ()( ) k = () + = or (rejected). Let (x, y) be the coordinates of P. Area of ABP = 8 sq. units (8 ) y = 8 y = 8 y = By substituting y = into y = x 7x +, we have x = x 7x + 0 ( x )( x ) 7x + x = or x = The coordinates of P are (, ) or (, ).
. The equation x + 7x + k has rational roots. We can let x + 7x + k = (x + m)( x + n), where m and n are integers. = x + ( m + n) x + mn By comparing the coefficients on both sides, we have 7 = m + n and k = mn The table below shows all the possible values of m and n, and the corresponding value of k. m n k The possible positive values of k are and. For k =, x + 7x + (x + )( x + ) x = or x =. The quadratic equation kx + 6x + k has equal roots. Δ 6 ( k )( k) 6 k k = 9 k = ± 6. The quadratic graph y = kx 6x does not intersect the x-axis. Δ < 0 ( 6) ( k )( ) < 0 6 + 0k < 0 9 k < The range of possible values of k is 9 k <. 7. (a) By substituting (a, 0) into y = x x + 9, we have 0 = a 0 = (a ) a = a + 9 y-intercept = 9 b = 9 (b) The axis of symmetry is x =.
(c) Let (c, 9) be the coordinates of C. B is reflected about the axis of symmetry to C. 0 + c = c = The coordinates of C are (, 9). 8. (a) g( = ( x )( x + ) = x + x = x + x f ( g( = x + x + ( x + x ) = x + x + x = x + x + By the remainder theorem, remainder = f ( ) = ( ) + ( ) + = 9. (a) p ( = 8x + x x + a By the remainder theorem, p = 7 8 + + a = 7 a = 9 (b) 8x + x p( = (x ) q( + 7 x + 9 = (x ) q( + 7 q( = (8x = x + x + 8x + 6 x 8) (x ) 0. (a) g() = () + () + 6 = 6 + 6 + 6 By the factor theorem, x is a factor of g (. (b) By long division, g( = x + x x + 6 = ( x )(x + x ) = ( x )( x + )(x )
. (a) x + is the factor of f (. ( ) + p( ) f ( ) + q( ) 8 x + is the factor of f (. ( ) + p( ) () (): we have p = p q = 9...() f ( ) + q( ) 8 p q = 8...() By substituting p = into (), we have q = 0. both correct (b) x + x + ( x + )( x + ) x + and x + are the factors of f (. f ( is divisible by x + x +.. (a) The equation of L is y ( ) ( 7) = x y + = x y + = x + x + y = 0 (b) Let (c, 0) be the coordinates of C. By substituting (c, 0) into the equation of L, we have c + 0 c = Coordinates of C =, 0. (a) L : x y + 7...() L :x y = 0...() () () : (9x y ) ( x y + 7) 8x 0 x = By substituting x = into (), we have () y = 0 y = Coordinates of P = (, )
(b) Slope of the straight line = = ( ) The equation of the straight line is y = ( x ) y = x x y + 7. Let (a, 0) be the coordinates of A. By substituting (a, 0) into the equation of L, we have a + 0 6 a = OA = PB = AP and PO AB OB = OA (prop. of isos. ) = Coordinates of B = (, 0) ( 6) y-intercept of L = = 6 Coordinates of P = (0, 6) 6 0 Slope of L = 0 ( ) = The equation of L is y = x + 6 x y + 6 SECTION A(). (a) f( x ax ( a + ) Δ = ( a) ()[ ( a + )] = a + a + = ( a + ) > 0 ( a ) The equation f( has two distinct real roots.
(b) Using the quadratic formula, ( a) ± ( a + ) x = () a ± ( a + ) = a + = or = a + or The roots of the equation f( are a + and. (c) The y-intercept of the quadratic graph y = f( is. By substituting x and y = into y = x ax ( a + ), we have = (0) a(0) ( a + ) a + = The x-intercepts of the graph are and. 6. (a) The axis of symmetry of the graph is x =. h = By substituting (0, 7) and h = into 7 = (0 ) 7 = 6 + k k = 9 (b) By substituting y, h = and k = 9 into ( x ) = 9 + k 0 = ( x ) + 9 y = ( x h) + k, we have y = ( x h) + k, we have x = ± x = or x = 7 The coordinates of A and B are (, 0) and (7, 0) respectively. + (c) y = ( x ) 9 Maximum value of y = 9 Maximum height of ABP = 9 Height of ABC = 7 Maximum height of ABP > height of ABC Also, base of ABP = base of ABC It is possible that the area of ABP is greater than that of ABC. 6
7. (a) y-intercept = c = By substituting (7, 9) and c = into 9 = 7 9 = 7b + 7 b = + b(7) + ( ) y = x + bx + c, we have (b) (i) When y, x x ( x + )( x 6) x = or x = 6 The x-intercepts are and 6. (ii) The axis of symmetry is + 6 x = x = (c) f () = () = 6 The coordinates of the vertex of the graph of y = f( are (, 6). k = 6 8. (a) g( = (x + )(6x x + 7) 6 = x 6x + x + g = 6 + + = + + By the factor theorem, x is a factor of g (. (b) g( x 6x + x + (x )(6x x ) (x )(6x + )( x ) x = or 6 or all correct 7
