Preparation for A Level Wadebridge School

Transcription

1 Preprtion for A Level Wdebridge School Bridging the gp between GCSE nd A Level Nme:

2 CONTENTS Chpter Removing brckets pge Chpter Liner equtions Chpter Simultneous equtions 6 Chpter Fctorising 7 Chpter Chnge the subject of the formul 9 Chpter 6 Solving qudrtic equtions Chpter 7 Indices Chpter 8 Surds 6 Chpter 9 Stright lines 8 Chpter 0 Sttistics Solutions to the exercises 7 The following book is covering prt of the topics needed AS-Level Mths Hed Strt Published by CGP Workbooks ISBN: Cost:.9

3 Chpter : REMOVING BRACKETS To expnd two brckets, we must multiply everything in the first brcket by everything in the second brcket. We cn do this in vriety of wys, including * the smiley fce method * FOIL (Fronts Outers Inners Lsts) * using grid. Exmple: (x - )(x + ) = x(x + ) - (x +) = x + x x - 6 = x x 6 or (x - )(x + ) = x 6 + x x = x x 6 or x - x x -x x -6 (x +)(x - ) = x + x - x - 6 = x - x - 6 Two Specil Cses Perfect Squre: Difference of two squres: (x + ) = (x + )(x + ) = x + x + (x - )(x + ) = x (x - ) = (x )(x ) = x x + 9 (x - )(x + ) = x = x 9 EXERCISE A. (x + )(x + ) Multiply out the following brckets nd simplify.. (t - )(t - ). (x + y)(x y). (x - )(x + ). (y - )(y + ) 6. ( + x)( x) 7. (7x - ) 8. (y - )(y + )

4 Chpter : LINEAR EQUATIONS Exmple : Solve the eqution x x Solution: Step : Find the lowest common denomintor: The smllest number tht both nd divide into is 0. Step : Multiply both sides by the lowest common denomintor Step : Simplify the left hnd side: 0( x) 0( x) 0 0 ( x ) 0 ( x ) 0 (x + ) + (x + ) = 0 Step : Multiply out the brckets: x + + x + 8 = 0 Step : Simplify the eqution: 9x + = 0 Step 6: Subtrct 9x = 7 Step 7: Divide by 9: x = Exmple : Solve the eqution x x x 6 Solution: The lowest number tht nd 6 go into is. So we multiply every term by : ( x) ( x) x 6 Simplify x ( x ) ( x) Expnd brckets x x 6 6 0x Simplify x 6 8 0x Subtrct 0x x 6 8 Add 6 x = Divide by x =.8

5 Exercise A: Solve these equtions ) ( ) x y y ) ) 7x x ) x x x FORMING EQUATIONS Exmple 8: Find three consecutive numbers so tht their sum is 96. Solution: Let the first number be n, then the second is n + nd the third is n +. Therefore n + (n + ) + (n + ) = 96 n + = 96 n = 9 n = So the numbers re, nd. Exercise B: ) Find consecutive even numbers so tht their sum is 08. ) The perimeter of rectngle is 79 cm. One side is three times the length of the other. Form n eqution nd hence find the length of ech side. ) Two girls hve 7 photogrphs of celebrities between them. One gives to the other nd finds tht she now hs hlf the number her friend hs. Form n eqution, letting n be the number of photogrphs one girl hd t the beginning. Hence find how mny ech hs now.

6 Chpter : SIMULTANEOUS EQUATIONS An exmple of pir of simultneous equtions is x + y = 8 x + y = In these equtions, x nd y stnd for two numbers. We cn solve these equtions in order to find the vlues of x nd y by eliminting one of the letters from the equtions. In these equtions it is simplest to eliminte y. We do this by mking the coefficients of y the sme in both equtions. This cn be chieved by multiplying eqution by, so tht both equtions contin y: x + y = 8 0x + y = = To eliminte the y terms, we subtrct eqution from eqution. We get: 7x = i.e. x = To find y, we substitute x = into one of the originl equtions. For exmple if we put it into : 0 + y = y = Therefore the solution is x =, y =. Remember: You cn check your solutions by substituting both x nd y into the originl equtions. Exmple: Solve x + y = 6 x y = Solution: We begin by getting the sme number of x or y ppering in both eqution. We cn get 0y in both equtions if we multiply the top eqution by nd the bottom eqution by : 8x + 0y = 6 x 0y = As the SIGNS in front of 0y re DIFFERENT, we cn eliminte the y terms from the equtions by ADDING: x = 69 + i.e. x = Substituting this into eqution gives: 6 + y = 6 y = 0 So y = The solution is x =, y =. 6

