WORKED SOLUTIONS Worked solutions Algera and functions : Manipulating algeraic epressions Prior knowledge page a (a + ) a + c( d) c cd c e ( f g + eh) e f e g + e h d i( + j) + 7(k j) i + ij + 7k j e l(l + ) (l + l ) l + l l l l l f m(m n) (m + n) m mn m n a ( + )( + ) + + ( )( + ) + c ( )( ) + d ( ) + 9 e ( + )( + ) + + f ( + )( + )( + ) + + + a a a a a a a 9 c d (c ) c e (d) d f ( e f ) ef ef a c 7 d e f Eercise.A a + + a + a + + 7 page HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 ( )( + ) ( )( + ) + ( + ) + + + + 9 (n) + (n + ) n + n + n + n + n + (n + n + ) Since is a multiple of, (n + n + )must e an even numer. ( + )( + ) ( + )( ) ( + ) + ( + ) ( ) + ( ) Should e: ( + ) + ( + ) ( ) ( ) + + + + + + Should e: + + + + + + + + Should e: + 9 7 Let p n + where n is an integer. Consequentl p is an odd numer. p (n + ) n + n + (n + n) + Since (n + n) is even, (n + n) + is odd. p(n +)(m + ) p(mn + n + m + ) mnp + pn + mp + p (mnp + pn + mp + p), an even numer Eercise.A page a ( + ) ()() () + ()() () + ()() () + ()() () + + + ( + ) ()() () + ()() () + ()() () + ()() () + ()() () + + + +
ALGEBRA AND FUNCTIONS : MANIPULATING ALGEBRAIC EXPRESSIONS a ( ) ()() ( ) + ()() ( ) + ()() ( ) + ()() ( ) + ()() ( ) + ()() ( ) + + + + + + ( ) + + + 7 + + ( ) ()() ( ) + ()() ( ) + ()() ( ) + ()() ( ) + ()() ( ) + + ( + )( ) ( + )( + + ) + + ()() ( ) ( + ) ()( ) ( ) + ()( ) ( ) + ()( ) ( ) + ()( ) ( ) ( ) ()( ) ( ) + ()( ) ( ) + ()( ) ( ) + ()( ) ( ) 9 + 7 ( + ) 9 + 7 + + + 9 ()()() (c) + ()()() (c) 9 c + c 9 c + c c + c (c )(c + ) c or c Eercise.B page a c a c d The are the coefficients for a cuic epansion from Pascal s triangle. C, C, C, C, C, C All the rows prior to the inde required have to e written in Pascal s triangle: a laorious and error-prone eercise. C 9! 79!! 7 Eercise.A page a ( + ) C + C () + C () + C () + C ()! + + (!!!! )(!!!! ) +!! ( )+ ( )!!!! + + + + ( + ) C + C () + C () + C ()! 7! 9!!!!!!! +!!!( ) + + 7 + + + a ( + ) )(! 7 ( + ) ( + ) ( ) )( (! ) a + C + C + C + C + +!!!!!!!!! +!!! + + + ( ) C () + C () ( ) a ( + ) ( + ) + C () ( ) + C () ( ) + C ( )! 7!!!!! +! ( 9!! )! + ( )!! 9 +! (!! ) + + 9 9 + 9 )(! 9 + + )(! HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS 7 7 ( + ) + 7 ( 77 )( 7 )! 7 7 C + C () () + C() () + C + C +!! 7!!!! +!! +!!!!! + + 9 + 9!! 9 ( ) + + + + + + 9 + 7 + 7 C 7 7 C +! +!! 9 7 9! 7!! C() + C () + C() () + C! 79!!!!! + +!!! +! 7!! 7 7 79 +..7 79. + 7.97.7 Eercise.A a ( + ) ( + ) c ( ) d + z + z doesn t factorise. a ( )( ) ( + )( + ) c ( + )( ) d ( 7)( + 7) 7( ) Should e: 7( ) page a ( ) ( )( ) c ( + ) d ( ) e (9 + 7z)(9 7z) f ( 7 + ) the quadratic cannot e factorised. + ( )( + ) or ( )( + ) (7 + )( ) 7 ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) ( + )( + ) p(q r ) r(p q) pq pr pr qr p (q r r) rq qr qr p or p q r r r + r q 9 ( ) + ( + ) Eercise.A page a + + f() + ( ) + so ( ) is a factor of f(). + + ( )( + )( + ) a + ( + ) ( ) + + ( + ) ( + ) + + ( + )( ) + ( + ) a f() + 7 + f() + 7 + so ( ) is a factor. + 7 + ( )( )( + ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : MANIPULATING ALGEBRAIC EXPRESSIONS f() + 7 + f( ) + 7 + so ( + ) is a factor. + 7 + ( + )( + + ) g() + + g( ) + 9 + 9 + so ( + ) is a factor. + 7 + ( + )( + )( )( ) c f() + f() + so ( ) is a factor. + ( )( + 9 ) g() + 9 g( ) ( ) + ( ) 9( ) so ( + ) is a factor. + ( )( + )( + )( ) a + + + (A + B + C)( ) + D : + + + D : (C)( ) + D C Equating coefficients: : A : A + B B 7 + + + ( + 7 + )( ) + + + + (A + B + C)( ) + D : + + + D : (C)( ) + D C 7 Equating coefficients: : A : A + B B + + + ( + + 7)( ) + c + + (A + B + C)( + ) + D : + D : (C)() C Equating coefficients: : A : A + B D B + + ( + )( + ) a f() + f() + f() + + 7 f( ) 7 9 c f() 9 + + f( ) + + 7 a + (A + B + C + D)( ) + E : + E E : (D)( ) + D Equating coefficients: : A : A + B B : B + C C + ( + + )( ) + + + 7 (A + B + C + D)( + ) + E : 7 E : (D)() 9 D Equating coefficients: : A : A + B B : B + C E 9 C 7 + + 7 ( 7 + )( + ) 9 f() 7 + + f() + + so ( ) is a factor. 7 + + ( )( + + 9) g() + + 9 g() 7 + 9 + 9 so ( ) is a factor. g() ( )( ) f() ( + )( )( ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Eercise.A page Eercise.7A page a a 9 c c d 7d e e f f a ± c 7 7 a a c c or c 7 d d e 9e f f g or f g a ( ) c ± ( ) Should e: 7 7 Area π a π or π 9 a 9 or a c c c c 9 Volume π 9 π and z z z z 7 or a a a e + e f f a 7c d d a g h + h gh j k k j a ( a) ( a)( a) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 a a + a a + a 7 ( a + )( a ) a ( a + a) ( a + a)( a + a) a+ a + a + a a + a + a 9 a a ( a a)( a + a) a ( a)( + a) Eercise.A page 7 a + c + + + 7 a + + + 7 + + 7 + 7 + + 7 7 + 7 c + a 7 7 7 + 7 + 7 + + + + c + + a c a + + + + + + + + c + 9
ALGEBRA AND FUNCTIONS : MANIPULATING ALGEBRAIC EXPRESSIONS a + + c + + + + + + + + + + + + + + + Eam-stle questions page 9 a a + a f() + + f() + 9 7 + so ( ) is a factor of f(). f( ) + + + so ( + ) is not a factor of f(). C 9 + ( 7) 9 + 7 9 7 + + 9 + ( 7)( + ) ( 7)( + )( ) 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 + ( ) a+ a ( a+ ) ( a+ )+ ( a ) a a + a + + a a + a + which is rational if a and are rational. 9 a + + + + + 9 f() 9 + f() 9 + hence ( ) is a factor of f(). + ( ) + 9 + 9 + + + 9 + ( )( + ) g() + g( ) + + hence ( + ) is a factor of g(). ( + ) + + f() 9 + ( )( + ) ( ) a ( ) The inde on the racket should e and the contents of the racket should e. The coefficient of < (it is ). + + + (A + B + C)( ) + D : + 7 + + D D 79
WORKED SOLUTIONS : (C)( ) + 79 C Equating coefficients: : A : A + B B + + + ( + + )( ) + 79 7 7 9 ± a a a ( ) a For : a a a 7 a ( ) C () + C () ( ) + C () ( ) + C () ( ) + C ( )! + +!!!!! ( )!!! ( ) +!!! ( ) +!!! ( ) + 9 + c. (.). +..9 +.. + 9 ( + )( ) ( + )( 7 ) 7 + + + + + + + So one numer is too high. As the numers are smmetrical, the should e a. 7 ( a+ ) ( a ) ( a+ ) ( a ) ( a ) a ( + ) a a a a a + a a + a a a + a which is irrational unless is a square numer. a HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 + + + 9 + 9 9 9 9 + + + + + + + + + + + 9 f() + + 7 f() + + 7 so ( ) is a factor of f(). + 9 7 ( ) + + 7 9 9 + 9 7 + 7 7 + 7 9 + ( )( + 9 7) g() + 9 7 g( ) + + 7 7 so ( + ) is a factor of g(). 9 ( + ) + 9 7 + 9 7 9 7 f() + + 7 ( )( + )( )( + ) 9 + + (A + B + C + D + E)( + ) + F : + + F F : (E)() E Equating coefficients: : A : 9 A + B 7
