BRIEF NOTES ADDITIONAL MATHEMATICS FORM 4

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BRIEF NOTESADDITIONAL MATHEMATICSFORM 4Symbol : f -1To find the inverse function, change f(x) toy and find x in tems of y.CHAPTER 1: FUNCTION1. f : x x 3x is the object, x 3 is the imagef : x x 3 can be written asf(x) x 3.To find the image for 2 meansf(2) 2 3 5Image for 2 is 5.Find the object for 8 means f(x) 8 what isthe value of x ?x 3 8; x 5The object is 5.If the function is written in the form ofordered pairs (x, y) , x is the object and y isthe image.E.g. (2, 4), (2, 6), (3, 7)Image for 2 is 4 and 6 whereas object for 7is 3.Given f : x x, find f -13 xLet f(x) yy x3 xy(3 – x) x3y – xy x3y x xy x(1 y)x 4.3x3y, thus f -1(x) 1 x1 yComposite FunctionGiven f : x 3x – 4 and g : x 2 – 3x,find(a) fg(x)(b) gf(x)(c) f 2(3)(d) gf -1(4)fg(x) f(2 – 3x) 3(2 – 3x) - 4 6 – 9x – 4 2 – 9x(b) gf(x) g(3x – 4) 2 – 3(3x – 4) 2 – 9x 12 14 – 9x(c) f 2(3) ff(3) f(9 – 4) f(5) 15 – 4 11.(a)In the arrow diagram, the set of object is{1, 2, 3} and the set of image is {4, 5}2.For f : x 5, x – 3 0, i.e. x 3x 3(d)5becauseis undefined.05Thus, if f : x , x k then k is 3.x 33.Function which maps into itself means f (x) x3If f : x , find the value of x whichx 2is mapped into itself.3 xx 23 x(x – 2) x2 – 2xThus, x2 – 2x – 3 0(x – 3)(x 1) 0x 3 or 13.Inverse Functionzefry@sas.edu.myLet y 3x – 4, x Thus f -1(4) gf -1(4) g(5.y 438388) 2 – 3 633To find f(x) or g(x) given the compositefunction.Given f(x) 2x 8 and fg(x) 6x 12,find g(x).f(x) 2x 8f[g(x)] 2g(x) 8 6x 122g(x) 6x 12 – 8 6x 4g(x) 3x 2Given f(x) 3x – 5 and gf(x) 9x2 – 30x 30,find g(x)1

gf(x) 9x2 – 30x 30g(3x – 5) 9x2 – 30x 30Let y 3x – 5, x 2.y 53Using SOR and POR and the formula x2 –(SOR)x POR 0Cari persamaan kuadratik denganpunca1dan 322y 5 y 5 g (y) 9 30() 30 3 3 1 3 21POR 3 2 y2 10y 25 – 10y – 50 30 y2 5Thus, g(x) x2 5SOR CHAPATER 2 : QUADRATIC EQUATION1. Find the roots of quadratic equation(a) FactorisationEquation is(b) formula x (a) b b 2 4ac2aSolve 6x2 – 7x – 3 0x2 73x 0222x2 – 7x 3 0 2,3.If ax2 bx c 0 is the general form ofthe quadratic equation,SOR α β POR αβ (2x – 3)(3x 1) 02x – 3 0, x (b)13 ( 4) ( 4) 4 2 ( 5)44 16 40 4 56 444 56x 2.87144 56x 0.87084SOR α 2α 3α x 2. Form equation form roots.Use the reverse of factorisatonFind the quadratic equation with roots2x 1, (2x – 1) 0x 3, (x – 3) 0The equation is(2x – 1)(x – 3) 02x2 – 7x 3 0 2,zefry@sas.edu.my1and 32 m m1POR α 2α 2α2 1821,2caThe roots are α and 2αIf it cannot be factorised, use theformula.Solve 2x2 – 4x – 5 0a 2, b 4 and c 5x baGiven that one root is twice the other rootfor the quadratic equation x2 mx 18 0,find the postive value of m.323x 1 0, x 72324.α2 9α 9 3When α 3, 3α 9 m, m 9 (notaccepted)When α 3, 3α 9 m, thus m 9Types of roots(a) 2 real and distinct roots.b2 – 4ac 0(b) 2 real and equal rootsb2 – 4ac 0(c)No real rootb2 – 4ac 0(d)Real root (distinct or same)b2 – 4ac 0Find the range of values of k in which theequation 2x2 – 3x k 0 has two real anddistinct roots.For two real and distinct rootsb2 – 4ac 02