7 Exercise A: Solve the pirs of simultneous equtions in the following questions: ) x + y = 7 x + y = 9 ) 9x y = x y = 7 ) + b = b = Chpter : FACTORISING Common fctors We cn fctorise some expressions by tking out common fctor. Exmple : Fctorise 9x y 8x y Solution: 9 is common fctor to both 9 nd 8. The highest power of x tht is present in both expressions is x. There is lso y present in both prts. So we fctorise by tking 9x y outside brcket: 9x y 8x y = 9x y(xy ) Exmple : Fctorise x(x ) (x ) Solution: There is common brcket s fctor. So we fctorise by tking (x ) out s fctor. The expression fctorises to (x )(x ) Exercise A Fctorise ech of the following ) pq - 9q ) 8 b b ) y(y ) + (y ) Fctorising qudrtics Simple qudrtics: Fctorising qudrtics of the form x bx c The method is: Step : Form two brckets (x )(x ) Step : Find two numbers tht multiply to give c nd dd to mke b. These two numbers get written t the other end of the brckets. 7

8 Exmple : Fctorise x 9x 0. Solution: We need to find two numbers tht multiply to mke -0 nd dd to mke -9. These numbers re -0 nd. Therefore x 9x 0 = (x 0)(x + ). Generl qudrtics: Fctorising qudrtics of the form x bx c The method is: Step : Find two numbers tht multiply together to mke c nd dd to mke b. Step : Split up the bx term using the numbers found in step. Step : Fctorise the front nd bck pir of expressions s fully s possible. Step : There should be common brcket. Tke this out s common fctor. Exmple : Fctorise 6x + x. Solution: We need to find two numbers tht multiply to mke 6 - = -7 nd dd to mke. These two numbers re -8 nd 9. Therefore, 6x + x = 6x - 8x + 9x = x(x ) + (x ) (the two brckets must be identicl) = (x )(x + ) Difference of two squres: Fctorising qudrtics of the form x Remember tht x = (x + )(x ). Therefore: x x x x 9 ( )( ) 6x ( x) (x )(x ) Also notice tht: nd x 8 ( x ) ( x )( x ) x 8xy x( x 6 y ) x( x y)( x y) Fctorising by piring We cn fctorise expressions like ) ) 0) 8 x xy x y using the method of fctorising by piring: x xy x y = x(x + y) (x + y) (fctorise front nd bck pirs, ensuring both brckets re identicl) = (x + y)(x ) Exercise B Fctorise ) x 6x 6 ) x x ) y 7y 0x x 0 x x x xy y

9 Chpter : CHANGING THE SUBJECT OF A FORMULA Exmple: ( F ) The formul C is used to convert between Fhrenheit nd 9 Celsius. We cn rerrnge to mke F the subject. ( F ) C 9 Multiply by 9 9C ( F ) (this removes the frction) Expnd the brckets 9C F 60 Add 60 to both sides 9C60 F Divide both sides by 9C 60 F 9C 60 Therefore the required rerrngement is F. Exmple : Mke the subject of the formul t h Solution: Multiply by Squre both sides Multiply by h: Divide by : t h t h 6t h 6t h 6th Exercise A Mke x the subject of ech of these formule: ) y = 7x ) ) x y ) ) P t g y y x (x ) 9 9