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES B : B + C C : C + D D 7 9 + + ( + 7 + )( + ) a + C + C + C + C! +! (! )+ (! )+ ( )!!!! + + +!!!! ( ) + Total numer of alls is (7 ) + (9 ) + P( RR) 7 + 7 + 9 9+ + + a + C + C + C + C + +!! +! +!...!!!!!!!! + + + + +.. +. +.7 +.. c It is onl an approimation ecause the epansion isn t complete. + + + 7 + + 7 When + 7 is divided + to get an epression for the area of the cross-section (width depth), then there is a remainder of. + Consequentl, this is not a possile correct epression for the volume of the cuoid. When + is divided + to get an epression for the area of the cross-section (width depth) then there is no remainder. Consequentl, this is a possile correct epression for the volume of the cuoid. Algera and functions : Equations and inequalities Prior knowledge page ( + )( ) or + ( ) ( ) + + 9 or a,, c ac ( ) ()()( ) so two real roots. ± ±. or. + Add and. 7, + + Sustitute into. + + + ( ) + ( ) < 7 < 7 < c + or a HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Eercise.A page a where. Intercepts and turning point at (, ). No -intercept, -intercept and turning point at (, ). c + where. -intercepts at (, ) and (, ), -intercept at (, ) and turning point at,. 9 7 9 9 7 9 7 a Line of smmetr at, -intercept at (, ), -intercept and turning point at (, ). 9 7 7 Line of smmetr at.7. -intercepts at (, ) and (, ), -intercept at (, ) and turning point at (.7,.). 9 7 7 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 9
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES c Line of smmetr at.. -intercepts at (, ) and (, ), -intercept at (, ) and turning point at (.,.). + 7 7 7 9 7 9 7 9 Line of smmetr at. (appro.). -intercepts at (.7, ) and (, ), -intercept at (, ) and turning point at appro. (., ). ( + )( ) Epand the rackets. 7 This equation has the approimate -intercepts and the -intercept at the turning point when.,. in the equation. 9 Rearranging. + + 9 Solutions for 9 can e found where the graphs intersect, hence and. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Eercise.A page a + + ac () ()()() ac < so no real roots. + + ac () ()()() ac > so two distinct real roots. c + + ac () ()()() ac so two equal real roots. + ac ( ) ()()() ac > so has two distinct real roots. a ac ( ) ()()( ) 9 ac > so two distinct real roots. + 7 ac ( ) ()()(7) ac < so no real roots. c + ac ( ) ()()() ac so two equal real roots. + (k + ) + (k ) ac so two equal real roots. (k + ) ()()(k ) Epand and simplif. k k + or k + k Both quadratic equations have the solutions k or k. For k, and for k,. a + ac () ()( )() ac > so two distinct real roots. 9 + ac ( ) ()()(9) ac so two equal real roots. c + ac ( ) ()()() 9 ac < so no real roots. + c a ac < () ()()( c) < + c < c < ac () ()()( c) + c c c ac > () ()()( c) > + c > c > Eercise.A a + ( + ) ( ) [( ) ] c + + 7 ( + ) + 7 ( + ) 9 + ( ) + ( ) a ( ) ( ) + 7 + ( ) + 9 7 7 c + + 9 + + 9 9 9 ( + ) + ( ) + + page HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES + 7 ( ) ( ) + + 7 At the turning point has a minimum value, i.e. when. When,, the coordinates of the turning point. [( + ) ] ( + + ) + + + a + + ( + ) ( ) + ( ) ut can go no further. c 7 + 7 7 9 ( + ) 7 ( ) + 7 + + ( ) ( ) 9 + 79 + 9 79 + > Consequentl ac < ecause the curve of this equation will not intercept the -ais and so there are no real roots. ( )[( + ) ] ( )( + + ) ( )( + + ) + + Eercise.A page a i 9 ( + )( ) or ± ()()( 9) ii or iii 9 or iv or i + + ( + )( + ) or ± 9 ii or iii ( + ) or iv or c i 7 Cannot e factorised. ± ( 7) ii ± iii ( ) ± iv ± d i ( ) or ± ii or iii ( ) 9 or iv or Factorisation: can onl e used on equations that factorise sometimes spotting factors can e difficult can solve: + + cannot solve: 7 Quadratic formula: can e used to solve an equation with real roots including ones that don t factorise cumersome and consequentl eas to make a mistake can solve: 7 cannot solve: + 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Completing the square: can e used to solve an equation with real roots including ones that don t factorise can e cumersome manipulations if is odd and a > ; consequentl eas to make a mistake can solve: 7 cannot solve: + 7 Calculator: eas if ou know how to use our calculator does not help ou to understand the underling methods can solve: + + and 7 cannot solve: + 7 a ( ) ± Completed the square. The equation was alread in form of a completed square. ± ± Does not factorise so used the quadratic formula. Alternativel, could have completed the square ut chose not to as a >. + ( )( ) + + 7 ( 7)( ) ± ± 7 or 7 a + + c c a + + a a c ( a ) a a ( a ) ( a ) ( a ) + + + a c a ac + a a ac + a + ± ac a a ± ac a c + ( )( + ) or Equation factorises so eas to do this. Alternativel, could have completed the square ut chose not to as a >. Eercise.A page a + + Sutract from. and d + 7 ( )( 7) or 7 Equation factorises so eas to do this. Alternativel, could have completed the square ut chose not to as a >. a Cannot e factorised if ac is not a square numer. ac ()()( ) not a square numer so does not factorise. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 + Multipl. + Add and. and c + Add and. and 9