9. (a) The remainder is when f ( is divided by x. f () = (a b) = a b = ± b = a + or a + (b) x is the factor of f (. f () ( a b) a b = ± a = b + or b (rejected) By substituting a = b + into the result of (a), we have When b = a +, b = ( b + ) + b = 8 b = Hence, a = + =. both correct When b = a, b = ( b + ) b = b = Hence, a = + =. both correct 0. (a) x is a factor of f (. f (0) mn + mn = m and n are integers and m < n < 0. m = and n = both correct 8
(b) f ( = ( x )( x )( x ) + = x 7x x 7x + x ( x + x f ( g( 9x + 0x + k) x + x k f ( g ( has real roots. Δ 0 () ()( k ) 0 k 8. (a) By long division, we have x + x x + x 7x + 6x x 8x x x + 6x x + x 8 Quotient = x + and remainder = x 8 + (b) (i) From the result of (a), we have f ( = (x + )( x x + ) + (x 8) g( = [(x + )( x = (x + )( x x + ) + (x 8)] ( rx + s) x + ) + (8 r) x (6 + s) g ( is divisible by x x +. 8 r and 6 + s, i.e. r = 8 and s = 6 + (ii) g( = (x + )( x x + ) = (x + )( x )( x ) 9
. (a) Slope of L = = ( ) Let m be the slope of OD. OD is the shortest distance from O to the line L. OD L m slope of L = m = m = The equation of OD is y = x x + y (b) (i) L : x y + () OD : y = x () By substituting () into (), we have x x + x + x = By substituting x = into (), we have y = ( ) = Coordinates of D = (, ) (ii) OD = = ( 0) units + ( 0) units = units The shortest distance from O to the line L is units. 0
. (a) Slope of 8 0 AB = 0 6 = The equation of AB is y 8 = ( x 0) y = x x + y CD AB Slope of CD slope of AB = Slope of CD = Slope of CD = The equation of CD is y 0 = [ x ( )] y = x + x y + (b) AB : x + y () CD : x y + () ( ) () :(x + 9y 7) (x 6y + 8) By substituting x + x = y = into (), we have y 0 y = Coordinates of D =,
Let (0, e) be the coordinates of E. e = y-intercept of CD = ( ) = Coordinates of E = (0, ) (c) Area of quadrilateral OBDE = area of AOB area of AED = (6)(8) (8 ) sq.units = ( 6) sq.units = 8sq.units. (a) E is the mid-point of AC. + 9 7 + Coordinates of E =, = (6, ) (b) (i) Slope of CD = = ( ) AB // DC Slope of AB = slope of CD = The equation of AB is y 7 = ( x ) x y + (ii) Let (b, ) be the coordinates of B. B(b, ) lies on AB. By substituting (b, ) into the equation of AB, we have ( b) + b = Coordinates of B = (, ) AB = ( ) + (7 ) units units = units Let G be a point on BC such that AG BC. AG = (7 ) units = units BC = (9 ) units = 8 units
By considering the area of parallelogram ABCD, AB CH = BC AG 8 CH = units 6 = units 6 = units. (a) Coordinates of E = (, ) Area of ABCD = 08 sq. units Area of BEA = 08 sq. units (9 )( b ) = 7 b = 9 b = E is the mid-point of BD. b + d = 0 = + d d = (b) Slope of AB = 9 = The equation of AB is y = ( x 9) y 0 = x + 7 x + y 7 CD // AB Slope of CD = slopeof AB = The equation of CD is y ( ) = ( x ) y + 8 = x + 9 x + y
6. (a) L : x + y...() L : x y...() () () : (x + y 0) (x y) y 0 y = By substituting y = into (), we have x + x = Coordinates of A = (, ) (b) (i) The equation of L is y = m( x ) y = mx m + Let (b, 0) and (0, c) be the coordinates of B and C respectively. By substituting (b, 0) into the equation of L, we have 0 = m( b) m + bm = m m b = m m Coordinates of B =, 0 m y-intercept of L = m + c = m + Coordinates of C = ( 0, m + ) (ii) P(, ) is the mid-point of BC. 0 + ( m + ) = = m + m = 7. (a) Slope of 0 BC = 0 = (b) P is the mid-point of BC. + 0 0 + Coordinates of P =, =,
Let m be the slope of PH. PH BC m slope of BC = m = m = The equation of PH is y = x 8y 6 = 6x 9 6x 8y + 7 Let (0, h) be the coordinates of H. 7 y-intercept of PH = ( 8) Coordinates of 7 h = 8 7 H = 0, 8 (c) O is the mid-point of AB and CO AB. CO is the perpendicular bisector of AB. Let Q be the mid-point of AC. + 0 0 + Coordinates of Q =, =, 0 Slope of AC = 0 ( ) Slope of = QH = 7 8 0 = Slope of AC slope of QH = = AC QH QH is the perpendicular bisector of AC. The three perpendicular bisectors PH, QH and CO in ABC pass through the same point H.