10 More difficult exmples Sometimes the vrible tht we wish to mke the subject occurs in more thn one plce in the formul. In these questions, we collect the terms involving this vrible on one side of the eqution, nd we put the other terms on the opposite side. W Exmple : Mke W the subject of the formul T W b Solution: This formul is complicted by the frctionl term. We begin by removing the frction: Multiply by b: bt bw W Add bw to both sides: bt W bw (this collects the W s together) Fctorise the RHS: bt W( b) Divide both sides by + b: W bt b Exercise B Mke x the subject of these formule: ) x bx c ) ( x ) k( x ) ) x y x x x ) b 0

11 Chpter 6: SOLVING QUADRATIC EQUATIONS A qudrtic eqution hs the form x bx c 0. There re two methods tht re commonly used for solving qudrtic equtions: * fctorising * the qudrtic formul Note tht not ll qudrtic equtions cn be solved by fctorising. The qudrtic formul cn lwys be used however. Method : Fctorising Mke sure tht the eqution is rerrnged so tht the right hnd side is 0. It usully mkes it esier if the coefficient of x is positive. Exmple : Solve x x + = 0 Fctorise (x )(x ) = 0 Either (x ) = 0 or (x ) = 0 So the solutions re x = or x = Note: The individul vlues x = nd x = re clled the roots of the eqution. Exmple : Solve x x = 0 Fctorise: x(x ) = 0 Either x = 0 or (x ) = 0 So x = 0 or x = Method : Using the formul Recll tht the roots of the qudrtic eqution b x x bx c 0 re given by the formul: b c Exmple : Solve the eqution x 7 x Solution: First we rerrnge so tht the right hnd side is 0. We get We cn then tell tht =, b = nd c = -. Substituting these into the qudrtic formul gives: x x 0 ( ) 0 x (this is the surd form for the solutions) If we hve clcultor, we cn evlute these roots to get: x =.8 or x = -.

12 EXERCISE 6A ) Use fctoristion to solve the following equtions: ) x + x + = 0 b) x = x ) Find the roots of the following equtions: ) x + x = 0 c) x = 0 ) Solve the following equtions either by fctorising or by using the formul: 6x - x = 0 ) Use the formul to solve the following equtions to significnt figures. Some of the equtions my not hve solutions. ) x +7x +9 = 0 c) x x = 7 e) x + x + = 0

13 Chpter 7: INDICES Bsic rules of indices y mens y y y y. is clled the index (plurl: indices), power or exponent of y. There re bsic rules of indices: ) ) ) ( ) m n m n e.g. m n m n e.g. m n mn e.g. 0 Further exmples y y y 7 6 (multiply the numbers nd multiply the s) c c 6 6c 8 (multiply the numbers nd multiply the c s) 7 7 d d d 8d d (divide the numbers nd divide the d terms i.e. by subtrcting the powers) Exercise 7A Simplify the following: ) ) ) ) b c bc = 6 n ( 6 n ) = 8 8n n = d d = 9 ) = 6) d =

14 More complex powers Zero index: Recll from GCSE tht 0. This result is true for ny non-zero number. 0 Therefore Negtive powers A power of - corresponds to the reciprocl of number, i.e. Therefore (you find the reciprocl of frction by swpping the top nd bottom over) This result cn be extended to more generl negtive powers: This mens: Frctionl powers: Frctionl powers correspond to roots: / / / In generl: /n n Therefore: / 8 8 / n. n / m / n / n A more generl frctionl power cn be delt with in the following wy: / 8 So / / / / m

15 Exercise 7B: Find the vlue of: ) ) / / 7 ) / 9 ) ) 6) 7) 8) 9) / 7 8 / 0) 0.0 / ) ) / / Simplify ech of the following: ) ) x / / x ) / xy