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES + Multipl. + Add and. Sustitute into. If there are n unknowns then ou need n distinct equations involving the n unknowns. + Sutract from. As neither the coefficient of nor is the same in oth equations, when one equation is sutracted from the other an unknown is not eliminated. + Add and. 9 and + ( ) Sutract from. and. ears old. 7 z z Multipl. z Add and. z Multipl. z Add and. Sustitute into : z Sustitute into : Eercise.B page a + + From. Sustitute into. and + From. Sustitute into. ( ) and c + From. + Sustitute into. + and 9 + Rearrange. Sustitute into. + ( ) Epand and simplif. Sustitute into. Solve. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS + 9 From. 9 Sustitute into. 9 and The graphs intersect at the point (, ). + From. + Sustitute into. + + and c + 7 7 c + From. Sustitute into. 7 7 c 7 7 + p and c 7p The simultaneous equations onl work if the cost of a small ag of sweets is 7p and the cost of a ar of chocolate is p. Although two equations can e formed and solved simultaneousl from the given figures, p is too little for a ar of chocolate so the student must e correct: one of the calculations is wrong. z z From. + Sustitute into. ( + ) z z From. z + Sustitute into. ( + ), z and Eercise.A page 7 a + Add and. + ( 7)( + ) 7 or or 7 + From. Sustitute into. + ( ) + ( )( ) or or c + ( + )( ) or or a Sustitution or elimination: for elimination the second equation would need to e multiplied so that can susequentl e eliminated. Sustitution onl: neither addition nor sutraction of the equations will eliminate a variale. c Sustitution onl: neither addition nor sutraction of the equations will eliminate a variale. a + Multipl. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES + Add and. + 9 ( + )( ) or 7 or + From. Sustitute into. ( ) ± or c From. + Sustitute into. ( + ) + ± ( ) ± ± and ± 7 From. 7 Sustitute into. ( 7) ( + )( ) or 7 or 9 So the coordinates of the points of intersection of this line and this curve are (, 7 ) and (, ). 9 + From. + Sustitute into. + ( + ) + ±. or.. or. So the coordinates of the points of intersection of this line and the circle are (.,.) and (.,.). + From. + Sustitute into. + ( + ) + + + + ( + ) and The line is a tangent to the circle. The intersect once at (, ). Eercise.7A page a > > > + 7 7 c + < < < d ( ) 7 7 a > > HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS + 7 7 c + < < > < d ( ) 7 7 a ( ) + + and > > > < < {,,, } 7 < < < or < + < < + < < < < < {, 9,, 7,,,,,,,,,,,,,, 7} c ( ) and < ( + ) < 7 < < 7 r πr 7 r. So. r miles 7 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 d cannot e oth less than and greater than or 7 equal to. 9 7 or + where {,,,,, } a cannot e oth less than and greater than in this wa. Can e rewritten as < or > cannot e oth less than or greater than in this wa. Can e rewritten as < or > 7
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES c cannot e oth less than and greater than., and 7 a c d + j < 7 j < j < 7 + j > 7 j > 7 7 < j < Eercise.A page a + > + > ( + )( ) > < or > + ( )( ) c 9 ( + )( ) < or > d < ( ) < < < a + 7 < + 7 + < ( + )( + ) < < < + > ( + )( + ) > < < c > d > ( )( + ) > < or > ( ) < < a > ( )( + ) > < < and + > > < {,,,,, } + 9 ( )( + ) or < < So < {all integers less than.} c ( + ) 9 + ± d and < < {,,, } < < ( ) < < < or > > < So < {all integers less than.} r πr r.9 So.9 r miles a + < 7 7 + < ( )( ) < < < {,,, } < < HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS ( + )( ) < < < {,,,, } c + 9 + > ( + ) + 9 + > + 9 + + > ( + ) > Inequalit true for all values of. d + > > + > ( )( + ) > < or > {all integers ecept,, } a + < ( + )( ) < < < and + < ( + )( ) < < < So < <, i.e. {,,, } + < or + < So < <, i.e. {,,,,, } c < and > < < and < or > So < <, i.e. { } d + > or + < < or > or < < {all integers ecluding,,, } Eam-stle questions page + Line of smmetr at. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 -intercepts at (, ) and (, ), -intercept at (, ) and turning point at (.,.) (appro.). 7 ac 9 ()()( 7) > so two distinct real roots ± 9 ( ) ±.79 or.79 > <.79 or >.79 + Multipl. + Add and. and 7 7 7 ( )( + ) or So or 7 7 -intercepts at (, ) and (, ) and -intercept at (, ); a < ( )( ) + + Satisfies the intercepts and a <. + + ac 9 ()()() 9 < so no real roots. The statement + + for all values of is correct. The discriminant is <, consequentl the equation has no real roots and so it does not intercept the -ais. As a > all of the curve will e aove the -ais. 9
ALGEBRA AND FUNCTIONS : EQUATIONS AND INEQUALITIES 9 7 7 7 9 ( ) 7 ( ) At the turning point will have a minimum value i.e. when 7. When 7, ( ) So coordinates of turning point are ( 7, ) + Multipl and multipl. + Sutract from. and So the coordinates of the point of intersection of these two lines are (, ). 7 Rearrange. 7+ Sustitute into. 7+ 9 + + 9 7 + ( )( ) ( 7, ) and (, ) a 9 > + < No solution (as no overlap of inequalities). > 7 + 7 > ( + )( ) > < or > a + + > ( + )( + 7) > < 7 or > and > > So >, i.e. {all integers greater than } + + > or > ( + )( + 7) > < 7 or > or > > So < 7 or > {all integers less than 7 or greater than } + > 7 + 7 > ± 7 ±.7 or.77 <.7 or >.77 a + a ac < ()(a)( ) < + a < a < a < ac ()(a)( ) + a a a c ac > ()(a)( ) > + a > a > a > HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS 7 a + + c a + + c a a + 9 c + a a a + c a 9 a a + 9 a c a a a + 9 a c a a a c + ± 9 a a ± 9 a c a Perimeter: ( + ) + ( + ) + + 9 ( + )( + ) + + Rearrange. 9 Sustitute into. + 9 + 9 9 ( 9)( + ) 9 or or So width 9 + 7 or + So length or + 9 7 9 πr r. πr > r >.. < r. + 7z 7z From z 7 Sustitute into and rearrange. From Sustitute into and solve. 7 + ( )( ) or or z or 7 7 One set of solutions is z,, 7 Another set of solutions is, z, 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : SKETCHING CURVES Algera and functions : Sketching curves Prior knowledge p 9 a ( 7)( + ) so or 7 ( + 7)( 7) so 7 or 7 c ( ) so d ( ) so or 7 e ( )( + ) so or f ( )( ) so or a ( ) 7 ± 7 ± 7 ( ) ± ± or c ( + ) 7 + ± 7 ± 7 a (, 7), minimum as a > (, ), minimum as a > c (, 7), minimum as a > a a,, c 9 ac ( 9) so one real root. a,, c ac ( ) so no real roots. c a, 7, c ac 9 ( ) so two real roots. Eercise.A p a 7 (, ) (, 9) Roots at and so factors are ( ) and ( ). ( )( ) + a 9 7 (/, 9/) The sketch is not correct. ( 7)( + ) So should cut -ais at (, ) and (7, ). a > so graph should e U-shaped. When, so -intercept at (, ). ( ) So turning point at (, ). HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS a 7 a.... ( /, 9/) a m s c Roots at and so factors are ( + ) and ( ), ut graph shape indicates a <, so ( + )( ) + + Eercise.A p a........... The sketch is not correct. Point of inflection should e at (, ) and curve should e a reflection in the -ais of the curve shown. The graphs are reflections of each other in the line............. C ( ) onl HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : SKETCHING CURVES Eercise.B p a a...... 7 a ( ).. a ( )( + )..... 7 ( 7) 7.............. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979........