( 3)2 – 4(2)k 09 – 8k 08k 93.9k 8CHAPTER 3: QUADRATIC FUNCTIONS1. To find the maximum/minimum value bycompleting the square.Find the range of value of x for which x2 –7x – 8 0x2 – 7x – 8 0Note: If the(x – 8)(x 1) 0coefficient of x2x 8, x 1is negative, theSketch the graphshape of thegraph is‘n’Given f(x) 2x2 – 6x 8, find themaximum or minimum value and state thecorresponding value of x.f(x) 2x2 – 6x 8 2[x2 – 3x] 82Quadratic Inequality(a) Factorise(b) Find the roots(c) Sketch the graph and determine therange of x from the graph.2 3 3 2[x – 3x ] 8 2 2 39 2[(x )2 ] 82439 2 (x )2 82237 2(x )2 222The minimum value (the coefficient of x2is positive and the graph is ‘u’ shaped) is733when x 0, or x .2222.To sketch quadratic function(a) Determine the y-intercept and the xintercept (if available)(b) Determine the maximum or minimumvalue.(c) Determine the third point opposite tothe y-intercept.Sketch the graph f(x) x2 – 8x 6(a)(b)(c)Y-intercept 6f(x) x2 – 8x 42 – 42 6 (x – 4)2 – 16 6 (x – 4)2 – 10Min value 10 when x – 4 0, x 4. Min point (4, 10)when x 8, f(8) 82 – 8(8) 6 6From the sketch, (x 8)(x 1) 0 1 x 84.Types of Roots(a) If the graph intersects the x-axis attwo different points 2 real anddistinct roots b2 – 4ac 0(b) If the graph touches the x-axis, 2equal roots b2 – 4ac 0(c) If the graph does not intersect the xaxis,(or the graph is always positiv oralways negative.) no real root b2– 4ac 0The graph y nx2 4x n 3 does notintersect the x-axis for n a and n b, findthe value of a and b.y nx2 4x n – 3 does not intersect thex-axis no real root b2 – 4ac 042 – 4n(n – 3) 016 – 4n2 12n 00 4n2 – 12n – 16 4n2 – 3n – 4 0(n – 4)(n 1) 0n 4, n 1From the graph, for (n – 4)(n 1) 0, n 1 and n 4zefry@sas.edu.my3

a 1 and b 4CHAPTER 4: SIMULTANEOUSEQUATIONSTo solve between one linear and one non-linearequation.Method : SubstitutionExample : Solvex 2y 4 --------(1)2x 2 y 5 -----(2)yxfrom (2), xy2x2 2y2 5xy ------------(3)from (1), x 4 – 2ysubstitute in (3)2(4 – 2y)2 2y2 5(4 – 2y)y2(16 – 16y 4y2) 2y2 20y – 10y28y2 10y2 2y2– 32y – 20y 32 020y2 – 52y 32 0 45y2 – 13y 8 0(5y – 8)(y – 1) 03.loga xn nloga x4.loga b 5.loga a 16.loga 1 0log c alog c bExample: Find the value of5log4 8 – 2log4 3 3log4 185log4 8 – 2log4 3 log4 18358 3 183232 18 log4 log4 64 log4 439 log4 3log4 4 3 1 3y 8or 158816 4y , x 4 – 2( ) 4 5555To solve index equations, change to the samebase if possible. If not possible to change to thesame base take logarithm on both sides of theequation.y 1, x 4 – 2 2Example: Solve 3.27x-1 93x48Thus, x 2, y 1 and x , y .55!Note Be careful not to make themistake(4 – 2y)2 16 4y2 wrongIf the equations are joined, you have toseparate them.22Solve x y x 2y 3x2 y2 3andx 2y 3CHAPTER 5: INDEX AND LOGARTHMIndex form:b axLogarithm formloga b x3.27x-1 93x3 33(x-1) 32(3x)31 3x – 3 36x1 3x – 3 6x 2 3xx 23Example: Solve 5x 3 – 7 05x 3 – 7 05x 3 7log 5x 3 log 7(x 3)log 5 log 7x 3 log 7 1.209log 5x 1.209 – 3 1.791Example: SolveLogarithm Law :1. loga x logay loga xy2.loga x – loga y logazefry@sas.edu.myxylog a 384 log a 144 log a 6 4loga384 6 41444

log a 16 4 a a416 x y 1a bby int erceptGraident m ax int erceptIntercept form:2a 4General form: ax by c 0CHAPTER 6: COORDINATE GEOMETRY1. Distance between A(x1, y1) andB(x2, y2)AB The equation of straight line given thegradient, m, and passes through the point(x1, y1) :y – y1 m(x – x1)( x2 x1 )2 ( y2 y1 )2Example: If M(2k, k) and N(2k 1, k – 3) aretwo points equidistant from the origin O. Findthe value of k.Equation of a straight line passing througtwo points (x1, y1) and (x2, y2) isy y1 y2 y1 x x1 x2 x1MO ON(2k )2 k 2 (2k 1)2 (k 3)2Square,4k2 k2 4k2 4k 1 k2 – 6k 90 2k 92k 9k 92Example: Find the equatioon of the straight line(a) with gradient 3 and passes through(1, 2)(b) passes through (2, 5) and (4, 8)(a)2.Point which divides a line segment inthe ratio m : n nx1 mx2 ny1 my2 n m , n m Equation of straight liney ( 2) 3(x – 1)y 2 3x – 3y 3x – 5(b) Equation of straight liney 5 x 2y 5 x 2Example: Given Q(2, k) divides the line whichjoins P(1, 1) and R(5, 9) in the ratio m : n. Find(a) the ratio m : n(b) the value of k8 54 2322(y – 5) 3(x – 2)2y – 10 3x – 62y 3x 4(a)n 5m 2n m3.n 5m 2n 2m5m – 2m 2n – n3m n(b)2.Parallel and Perpendicular LineParallel lines,m1 m2Perpendicular lines,m1 m2 1m 1 thus, m : n 1 : 3n 33 1 1 9 k1 312 3 k4Example: Find the equation of the straight linewhich is parallel to the line 2y 3x – 5 andpasses through (1, 4)Equation of a straight lineGradient form: y mx cm 2y 3x – 5 , y 35x223, passes through (1, 4)2Persamaan garis lurus ialahzefry@sas.edu.my5