16 6 Chpter 8: SURDS Surds re squre roots of numbers which don t simplify into whole (or rtionl) number: e.g.... but it is more ccurte to leve it s surd: Generl rules b b b b But you cnnot do: b b These re NOT equl b b b b b b ) )( ( Simplifying Surds Find the lrgest squre numbers nd simplify s fr s possible Worked Exmples Creful - this is times the squre root of NOT the cube root of Rtionlising the Denomintor This is fncy wy of sying getting rid of the surd on the bottom of frction. We multiply the frction by the denomintor (or the denomintor with the sign swpped) Worked Exmples. e Rtionlis we multiply by which is the sme s multiplying by, which mens we don t fundmentlly chnge the frction.. 0 e Rtionlis. 0 0 ) ( ) ( ) ( ) ( Rtionlis e ) ( ) ( ) ( ) ( e Rtionlis

17 Exercise 8A: Simplify the surds ) ) ) 8 Exercise 8B: Expnd nd simplify ) ) 6 8 ) ( )( ) ) ( )( ) ) ( 8 )(8 ) 6) ( )( ) Exercise 8C: Rewrite the following expressions with rtionl denomintors ) 9 ) 8 ) ) 6) 7) 7 7 7

18 Chpter 9: Stright line grphs Liner functions cn be written in the form y = mx + c, where m nd c re constnts. A liner function is represented grphiclly by stright line, m is the grdients nd c is the y- intercept of the grph. Exmple : Drw the grph of y = x + Solution: Step : Mke tble of vlues Step : Use your tble to drw the stright line grph Exmple : Plot the stright line using the grdient nd y intercept Solution: Step : Mrk on the y xis the y-intercept = Step : The grdient= so strt from the y intercept for every unit cross to the right go down by hlf unit nd mrk second point there. Step : Join the y intercept with the new point with line nd extend form both sides.. Here re some exmples of liner functions not ll of them in the form y = mx + c. You need to be confident into rerrnging the functions mking y the subject in order to identify the grdient nd y- intercept. y = x + x - y + = 0 y - x = so y x so y x grdient= grdient= grdient= y-intercept= y-intercept= y-intercept= 8

19 To find the y-xis crossing, substitute x = 0 into the liner eqution nd solve for y. To find the x-xis crossing, substitute y = 0 into the liner eqution nd solve for x. Exmple : Rewrite the eqution y - x = into the form y = mx + c, find the grdient nd the y- intercept Solution: Step : Add x to both sides (so tht the x term is positive): y = + x Step : Divide by both sides: y x Step : Identify the grdient nd y-intercept grdient= y-intercept= Exmple : Find the grdient of the line which psses through the points A (, ) nd B (-, ) Solution: Step : Use the x nd y vlues of A x, ) nd B x, ) Step : find the grdient m y x y x ( y ( y m Finlly you need to be ble to find the eqution of line from grph. 9

20 Exmple : Find the eqution of the stright line which psses through the point (, ) nd hs grdient Solution: Step : Find where the line crosses the y xis. This is the y intercept, c. Line crosses y xis t, so y-intercept c= Step : Drw tringle below the line from the intercept to point you know y y And work out the grdient between the two points m x x Grdient tringle from (-6,) to (0,) so m 0 6 Step : Write in the form y = mx + c y x Exercise 9A: Plot the grph of ech function tking the given vlues ) y= x - ( x = - to ) b) y=- x + ( x = - to ) c) y = x ( x = - to ) d) y= -x + ( x = - to ) 6 Exercise 9B: Rewrite the equtions below into the form y = mx + c, find the grdient nd the y-intercept )x y = 0 b) x + y 8 =0 c) = x y Then plot the grph of ech eqution Exercise 9C: Work out the grdient between the sets of coordintes ) A ( 0, ) nd B(, 6) b) A (, 0) nd B(, -) c) A (, -) nd B(, -) d) A ( -, ) nd B(, ) e) A (, 0.) nd B(, -) f) A ( -7, -) nd B( -, -6) 0

21 Exercise 9D: Find the eqution of these lines in the form...

22 Chpter 0: Sttistics Clcultions Mens from frequency distributions Exmple Mens from grouped dt Find the mid-point of ech group nd then multiply by frequency. Sum nd then divide by totl frequency Exmple Averges Rnge = difference between highest nd lowest observed vlues Interqurtile rnge = difference between upper qurtile (Q) nd lower qurtile (Q)