WORKED SOLUTIONS ( ) Using the quadratic formula to solve. or +......... Eercise.A p 9 a............ The curve is the wrong shape: it should not e a paraola. The curve should e a U-shape sitting on the -ais at (, ) and intercepting the -ais at (, ). The do not generate the same curve ecause the are different functions. ( + ) ( + )( + )( + )( + ) ( ) ( )( )( )( ) ( ) (+) a....... B ( ) and C ( ) Eercise.B p 7 a............. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 7
ALGEBRA AND FUNCTIONS : SKETCHING CURVES a........ 7.... ( )( + )( + ) a..... 7....... Eercise.A p 7 a Asmptotes at and........... Asmptotes at and........... c In oth graphs the asmptotes are in the same locations. The ke difference is that the curves of the graph of are much closer to the origin than those of. a -intercept at (, ); asmptotes at and. -intercept at (,.7); asmptotes at and. a + ( + ) a iv ii c iv d i a Asmptotes at and ; -intercept at (, ). ( ) ( + )....... HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Asmptotes at and ; -intercept at (, )........ No. The asmptotes are in the correct locations at and. However, the graph drawn is for a reciprocal with a positive numerator whereas the given equation has a negative numerator. Eercise.B p 7 a Asmptotes at and....... Asmptotes at and....... a Asmptotes at and ; intercept at (,.). a 7 Asmptotes at and ; -intercept at (, ). 7 ( ) + ( ) a iii i c ii d i a Asmptotes at and ; -intercept at (, ). 7....... c In oth graphs the asmptotes are in the same locations. The ke differences are (i) that the curves of the graph of are much closer to the origin than those of and (ii) the graph of is a reflection in the -ais. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 Asmptotes at and ; -intercept at (, )....... 7
ALGEBRA AND FUNCTIONS : SKETCHING CURVES Yes, the graph is for the given function. Correct shape with asmptotes at and ; -intercept at (, 7 9 ). Eercise.A p ( + ) ( ) rearranged is ( + )( ) ( ) ( + ) ( ) ( ). There are two solutions to There are three points of intersection. There are no solutions to ecause the graphs do not intersect. ecause the graphs intersect twice......... ( + ) rearranged is ( + ) ( + ). There are no solutions to ( + ) ecause the graphs do not intersect. Also, there are no real solutions to the fourth root of a variale with a value of. 9 7 There are two solutions to ( )( ) ecause the graphs intersect twice. 7 At points of intersection, the equations are equal. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 7 ( ) ( ) ( )( + ), or Sustituting into one of the original equations:, or 9
WORKED SOLUTIONS 7 There are two points of intersection. There are two points of intersection. 7 Eercise.A p m (g) m......... V (cm ) Using the graph, when m, V.7. The volume of the ring is.7 cm. a C, iii A, ii c B, i V r V kr k π π 7 V π r π 7 7π When r 7, V V V a k k When, () V (cm ) d t d kt k. Constant speed is mph. d d (miles) 7 9 t t (hours) Using the graph, when t., d 7. The distance travelled is 7 miles. m V m kv 9. k 9.. When V.7, m 9..7.9 g HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 7 9 r r (cm) a Using the graph, when r, V cm. Using the graph, when V π, r cm. c Sustituting r into the equation gives V. Likewise sustituting V π into the equation gives r. a k From the graph, when,. k k c 9
ALGEBRA AND FUNCTIONS : SKETCHING CURVES Eercise.7A p 9 a f( + ): one solution. 9 7 f() + : asmptotes at and. f() + : no solutions. c f( ): one solution. 7 9 d f( + ): one solution. 9 7 a f( + ): asmptotes at and. 7 c f( ): asmptotes at and. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 d f() : asmptotes at and. f( ): Q at (, ); one solution. f() + : Q at (, ); one solution.
WORKED SOLUTIONS a f( + ): asmptotes at and. 7 f ( + ) ( + ) c (, ) f( + ): P at (, ); asmptote at. f() + : P at (, ); asmptote at. f( + ): f() + has one solution (one point of intersection). f() + f( + ) a, f( ) where : asmptotes at 9 and 9. Ais intercepts at (, ), (, ) and (, ). 7 a f( + ) ( + )( + ) : two solutions. f() + ( + )( ) + : two solutions. c f ( ) ( ) : two solutions. d f() ( + )( ) : two solutions. 9 9 c, d f() + : asmptotes at 9 and 9. a, f( + ) + : no solutions. f() + + + : no solutions. f( + ) f()+ 9 9 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : SKETCHING CURVES c If f() then f( + ) + and f( + ) + + + 9 f() + + f( ) ( ) ( ) + ( ) + ( ) + + + + + + + + + 9 + Eercise.A p 9 a f(): one solution. 7 f(): one solution. 7 c f(): one solution. 7 f : one solution. d 7 a f(): asmptotes at and. f(): asmptotes at and. c f( ): asmptotes at and. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS d f(): asmptotes at and. f( ): P(, ), R(, ), Q(, ). f( ) has two solutions. f(): P(, ), R(, ), Q(, ). f() has two solutions. f( ) f(): has no solutions (the do not intersect). f( ) f() a, f() where : asmptotes at,, and. Ais intercepts at (, ), ( 9, ), (, ), (9, ) and (, ). 9 9 c, d f() where : asmptotes at 9 and 9. Ais intercepts at (, ), (, ) and (, ). a f(): asmptotes at and. 7 f ( + ) c (, ) f( ) and f(): in oth cases asmptote at. For f( ), P is at (, ). For f(), P is at (, ). 9 9 7 a f() ( + )( ) : two solutions. 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : SKETCHING CURVES f() ( + )( ) : two solutions. 7 c f ( + )( ) : two solutions. c If f(), f( + ) + and f( + ) ()( + ) 9 f() + + f + + + + 7 9 Eam-stle questions p 9 a f( + ) f( ) c f() + d f() The cale will touch the rollercoaster at two points in the given interval (two points of intersection)....... f d ( + )( ) : two solutions. a,, d f() + : no solutions. f() ()( + ): no solutions. 7 r p r kp k k r p The echange rate is reals to. r p 7 f() f() Using the graph: a r 9 r p p HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS a f( + 9) 7 9 9 7 f() k π A πl When l 7, A 9π A 7 9 / c f() The shape is a circle 7 a Asmptotes at and. There are two points of intersection. 7 9 9 7 d f ( ) 7 9 + and + Asmptotes at and. There are two points of intersection. 7 9 7 9 The rait and mice populations are onl the same when the are oth first introduced to the island. A l A kl. k a f() + f() c f( ) d f( ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
