Stage 11 Prompt Sheet
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- Eric Nicholson
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1 Stge 11 rompt Sheet 11/1 Simplify surds is NOT surd ecuse it is exctly is surd ecuse the nswer is not exct surd is n irrtionl numer To simplify surds look for squre numer fctors 7 = = 11/ Mnipulte expressions in surds dd & sutrct m ± n = (m±n) Exmple 1 + = (+) = Multiply & divide x = Exmple x 1 = 4 = Exmple (4 + )( - ) = = - = Exmple = = 11/ Rtionlise surd denomintors To remove surd from the denomintor multiply the numertor & the denomintor y tht surd Exmple 1 1 = x = 1 1 = = = (Multiply oth top & ottom y 1) (ncel y ) = 11/ lculte with upper & lower ounds If is rounded to nerest x Upper ound = + ½x Lower ound = ½x Exmple: if 1.8 is rounded to 1dp Upper ound = ½(.1) = 1.8 Lower ound = 1.8 ½(.1) = 1.7 lculting using ounds dding ounds Mximum = Upper + upper Minimum = Lower + lower Sutrcting ounds Mximum = Upper - lower Minimum = Lower upper Multiplying Mximum = Upper x upper Minimum = Lower x lower Dividing Mximum = Upper lower Minimum = Lower upper 11/4 lgeric frctions dding & sutrcting lgeric frctions Exmple 1 x + + x (common denomintor is 1) 4 = (x + ) + 4(x ) 1 1 = x + + 4x 1 = 7x 11 1 Exmple - (common denomintor is (x+1)(x+) (x + 1) (x + ) = (x + ) (x + 1) (x+1)(x+) = x + 1 x (x+1)(x+) = x + 7 (x+1)(x+)
2 Simplifying lgeric frctions Exmple: x + x + 1 (fctorise) x -x 4 = (x + 1)(x + 1) (cncel) (x 4)(x + 1) = (x + 1) (x 4) 11/ Solve equtions with frctions x + 4 = 1 ommon denomintor (x-)(x+1) x x + 1 x(x+1)+ 4(x-) =1 (x-)(x+1) x + x + 8x - 1 =1 (x-)(x+1) x + x 1 =1(x-)(x+1) x + x -1 = x x - (-x from oth sides) x -1 = x x - (-x from ech side) -1 = x 1x - (+1 to ech side) = x 1x + (fctorise) (x + )(x + 1) = x = - or x = -1 11/ Solve qudrtic eqution y fctoring ut eqution in form x + x + c = x =x + x x = Fctorise the left hnd side (x )(x + 1) = Equte ech fctor to zero x = or x + 1 = x =. or x = -1 11/7 Interpret expressions s functions function is rule tht tkes numers s inputs nd ssigns to ech input exctly one numer s output. The output is function of the input. Simple expressions s functions Exmple: y = x + f(x) = x + (Replce y with f of x ) f(4) = (4) + = 17 Inverse function This is the reverse process tht tkes you ck to the originl vlues We write the inverse of f(x) s f -1 (x) Exmple: if f(x) = x + We sy y = x + x + = y x = y Rerrnge in terms of x f -1 (x) = x hnge y ck to x x Inverse function using flow digrm x x f -1 (x) = x + x- omposite function pplying one function to the results of nother Exmple 1: To comine these two functions f(x) = x nd g(x) = x - 1 gf(x) mens g(x) = (x) 1 = x - 1 Replce x in the function g(x) with x x+ fg(x) mens f(x - 1) = (x - 1) = x Replce x in the function f(x) with x-1 Exmple : To evlute the composition of functions If f(x) = x - 1 nd g(x) = x +, work out fg() Find g() = ()+ = Then f() = -1 = -1 - x This is the input into the function This is the output of the function
3 11/8 Deduce turning point of qudrtic functions y completing the squre To complete the squre x + 4x + 4 = (x+) (perfect squre) x + 4x + = (x+) 1 (completed squre form) Rules to complete the squre Exmple 1: x +4x+ (x+) x in rcket with ½ of +4 (x+) put squred sign on rcket (x+) - 4 sut the squre of the new end numer (x+) -4+ dd /sutrct the originl end numer (x+) -1 simplify Exmple : x + x + divide ll terms y (x + x +.) ((x + 1.) -.) + (x + 1.) (x+1.) +. Deduce turning point of qudrtic Exmple: y = x +4x+ x + 4x + = (x+) 1 complete the squre Turning point is (-,-1) identify the coordintes / Solve qudrtic eqution y completing the squre 11/1 Solve qudrtic equtions y formul x + x + c = Formul (to lern): x = - ± 4c Exmple To solve: x + 4x = x = - ± 4c x = -4 ± (-4) 4()(-) () = -4± 1+4 = -4± 4 x = OR -4-4 x =.(dp) OR -1.7 (dp) 11/11 Solve qudrtic inequlities Exmple: x + x > 1 x + x 1 = Replce inequlity symol with = (x - )(x + ) = Fctorise x = nd x = - Solve repre the numer line Test x=- in inequlity (-) +(-) =18 18>1 TRUE Test x= in inequlity () +() = >1 FLSE - Solution set on numer line = = 4 c = - These re EXT vlues Test x= in inequlity () +() =18 18>1 TRUE Mke the coefficient of x squre x + 1x + = (mult y ) 4x + x + 1 = dd numer to oth sides to mke perfect squre 4x + x + 1 = (dd 1) 4x + x + = 1 (x + ) = 1 Squre root oth sides x + = ± 1 (- from oth sides) x =- + 1 OR x = -. OR (dp) - Set nottion: x<- nd x> Solution on grph