23 Chrts nd Grphs Box plots Lowest vlue Lower qurtile Medin Upper qurtile Highest vlue Outliers Any vlues. x IQR bove the UQ or below the LQ re considered to be n outlier Histogrms The frequency density is the frequency of vlues divided by the clss width of vlues. The re of ech block/br represents the totl of frequencies for prticulr clss width. The width of the block/br(long the x-xis) reltes to the size of the clss width. So the width of block/br cn vry within histogrm. The frequency density is lwys the y-xis of histogrm. Histogrms re only used for numericl continuous dt tht is grouped. Exmple Here is tble of dt similr to the lst one but with vlues of height grouped differently using inequlities. Note: becuse the clss is grouped using inequlities, one 'equl to nd greter' nd the other 'less thn, the clss width is stright subtrction of the two numbers mking up the clss group. clss clss width frequency frequency density (height - h) cm (f) 6 h < 7 0 /0 = 0. 7 h <80 7 7/ =. 80 h <90 0 /0 =. 90 h <0 / =.0 0 h <0 / =.

24 Significnce of re: The re on histogrm is importnt in being ble to find the totl number of vlues/individul results in the dt. In our histogrm(from the tble), the 6 to 7 block represents children, the 7 to 80 block represents 7 children, nd so on. So one block squre represents one child. If we count the squre blocks in the whole smple we get 7 - the sum of ll the frequencies i.e. the totl number of children tking prt - the number of individul results. Probbility Mutully Exclusive events Events tht cnnot hppen t the sme time Independent events The probbility of one event is not ffected by the probbility of nother event. Exhustive events A set of events is exhustive if the set contins ll possible outcomes. Rules of probbility P( or b ) = P ( ) + P(b) P( nd b) = P( ) x P(b) Tree digrms When completing tree digrm remember ech pir of brnches must dd to mke. As you trvel long the brnches to find possible outcomes you multiply the probbilities. If the is more thn one possible outcome, sum them.

25 Exmple Exercise 0A. A botny student counted the number of disies in ech of rndomly chosen res of m by m in lrge field. The results re summrised in the following stem nd lef digrm. Number of disies mens (7) (7) 0 0 (8) (7) 0 0 (6) () () () (b) (c) Write down the modl vlue of these dt. Find the medin nd the qurtiles of these dt. Showing your scle clerly, drw box plot to represent these dt. () () (). The probbility of collecting key t the first stge is, t the second stge is, nd t the third stge is. () (b) (c) (d) Drw tree digrm to represent the stges of the gme. Find the probbility of collecting ll keys. Find the probbility of collecting exctly one key in gme. Clculte the probbility tht keys re not collected on t lest successive stges in gme.. A keep-fit enthusist swims, runs or cycles ech dy with probbilities 0., 0. nd 0. respectively. If he swims he then spends time in the sun with probbility 0.. The probbilities tht he spends time in the sun fter running or cycling re 0. nd 0. respectively. () Represent this informtion on tree digrm. () () () ()

26 (b) Find the probbility tht on ny prticulr dy he uses the sun. () (). The totl mount of time secretry spent on the telephone in working dy ws recorded to the nerest minute. The dt collected over 0 dys re summrised in the tble below. Time (mins) No. of dys Drw histogrm to illustrte these dt. (Totl mrks) 6 Frequency Density Histogrm of times t The digrm bove shows histogrm for the vrible t which represents the time tken, in minutes, by group of people to swim 00m. () Complete the frequency tble for t. t Frequency 0 6 (b) Estimte the number of people who took longer thn 0 minutes to swim 00m. (c) Find n estimte of the men time tken. () () () 6. Children from school A nd B took prt in fun run for chrity. The times to the nerest minute, tken by the children from school A re summrised in the figure below. School A () Time (minutes) Write down the time by which 7% of the children in school A hd completed the run. For school B the lest time tken by ny of the children ws minutes nd the longest time ws minutes. The three qurtiles were 0, 7 nd 0 respectively. () 6 (b) Drw box plot to represent the dt from school B. () (c) Compre nd contrst these two box plots. ()