ALGEBRA AND FUNCTIONS : SKETCHING CURVES 9 a d a f() + 9 c i f() + a where a >. ii f( a) where a > 9 or a < HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Coordinate geometr : Equations of straight lines Prior Knowledge (, ), (, ) a A:, B:, C:, D:, E:, F:, A, E D, F c B, C Eercise.A p The lines not in the form a + + c, where a, and c are integers are: c ecause the right-hand side is not zero d ecause the right-hand side is not zero e ecause c is not an integer f ecause a, and c are not integers. a Rearrange +. + where a, and c Rearrange. + where a, and c a Multipl :, and rearrange. where a, and c Multipl : +, and rearrange. + where a, and c c Multipl : + Multipl : +, and rearrange. + where a, and c Rearrange: + Rearrange and divide : + Gradient and intercept ; coordinates (, ) Equation written down incorrectl. It should e. p Not all epressions in the equation have een multiplied. It should e. Although the values are wrong this step is actuall correct. With the correct values it should e +. Although the values are wrong this step is actuall correct. With the correct values it should e +. ( ) where a, and c 7 When the line intercepts the ais,. Sustituting : + So the coordinates of the intercept are (, ). When the line intercepts the ais,. Sustituting : + So the coordinates of the intercept are (, ). Eercise.A p a m m + c m d m + + HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 7
COORDINATE GEOMETRY : EQUATIONS OF STRAIGHT LINES a m + m + + m + m + + First find the point of intersection. + 7 + m and (, ) (, ) m + Find the equation of the first line. m and (, ) (, ) m Find the equation of the second line. m and (, ) (, ) m + + Find the point of intersection. + + 7 Point of intersection is (, 7). 7 m and (, ) (, ) m When ou sustitute in, will equal if the line goes through the origin. ( ) ( ) The line does not go through the origin. Eercise.A p a m (, ) (, ) (, ) ( 7, ) m 7 m m m (, ) ( 9, ) (, ) (, ) m 9 m m c m (, ) (, ) (, ) (, 9) m 9 m m HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS a m (, ) ( a, a) (, ) ( a, a) m a a a a m a a m m (, ) ( a, a) (, ) ( a, a) m a a a a m a a m Should e m coordinate incorrect in (, ). It should e (, ). Numerator and denominator confused. Should e: m m m a m (, ) (, ) (, ) (, ) m m 9 m Lie on a straight line with a gradient of. m (, ) (, ) (, ) ( 9, ) m 9 m m Do not lie on a straight line with a gradient of. c m (, ) (, 7) (, ) (, ) m 7 m 9 m Lie on a straight line with a gradient of. d m (, ) ( 7, ) (, ) (, ) m 7 m m Lie on a straight line with a gradient of. m (, ), ( ) (, ) (, ) m m m m (, ), (, ) (, ) m 9 m 9 m The gradient, m, is the same etween oth of the pairs of points, so we can conclude that all the points lie on a straight line. Ascent: m (, ) (, ) (, ) (, ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 9
COORDINATE GEOMETRY : EQUATIONS OF STRAIGHT LINES m m m Descent: m (, ) (, ) (, ) (, ) m m m The descent is the steepest part ecause >. 7 m (, ), (, ), m m m Eercise.A p a (, ) (, ) (, ) ( 7, ) 7 + (, ) (, ) (, ) ( 9, ) 9 9 a (, ) (, ) (, ) (, 9) 9 + + + + (, ) (, ) (, ) ( 7, ) 7 + + + + + Let (, ) (, ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 Let (, ) (, ) and and and confused. The net line should e: With this correction, the remainder of the calculation should e: ( ) ( ) + + So + 9 intercept when, so when (, ) (, ) (, ) (, )
WORKED SOLUTIONS + + intercept when so (, ) (, ) (, ) (, 7) 7 7 7( ) 7+ 7+ Gradient 7 and -intercept (, ) Line A: (, ) ( 7, ) (, ) (, ) 7 7 7 7 m Line B: (, ) (, ) (, ) ( 7, ) 7 + + + m HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 Line B is steeper ecause >. (The negatives can e ignored ecause the just indicate the direction of the line.) 7 a (, ), ( ) ( ) (, ), + 7 + (, ), 7 (, ), 7 7 + 7 + 7 + 7 + + 7 + 9 + 7 7 intercept when + 7 so 7 + intercept when + so (, ), 7 (, ), 7 7 7 7
COORDINATE GEOMETRY : EQUATIONS OF STRAIGHT LINES 7 7 ( ) 7 + 7 Rearranging: + 9 Line A: (, ) (, ) (, ) (, ) 9 9 Line B: 7 (, ) (, ) (, ) (, ) + + 7 + 7+ 7 7 At the point of intersection: 7 7 Sustituting into either equation gives. So point of intersection is (, ). Point A: 9 7, Point B: 7 +, (, ) ( 7, ) (, ), 7 7 + 7 7 ( + ) ( ) + 7 ( ) + + + When, so this line does not pass through the point (, ). Eercise.A p a m in oth equations so the lines are parallel. m in oth equations so the lines are parallel. Line A: (, ) ( 7, ) (, ) (, ) m m 7 m Line B: (, ) (, ) (, ) (, ) m m m m for oth, so the lines are parallel. From 7, m 7. The intercept of is when so is at (, ). (, ) (, ) m 7( ) 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS a Arrange the equations in the form m + c: +, m is not equal so the lines are not parallel. Arrange the equations in the form m + c: +, Line A: m in oth equations so the lines are parallel. (, ) (, ) (, ) (, ) m m m Line B: (, ) (, ) (, ) (, ) m m m m for oth, so the lines are parallel. From + + 7, 7, so m. The intercept of + is when so is at (, ). (, ) (, ) m ( ) + 7 a Arrange the equations in the form m + c: 7, + m in oth equations so the lines are parallel. Arrange the equations in the form m + c: 7, m is not equal so the lines are not parallel. Arrange the equations in the form m + c: +, +, m +,,, m,, m +,,, m So + and are parallel lines as m for oth lines. And + and + are parallel lines as m for oth lines. Eercise.B p 7 a m and m so m m so the lines are not perpendicular. m and m so m m so the lines are perpendicular. m for the given line ut the gradient of a line perpendicular to this will e m. Let (, ) (, ). m (, ) incorrectl sustituted. Correcting this and using correct value of m gives: So a m and m so m m so the lines are perpendicular. m and m so m m so the lines are not perpendicular. Line A: (, ) (, ) (, ) ( 9, ) m m 9 m Line B: (, ) (, ) (, ) (, ) m m m m and m so m m so the lines are perpendicular. Arrange the equations in the form m + c: +, +, m +,,, m,, m +,,, m HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