4 11/1 Rerrnge more complex formule (inc where suject ppers twice) ollect ll the terms with the new suject Fctorise to isolte the new suject 11/14 Grphs of trigonometric functions LERN THE SHES OF THE GRHS Grph of y=sin x Exmple: to mke the new suject = 7 (multiply oth sides y ( ) - ( ) = 7 (Expnd the rcket) = 7 (ollect terms in new suject) 7 + = (+ to oth sides) 7 + = + (fctorise to isolte ) (7 + ) = + ( (7 + ) oth sides) (7 + ) (7 + ) = + (7 + ) Grph y = cos x -1 sin x 1 11/1 Exponentil grphs The grph of the exponentil function is: y = k x Exmple: y = x It hs no mximum or minimum point It crosses the y-xis t (,1) It never crosses the x-xis Grph y = tn x -1 cos x 1 tn x is undefined t, 7... Solutions to trigonometric equtions cn e found on the clcultor nd y using the symmetry of these grphs Exmple: If sin x =. x =, 1, (See the solutions on sin grph ove or from clcultor)
5 11/1 Trnsformtion of functions For ny grph y = f(x) LERN the trnsformtions y=f(x) ± Trnsltion ( ) moves up(+)/down(-) ± y=f(x± ) y=-f(x) y=f(-x) Trnsltion ( ± ) moves right(-)/left(+) Reflection in the x-xis (horizontlly) Reflection in the y-xis (verticlly) 11/17 Grph of the circle The grph of circle is of the form: x + y = r where r is the rdius nd the centre is (,) 11/1 Grdient of curve Exmple: To find grdient t point x=4 Drw tngent t x=4 to the curve ick points on the tngent (x 1,y 1) & (x,y ) Work out rise & run or use y y 1 x x 1 heck if positive or negtive run Rise = 4 Run =.8 Grdient = 4.8 = 4 8 = Its slope is positive 11/1 re under curve Split into trpeziums Find the sum of their res 14 rise 1 tngent 1 4 re = ½ x 1 x( +( ) + 1) = ½ x 1 x ( ) = 4 units The curve is concve, so it will e slight over-estimte onvex curves give n over-estimte This circle of rdius nd centre (,) The grph of this circle is x + y = x + y = 11/17 Eqution of tngent to circle Eqution of tngent: y y 1 = m(x x 1) m= grdient of tngent t the point It is perpendiculr to the rdius so m rdius x m tngent = -1 (x 1, y 1) = point on circle where tngent meets Exmple entre (,) (,1) Grdient of rdius = ½ Grdient of tngent = - (m rdius x m tngent = -1) y y 1 = m(x x 1) y 1 = -(x - ) y 1 = -x + 4 y =-x + (eqution of tngent)
6 Wter level in cm Wter level in cm 11/1 Solve simultneous equtions~one liner, one qudrtic lgericlly Rewrite the liner with one letter in terms of the other Sustitute the liner into the qudrtic Solve the qudrtic y fctorising Exmple: To solve y=x- nd y=x -x- Sustitute y=x- into y=x -x- x- = x -x- x -x 4 = (fctorise) (x 4)(x + 1) = x = 4 or x = -1 when x= 4, y = (4)- = when x= -1, y = (-1)- = -4 See points of intersection of grphs for solutions y=x- 7 Grph of y=x -x Solutions re: (4, ) nd (-1, -4) 11/ Interpret grdient of tngent & chord The instntneous rte of chnge of quntity t given time is the grdient of the tngent to the grph t tht time e.g. min Time in minutes Instntneous rte of rise of wter level t min 4cm. min.7cm/min The verge rte of chnge is the grdient of the chord etween the two given times Time in minutes verge rte of chnge etween & 4min 1cm min = cm/min
7 11/ Use itertion to solve equtions Itertion mens repeting process. Ech repetition is clled itertion. The result of n itertion is used s the strting point of the next itertion e.g. x x + 1 = Write x in terms of x (here is one wy) x = x + 1 x = x + 1 Then write s the itertion formul x n+1 = x n + 1 (n=previous term; n+1=next term) hoose vlue for x 1 (it my e given/found from grph e.g. x 1 =. Find x y sustituting x1 into the itertion formul x = (.) + 1 =. Find x y sustituting x into the itertion formul X = (.) + 1 =.4... ontinue until nswer converges to given numer of d.p. (in this cse.(dp)) Quick method with clcultor.