27 SOLUTIONS TO THE EXERCISES CHAPTER : Ex A ) x + x + 6 ) t t 0 ) 6x + xy y ) x + x ) y 6) + 7x x 7) 9x 8x + 8) y 9 CHAPTER Ex A ) 7 ) /7 ) ) 9/ Ex B ), 6, 8 ) 9.87, 9.6 ), 8 CHAPTER ) x =, y = ) x =, y = ) = 7, b = - CHAPTER Ex A ) q(p q) ) b ( b ) ) (y )(y + ) Ex B ) (x + 8)(x ) ) (x - )(x + ) ) (y + )(y + 7) ) (x )(x + ) ) (x + )(x ) 6) (x )(x y) CHAPTER Ex A y ) x 7 ) xy ) x( y ) ) Ex B c k y ) x ) x ) x b k y ) x CHAPTER 6 ) ) -, - b) -, ) ) 0, - b), - ) -/, / ) ) -.0, -.70 b) -.0,. c) no solutions CHAPTER 7 Ex A ) b c ) -n 8 ) n ) d ) 6 6) -d 9y 0 x ) b b t Pg Ex B ) ) ) / ) / ) 6) /7 7) 9 8) 9/ 9) ¼ 0) 0. ) /9 ) 6 ) 6 ) x ) xy 7

28 8 CHAPTER 8 ExA ) ) ) ExB 8 6) ) 6 ) ) $ 8 ) 0 ) Ex C ) ) 8 ) )( ( 6) ) ( ) ) ( ) ) 7 7( 7 7 ) CHAPTER 9 ExB : ) ) ) x y c x y b x y ExC : ) 8. ) 7 ) ) ) ) grdient f grdient e grdient d grdient c grdient b grdient

29 Ex D : ). y = - x- b. y = -6x c. y = -x ). y = -0.x - b. y = - x + c. y = -x + ). y = - x b. y = 0.x + c. y = -x - CHAPTER 0. () Mode = B (b) n For Q : = 0. th observtion Q = 7 B n For Q : = = (st & nd) observtions (c) Q = =. M A n For Q : =. nd observtion Q = B Number of Disies Box plot M Scle & lbel M Q, Q, Q A, A. () Tree with correct number of brnches, A,,, A,... A () M K K K K K K K K K K K K K (b) P (All Keys) = M A K ; 0.08 ; 0.08 (c) P(exctly key) = = 0 Ech correct triples dded M 9

30 0 ; ; 0. 6 ; 0.7 A A A A (d) P (Keys not collected on t lest successive stges) triples dded M Ech correct A A A 0 0 A 0 ; ; 0.6 ; 0.7 Alterntive: P (Keys collected on t lest successive stges) M = A A A = 8 A. () Tree with correct number of brnches M 0., 0., 0. A All correct A (b) P(used sun) = (0. 0.) + (0. 0.) + (0. 0.) M A = Sun A. Frequency densities: 0.6,.0,.0, 0., 0., 0.08 M, A Histogrm: Scle nd lbels Correct histogrm. () 8- group, re = 7 = B -0 group, re = = B (b) ( 0) + (0 ) = 0 MA is enough evidence of method for M. Condone 9., 0. insted of 0 etc. Awrd if 0 seen. (c) Mid points re 7.,, 6,.,. M f = 00 B f f 00 MA Swim Run Cycle Use of some mid-points, t lest correct for M No Sun Sun No Sun Sun No Sun B B []

31 6. Their ft f for M nd nything tht rounds to 8.9 for A. () 7 (minutes) B (b) Time (School B) Box & medin & whiskers M Sensible scle B 0, 7, 0 B, B (c) Children from school A generlly took less time B 0% of B 7 mins, 7% of A < 7 mins (similrly for 0) B Medin / Q / Q / of A < medin / Q / Q / of ( or more) A hs outliers, (B does not) B Both positive skew IQR of A < IQR of B, rnge of A > rnge of B Any correct lines B