COORDINATE GEOMETRY : EQUATIONS OF STRAIGHT LINES So + and + are perpendicular lines as mm. And + and + are perpendicular lines as mm. And and + are perpendicular lines as mm. And and + are perpendicular lines as m m. From 7, 7, 7, m mm So m The intercept of + is when so is at (, ). (, ) (, ) m 7 a m and m so m m perpendicular. m 9 and m so mm 9 not perpendicular. Descrie as line A: (, ) (, ) (, ) (, ) m m m m Descrie as line B: (, ) (, ) (, ) (, ) m m m m Descrie as line C: (, ) (, ) (, ) (, ) so the lines are so the lines are m m m m Descrie as line D: (, ) (, ) (, ) (, ) m m m m So A is parallel to C, B is parallel to D, A is perpendicular to B and D, and C is perpendicular to B and D. So the quadrilateral can onl e a square or a rectangle. Eercise.A p (, ) (, ) (, ) (, ) a False: m and m so m m so the lines are not perpendicular. False: c and c so the lines do not share the same intercept. c False: negative gradient means the oject is travelling ack to the start. d True: speed is the gradient and the gradients are the same (the difference in the signs just indicates direction) so the are travelling at the same speed. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS e False: m m so the lines are not parallel. f False: positive gradient means the oject is travelling awa from the start. We have een given two pairs of coordinates for each printer. The cost of the printer is actuall the start of the costs when, i.e. the intercept. We have also een given the coordinates when. Anna s printer: (, ) (, ) (, ) (, 9) 9 9 + Bhavini s printer: (, ) ( 7,. ) (, ) (, 7. ) 7. 7. 7. 7. 7. + 7. m is the cost per ear to run the printer and is in Anna s case and in Bhavini s case. c is the cost to u the printer and is in Anna s case and 7. in Bhavini s case. The cost of one ink cartridge for Anna s printer is. The cost of one ink cartridge for Bhavini s printer is. Nothing is significant aout the intercepts. Although ou can mathematicall work out the intercept in each case, this would actuall e a negative value which would not make sense as ou cannot have a negative numer of ears. Tai firm A: (, ) (, ) (, ) (, ) When,. When 7,. Tai firm B: (, ) (, ) (, ) (, ) + When,. When 7,. Tai firm C: (, ) (, ) (, ) (, ) + When,. When 7, 9. Tai firm A is the cheapest on a -mile journe. Tai firm C is the cheapest on a 7-mile journe. (, ) ( 7., ) (, ) ( 7,. ) 7. 7.. 7 7. 7.. 7. 7 7 7 7 + 7 HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
COORDINATE GEOMETRY : EQUATIONS OF STRAIGHT LINES Eam-stle questions p m, (, ) (, ) m ( ) + a Line A: (, ) ( 7, ) (, ) (, ) m m 7 m m Line B: (, ) (, ) (, ) (, ) m m m Line A is steeper ecause >. m mm so m (, ) (, ) m + a Runner A starts miles awa from runner B s home. The two runners meet 9 miles awa from runner B s home so runner A will have run miles. Runner A: (, ) (, ) (, ) ( 9, ) 9 + Runner B: (, ) (, ) (, ) ( 9, ) 9 9 c Runner B runs faster. Gradient speed and >. a m mm so m (, ) (, ) ( ) m + + When, intercept (, ) When, intercept (, ) c Area of triangle h units Line AB: (, ) (, ) (, ) (, ) m m m Line BC: (, ) (, ) (, ) (, ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS m m m mm So AB and BC are perpendicular to each other and ABC is a right-angled triangle. 7 Pa as ou go: (, ) (, ) (, ) (, ) 7 7 + For this option: rides: so 7 rides: 7 so 9 For other option (ook of tickets): Cost of rides + so pa as ou go option is cheaper. Cost of 7 rides + ( ) so ook of tickets option is cheaper. Line A: (, ) (, ) (, ) (, a) m a a a m, (, ) (, ) m m m Gradient perpendicular to this: m (, ) (, ) m( ) ( ) + + + ( )+ + Solving gives 9 7 Sustituting this value ack into the original equation gives 7 Point of intersection (, ), 9 ( 7 7 ) + + Rearranging gives m So m m( ) 9 7 a Line A: + 7 (, ) (, ) (, ) (, ) 9 9 + 7 9 (, ) (, ) (, ) (, ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 7
COORDINATE GEOMETRY : CIRCLES Line B: (, ) (, ) (, ) (, ) 9 ( ) 9 The two equations are the inverses of each other. At the point of intersection: 9 + ( ) 9 Rearranging gives Sustituting ack into one of the original equations gives. So the point of intersection is (, ). c The transformation linking the two lines is a reflection in the line. 9 7 Coordinate geometr : Circles Prior knowledge p 7 Gradient 9 ( ) Gradient of first line Perpendicular gradient Equation of perpendicular line through (, ) is given ( ). + 9 Sustitute into + 7. + ( ) 7 + 7 7 7 9 When 9, 9 Hence point of intersection (9, ). Complete square on +. + ( 7) 9 + ( 7) Eercise.A p 7 9 d The point of intersection is the onl point where the temperature in degrees Celsius and the temperature in degrees Fahrenheit have the same numerical value, i.e. at (, ). a Centre (, ), radius Centre (9, ), radius c Centre (, ), radius d Centre (, ), radius 7 a ( + ) + ( 9) 9 and + + + 7 ( + ) + ( + ) 9 and + + + 7 c ( ) + and + 9 d ( ) + ( ) and + + a ( ) + ( + ) ( + ) + ( ) 9 c ( + ) + ( ) a ( 9) + ( + 7) 9 ( 9) + ( + 7) Centre (9, 7), radius ( + ) + 9 ( + ) + Centre (, ), radius HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS c + + + + 79 ( + ) + ( + 9) + 79 ( + ) + ( + 9) 7 Centre (, 9), radius d + + + ( + ) + ( ) 9 + d ( ) + ( ) r ( ) + (7 ) ( + ) + ( ) Centre (, ), radius 9 r ( ) + () The circle has the equation ( + ) + ( 7). ( 9) + ( ) 7 d ( ) + ( ) d ( ) + (7 ) d () + () The circle has the equation ( ) + ( ). The locations on the -ais will lie on the circumference of the circle. When, ( ) + ( ) ( ) + 9 ( ) ± ± or The circle cuts the -ais at (, ) and (, ). Rewrite the equation of the circle completing the square. ( + p) p + ( + ) 9 9 ( + p) + ( + ) p + 9 + 9 Since r, p + 9 + 9 p + 9 + 9 p p Since p is a positive constant, p. The centre of the circle has the coordinates ( p, ). Since p, the centre has the coordinates (, ). The centre is units from the origin. Eercise.B p d ( ) + ( ) d ( 7 7) + ( ( )) d ( ) + () d HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 a The centre is the midpoint etween D and E. Centre + 9 7 (, ) (, ) The radius is the distance etween the centre and a point on the circumference. Let the centre (, ) (, ) and D(, 9) (, ). d ( ) + ( ) r ( ) + ( 9) r () + ( ) r c The circle has the equation ( ) + ( ). Centre of circle + (, ) (9, ) r (9 ( )) + ( ) r () + ( ) 9 The circle has the equation ( 9) + ( ) 9. a The student has correctl worked out that AB 9, ut the equation of the circle needs the square of the radius, not the square of the diameter. The diameter is 9 7, so the radius is 7. ( +.) + ( 7) ( 7 ) a + 9 +, (, ) + 9 + d ( ) + ( ) r (9 ) + ( ( )) r () + () r c ( ) + ( + ) d Sustitute the coordinates of H into the equation of the circle. ( ) + ( + ) ( ) + () Since oth sides of the equation are satisfied, H lies on C. a The centre of the circle is the midpoint of ST 9+ + (, ) (, 7). The midpoint of UV will also e the centre of the circle, (, 7). Hence + p + q, (, 7) + p p + q 7 q 9