= (NS +1) = = = etc till it converges 11/ re of tringle height not known Exmple re = ½ sin re = ½ c sin re = ½ c sin Formul NOT provided 11/ ircle Theorem proofs ngle in semicircle = Drw in rdius s shown ngle t centre = x ngle t circumference Drw in rdius s shown ngles in the sme segment re equl Drw ngle t centre from the chord s shown x y z c In the old tringle: + = 18 ( ) => + = x=18 - y=18 - z= - (x + y) =-(--) = + =( + ) Opposite ngles of cyclic qudrilterl = 18 Drw in two rdii s shown x y c= (lredy proved) c= (lredy proved) = = x= (lredy proved) y= (lredy proved) x + y = + = + = re = ½ sin = ½ x 8 x x sin8 = 14.8 cm (1dp) =8cm =cm 8 ngle etween tngent nd its chord is equl to the ngle in the 'lternte segment' Drw tngent nd ngle in semicircle s shown + = x + = = x x
8 11/ ircle Theorem proofs (continued) Equl tngents from point to the circumference rdius, perpendiculr to chord isects the chord 11/4 Use circle theorems See Stge 1 rompt Sheet 11/ ythgors Theorem in D 1 cm 4 cm cm Q Exmple: Identify the tringle in the D shpe contining the unknown side Q R Use ythgors in ΔRQ to find RQ RQ = ( + 4 ) = cm 4cm R O O Q M In ΔO & ΔO O =O (rdii) O is common ngles etween tngent & rdius = ΔO & ΔO re congruent (RHS) = In ΔOM & ΔQM O =OQ (rdii) OM is common ngles = ΔOM & ΔQM re congruent (RHS) QM = M Leve in surd form when needing the EXT vlue 11/ Trigonometry in D Identify the tringle in the D shpe contining the unknown ngle QR Use ythgors in ΔRQ to find RQ RQ = ( + 4 ) = cm Use Trigonometry in ΔRQ to find QR Tn QR = 1 =.4 Tn -1 QR = /7 Sine Rule (non-right ngled tringles) Use SINE RULE when given: two sides nd non-included ngle ny two ngles nd one side To find n ngle, use: sin = sin = sin c Exmple: To find ngle =7.1cm 1cm sin = sin c sin = sin 7.1 sin = sin x 7.1 sin = = sin -1 (.44...) = 8. (1dp) R =cm cm Q Formul NOT provided Use ythgors in ΔRQ to find Q Q= (1 + ) = 1cm OR = 1 = 1cm 1cm R Q cm Q cm
9 11/7 Sine Rule (continued) To find side use: = = c sin sin sin Exmple: To find side Formul NOT provided To find side given sides & included ngle use: = + c c cos = + c c cos c = + cos Exmple: To find side Formul NOT provided =cm c=4.cm 11 =.7cm 4 = sin sin = sin sin 4 = x sin sin 4 =.8 cm (1dp) = + c c cos = x.7x4. cos11 = 41. =.48(dp) 11/ Vectors =? 11/8 osine Rule (non-right ngled tringles) Use OSINE RULE when given: sides sides nd the included ngle To find n ngle, given sides use: Formul NOT provided Vector nottion This vector cn e written ( ) or or Exmple: To find ngle c=cm cos = + c c cos = + c c cos = + c =8cm =cm vector hs mgnitude(length) & direction(shown y n rrow) Mgnitude cn e found y ythgors Theorem = + = 1 =. prllel vector with sme mgnitude ut opposite direction cos = + c cos = 8 + x8x cos = = cos -1 ( ) = 8. (1dp) Q is equl in length to Vector Q direction so we sy: Q = - ut opposite in
10 11/ Vectors (continued) prllel vector with sme direction ut different mgnitude Vector sutrction = ( ) = ( ) D - Vector D is twice (sclr ) the mgnitude ut sme direction so we sy: D = negtive sclr would reverse the direction Vector ddition dding grphiclly, the vectors go nose to til = ( ) = ( ) + - The comintion of these two vectors: - = = - =( ) - ( ) = ( ) is clled the RESULTNT vector The sum of vectors M N = + M + M The comintion of these two vectors: = + + = = ( ) + ( ) = ( 1 4 ) The vector is equl to the sum of these vectors or it could e different route: = N + N go vi N Strt point End point olliner points (in sme stright line) To prove points re colliner: hoose two line segments, e.g. nd. rove tht they hve: (i) ommon direction (equl grdients) nd (ii) common point (e.g. )
11 frequency density (no. ooks per ) Frequency density 11/ Histogrms lss intervls re not equl Verticl xis is the frequency density Frequency is re of r not the height Frequency = clss width x frequency density Frequency density = frequency clss width To drw histogrm lculte the frequency density Exmple rice () in ( ) f = width x height(fd) < x 8 = 4 < 1 x 1 = 1 < 1 x = < 4 x = 4 ge (x yers) lss width f Frequency density < x 8 8 = 1.4 < x 1 1 =.4 < x = 4 < x = 1. Scle the frequency density xis up to.4 Histogrm of ge groups ge in yers To interpret histogrm Histogrm to show the numer of ooks sold rice () in
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