COORDINATE GEOMETRY : CIRCLES r ( 9 ( )) + ( 7) r ( 7) + ( ) The circle has the equation ( + ) + ( 7). 7 The equation of the circle can e rewritten as ( + ) + ( ), so (, ) is the centre of the circle. Sustitute + into + +. + ( + ) + ( + ) Epand and simplif. + + + 9 + + + Solve. ( + )( ) or M and N are the two intersection points, it doesn t matter which is which. When, +, so M (, ) When, +, so N (, ) The midpoint of MN + + (, ) (, ) Since the midpoint of MN is the centre of the circle, MN is a diameter of the circle. Eercise.A p 7 a d ( ) + ( ) DE ( ) + ( ) DE ( ) + () EF ( ) + ( 9) EF ( ) + ( ) DF ( ) + ( 9) DF ( ) + () Since +, DE + EF DF. Gradient of DE Gradient of EF 9 Since m m, DE and EF are perpendicular. c DF is a diameter of the circle. Centre + + (, 9 ) (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r () + () The equation of the circle is ( ) + ( ). a Prove using Pthagoras theorem (could compare gradients). PQ ( ) + ( ) ( ) + ( ) PR ( ) + ( ) ( ) + ( ) QR ( ) + ( ) ( ) + () Since PR + QR PQ, Pthagoras theorem is satisfied and the triangle PQR is right-angled. Centre + + (, ) (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r ( ) + ( ) Area of triangle h Area of circle πr π π Green area (π ) m c π. m.., so tins are required. Since AB is a diameter, ACB 9 and the lengths of the sides satisf Pthagoras theorem. AC + BC AB ( 9) + ( ) ( + ) 7 + + 9 + + + + ( )( ) or When, d + 9 Radius 9 9 When, d + 7 Radius 7 7 a Sustitute and into ( + ) + ( ) r. ( + ) + ( ) r () + ( ) r r Sustitute and q into ( + ) + ( ) ( + ) + (q ) ( ) + (q ) (q ) 9 q ± q ± q or Since q >, q HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS c Gradient of AB Gradient of AC ( ) Gradient of BC ( ) d m m so AC and BC are perpendicular. Since the angle in a semicircle is a right angle, AB is a diameter of the circle. a Gradient of L ( ) Gradient of L ) 9 m m so L and L are perpendicular. Because L and L are perpendicular and intersect at point W on the circumference, SV is a diameter. Centre + (, ) (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r ( ) + () The equation of C is given ( ) + ( ) Epanding and simplifing: + + a PR PQ + QR PR ( (t + 7)) + ((t + 7) ) ( t) + (t + ) t + t + 7 PQ ( ) + ((t + 7) (t + )) ( ) + ( ) QR ( (t + 7)) + ((t + ) ) ( t) + (t + ) t + t + Hence t + t + 7 t + t + + t t Hence the coordinates of P and R are (, ) and (, ). Centre of circle + + (, ) (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r (7) + ( ) The equation of C is given ( ) + ( ). JL JK + KL JL ( 7) + ( ) ( ) + ( ) JK ( ( )) + ( (p )) ( ) + ( p) p p + 9 KL ( 7) + ((p ) ) ( ) + (p ) p p + 9 Hence p p + 9 + p p + 9 p p + p p + (p )(p ) p or Eercise.A p a 9+ 7, (, ) Gradient of DE ( ) 9 7 c Gradient of FG d Equation is given ( ). +. e When, ( ) + as required. f For the radius: d ( ) + ( ) r (9 ) + ( ) r () + () The equation is given ( ) + ( ). Gradient of JK 7 7 Gradient of isector 7 Midpoint + 7+ (, ) 9 (, ) Equation of isector is given 9 7 ( ) When, 9 7 ( ) 7 7 9 7 9 Centre of circle (9, ) For the radius: d ( ) + ( ) r (9 ) + ( 7) r () + () 7 The equation is given ( 9) + ( ) 7. a Midpoint + +, (, ) + HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
COORDINATE GEOMETRY : CIRCLES + Coordinates of V are (, ). Gradient of UV Gradient of isector Equation of isector is given ( ). When, ( ) Centre of circle (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r ( ) + ( 9) The equation is given ( ) + ( ). d 7 d + m a Equation of circle can e rewritten as ( + ) + ( + ) Hence centre of circle (, ) When, ( + ) + ( + ) ( + ) + ( + ) + ± or Length of chord ( ) c Height of triangle ( ) Base of triangle PQ Area of triangle h a Midpoint of AB + + (, 7 ) (, ) 7 Gradient of AB Gradient of isector Equation of isector is given ( + ) + 7 Midpoint of BC + 7 (, ) (, ) Gradient of BC 7 ( ) Gradient of isector Equation of isector is given ( ). + 9 c 9 ecause gradients are perpendicular. d Gradient AB from a Gradient BC from Hence AB is perpendicular to BC and angle ABC 9. e AC is a diameter. 7 For each circle, let the first point A, the second B and the third C. a A(, ), B(, ) and C(, ) Midpoint of AB + + (, ) 9 (, ) Gradient of AB 7 Gradient of isector 7 Equation of isector is given 9 7 ( ) 7 + 7 Midpoint of AC + + (, ) (, ) Gradient of AC Gradient of isector Equation of isector is given ( ) For point of intersection, 7 + 7 7 7 When, Centre of circle (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r () + ( ) The equation is given ( ) + ( ). A(, ), B(, ) and C(, ) Midpoint of AB + + (, ) (, ) Gradient of AB Gradient of isector Equation of isector is given ( ) + 7 Midpoint of AC + (, ) (, ) Gradient of AC ( ) Gradient of isector HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS Equation of isector is given ( ( )) + For point of intersection, + 7 + When, + Centre of circle (, ) For the radius: d ( ) + ( ) r ( ) + ( ) r ( ) + () The equation is given ( + ) + ( ). c A(9, ), B(7, ) and C(, ) Midpoint of AB 9 + 7 + (, ) (, ) Gradient of AB 9 7 Gradient of isector Equation of isector is given ( ) + Midpoint of AC 9 + (, ) (, ) Gradient of AC ( ) 9 Gradient of isector Equation of isector is given ( ) + For point of intersection, + + When, + Centre of circle (, ) For the radius: d ( ) + ( ) r ( 7) + ( ) r ( 7) + () Equation of isector is given + ( + ) For point of intersection, + When, + Centre of circle (, ) For the radius: d ( ) + ( ) r ( + ) + ( + ) r ( 7) + ( 9) The equation is given ( + ) + ( + ). a Circle A: + + 7 Circle B: + + Sutract from. 9 7 Hence the line on which the circles intersect is given 7. Sustitute 7 into the equation for Circle A. + ( 7) + ( 7) 7 + + ( )( ) or When, 7 When, 7 Points of intersection are (, ) and (, ). Length of chord: d ( ) + ( ) d ( ) + ( ( )) d () + () The length of the chord is is The equation is given ( ) + ( ). Eercise.A p d A(, ), B( 7, 9) and C(, ) Midpoint of AB 7 (, 9) (, ) Gradient of radius Gradient of AB ) 7 9 7 ( ) Gradient of the tangent The equation of the tangent is given Gradient of isector ( 7) Equation of isector is given + ( + ) + + + 7 Midpoint of AC (, ) (, ) Gradient of AC ( ) a + ( ) + 9 Since oth sides of the equation agree, T lies on ( ) Gradient of isector the circle. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
COORDINATE GEOMETRY : CIRCLES Centre of circle (, ) Gradient of radius ( ) Gradient of the tangent The equation of the tangent is given + ( ) + Hence + a Centre (, ) Radius c Let centre X, point P (, 7) and point where tangent meets circle T. The length of PT is required, where PT + XT PX For PX: d ( ) + ( ) PX ( ) + ( 7) PX ( ) + ( ) PT + XT PX PT + PT PT 9 m For A: Gradient of radius 7 7 Gradient of the tangent The equation of the tangent is given 7 ( ) Hence ( ) + 7 For B: Gradient of radius 7 7 Gradient of the tangent The equation of the tangent is given 7 ( ) Hence ( ) + 7 Put tangents equal. ( ) + 7 ( ) + 7 ( ) ( ) ( ) + 7 (this could also e deduced smmetr since AB is horizontal) When 7, (7 ) + 7 ( ) + 7 AXBQ is a kite. Split into congruent triangles AXQ and BXQ. Base of triangle AXQ 9 Perpendicular height of triangle AXQ Area of triangle AXQ 7 Area of AXBQ 7 7 Rewrite the equation of the circle as ( + ) + ( ) Centre (, ), radius 9 Let centre X and point where tangent meets circle T. PT + XT PX For PX: d ( ) + ( ) PX (7 ( )) + ( ) PX () + () PT + XT PX PT + 9 PT 9 PT 9 7 a Rewrite the equation of the circle as ( ) + ( ) Centre (, ), radius Gradient of radius Gradient of the tangent The equation of the tangent is given ( ) + + 9 When, q + q 9 q 9 q Use d ( ) + ( ) For PT: PT ( ) + ( ) PT ( ) + () For PX: PX ( ) + ( ) PX ( ) + () TX is the radius, so TX Hence PT + TX PX ecause +. 9 Hence the coordinates of Q are (7, 9 ). HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS c Since triangle is right-angled, area TX PT. TX PT Area 7 a Gradient of tangent Gradient of radius Equation of line with gradient through (, ) is given ( ) Find point of intersection of and +. + 7 7 + 9 Hence the point (7, 9) lies on the circumference. For the radius: d ( ) + ( ) r (7 ) + (9 ) r () + ( ) The equation of the circle is given ( ) + ( ) Gradient of tangent Gradient of radius Equation of line with gradient through (, ) is given ( + ) ( + ) + Find point of intersection of ( + ) + and +. ( + ) + + ( + ) + 9 9 + + 9 9 +.. +. Hence the point (.,.) lies on the circumference. For the radius: d ( ) + ( ) r (.) + (.) r (.) + (.) 7. The equation of the circle is given ( + ) + ( ) 7. Eercise.B p a Sustitute into +. + ( ) + + + + ( ) The equation has repeated roots so is a tangent of the circle. Sustitute 7 into +. (7 ) + 9 7 + + 7 + + 9 ( 7) The equation has repeated roots so 7 + is a tangent of the circle. Sustitute into + + +. + ( ) + ( ) + + + + 7 + 9 + 7 + + 9 ( ) The equation has repeated roots so is a tangent of the circle. Sustitute + c into +. + ( + c) ( + c) + 9 + c + c c + (c ) + (c c ) For ac : (c ) (c c ) c c + 7 c + c + c 9c + c + c (c )(c + ) c or Sustitute + c into + +. + ( + c) ( + c) + + c + c + c + + ( c) + (c c + ) + ( c) + (c c + ) HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
COORDINATE GEOMETRY : CIRCLES For ac : ( c) (c c + ) c + c c + c c + c 99 c c + (c )(c ) c or The tangents are + and +. Sustitute m + into ( ) + ( ). ( ) + (m + ) ( ) + (m + ) + + m + m + 9 ( + m ) + (m ) + For ac : (m ) ( + m ) 7m m + 9 7 7m m m 7 m + m + (7m + )(m + ) m 7 or For the tangent 7 + : 7 + + 7 For the tangent + : + + Eam-stle questions p a The equation of the circle is given ( a) + ( ) r, so centre (9, ). Gemma has copied the signs from the equation ut needed to change them oth. Olivia has stated the length of the radius, not the diameter. Radius Diameter Radius Equation is given ( 7) + ( + ) Epand. + 9 + + + 9 + + On the -ais,. ( + ) + ( 7) ( + ) + ± ± or Coordinates are (, ) and (, ). On the -ais, ( + ) + ( 7) () + ( 7) ( 7) 7 ± 7 ± or Coordinates are (, ) and (, ). Dilgusha has not halved the coefficients of and when completing the square. She has added rather than sutracting. When square-rooting, which shouldn t have een possile in this situation, she has squarerooted + and made it negative. Correct solution: Complete the square on and. ( + ) 9 + ( 7) 9 + ( + ) + ( 7) Centre (, 7) Radius a d ( ) + ( ) r ( ) + ( ) r () + ( 9) Radius cm The circle has the equation ( ) + ( ) c Gradient of radius Gradient of the tangent The equation of the tangent is given ( ) 7 + d d ( ) + ( ) d (9 ) + ( ) d (7) + () Distance 7 cm 7 m e Unlikel athlete will remain at same point whilst turning. Hammer s motion will not e horizontal ecause it will e affected gravit. HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979
WORKED SOLUTIONS a Midpoint of AB 9+, (, ) For the radius, d ( ) + ( ) r ( 9) + ( ) r () + ( ) The circle has the equation ( ) + ( ). Gradient of radius 9 Gradient of the tangent The equation of the tangent is given ( 9) 9 + 7 Centre of circle, X ( 9, ) Radius 9 7 PX ( ( 9)) + ( ) PX (7) + (7) 7 PT PX XT PT 7 7 9 PT 9 a r ( 7) + ( ) r () + () 7 Equation is given ( 7) + ( ) 7. Midpoint of XA 7+ + (, ) (9, 7) Gradient of XA 7 Gradient of the tangent The equation of the tangent is given 7 ( 9) or +. c Let P e the midpoint of XA. XM XA AM ecause XM and XA are radii and since MN is the isector of XA it is a line of smmetr for triangle MAX (hence XM AM). MP MX XP MP 7 MP 7 MN 9 a ( + ) + ( ) 7 ( + ) + (7 ) ( + ) + ( ) + + + + + + + 9 ( + 9)( ) 9 or HarperCollinsPulishers 7 Edecel A-level Mathematics Year and AS 979 When 9, 7 ( 9) When, 7 Since the -coordinate of A is less than the -coordinate of B, A is ( 9, ) and B is (, ). c Tangent at A: Gradient of radius 9 ( ) Gradient of the tangent (Note that since the centre of the circle lies on + 7, AB is a diameter, so the tangent is perpendicular to + 7.) The equation of the tangent is given ( + 9) Hence + Tangent at P: Gradient of radius ( ) 7 Gradient of the tangent 7 The equation of the tangent is given 7 ( + ) Hence 7 ( + ) + From which, + 7 ( + ) + + 7 7 + 7 + 7 + When, + 7 Coordinates of T are (, 7 ). a Equation is given ( ) + ( ) 9 The centre is (, ). The radius is. Method : Similar triangles PTX and PAO: XP OP TX OA Let length of XP a a + ( +a) a + ( + a) a + ( + a) a + 9 + a + a a a a a (a )(a + ) a (can t e ) Method : Since m at T, ( ) + (m ) 9 + + m m + 9 9 ( + m ) + ( m) + 7