y–4 3(x – 1)2Standard deviation varianceExample: For the data3, 5, 5, 6, 7, 8 find the(a) mean(b) variance(c) standard deviation2y – 8 3x – 32y 3x 5Example: Find the equation of the straight linex x(a)passes through (2, 3)(b) variance, 2 N 5. 6679 25 25 36 49 64 34 6 6 x y ( 4) 4 1 , m1 3 4334 m2 133m2 , passes through (2, 3)422 The equation of the straight line isy–3 3 5 5 6 7 8 6x ywhich is perpendicular to the line 1 and3 43(x – 2)4208 34 2.556 6 6 (c)standard deviation 2.Grouped DataMean, x 4y – 12 3x 64y 3x 18 fx fi2.556 1.599xi mid-pointf frequency4. Equation of LocusExample: Find the equation of the locus for Pwhich moves such that its distance from Q(1, 2)and R( 2, 3) is in the ratio 1 : 2Median,12M L N Fcu cfmL lower boundary of the median classN total frequencyFcu cumulative frequency before themedian classfm frequency of median classc class interval sizeLet P(x, y), Q(1, 2), R( 2, 3)PQ : PR 1 : 2PQ 1 PR 2PR 2PQ( x 2)2 ( y 3)2 2 ( x 1)2 ( y 2)2Mode is obtained from a histogramfrequencySquare,x2 4x 4 y2 – 6y 9 4(x2 – 2x 1 y2 – 4y 4) 4x2 4y2 – 8x – 16y 200 4x2 – x2 4y2 – y2 – 12x – 10y 73x2 3y2 – 12x – 10y 7 0CHAPTER 7: STATISTICS1. Ungrouped DataMean, x xNVariance, 2 Mode ( x x) 2Nx2NStandard deviation, fx fi x2class2 ( x) 2Example:zefry@sas.edu.my6

The table shows the marks obtained in a test.MarksFrequency10 – 14215 – 19520 – 24825 – 291230 – 341035 – 39740 – 446Find,(a) mean mark(b) median(c) mode(d) standard devitionMark10 – 1415 – 1920 – 2425 – 2930 – 3435 – 3940 – 44(a)(b)f258121076Mean x xi12172227323742fxi2485176324320259252 fx fi fxi228814453872874810240958310584(d) i2 ( x) 244760 28.8250 65.76 8.109CHAPTER 8: DIFFERENTIATIONdyrepresents the gradient of a curve at a point.dxdy f (x) first derivativedx gradient function.C.F.271527374450d(ax n ) anx n 1dxDifferentiation of Polynomials1. Differentiate with respect to x:(a) y 3x4 2x3 – 5x – 2x2(c) y 2x(b) y 1440 28.850(a)11N 50 2522y 3x4 2x3 – 5x – 2dy 12x3 6x2 – 5dx1(b) y x x 21dy 1 12 1 1 12 x x dx 222 x2(c) y 2 2x-2x 4dy 4x-3 3dxxMedian class 25 – 29M 24.5 fx f25 15 5 28.6712(c)FrequencyDifferentiation of Productddvdu(uv) u vdxdxdx2.Differentiate with respect to x:y (3x 2)(4 – 5x)dy (3x 2) 5 (4 – 5x) 3dxFrom the graph, mode 28 mark 15x – 10 12 – 15x 2 – 30xDifferentiation of Quotientzefry@sas.edu.my7

dv u dxd u v dudx dx v v23x 43. Differentiatewith respect to x2x 53x 4y 2x 5dy (2 x 5)3 (3x 4)2 dx(2 x 5) 26 x 15 6 x 823 2(2 x 5)(2 x 5) 2Differentiation of Composite Functiond(ax b)n n(ax b)n-1 adx4.Differentiate with respect to x :(a) (3x 5)8(b) (2x – 1)4(3x 2)5(a)Note:you mustdifferentiatethe function inthe brackets.y (3x 5)8dy 8(3x 5)7 3dxy (2x – 1)4(3x 2)5dy (2x – 1)45(3x 2)4 3 (3x dx2)54(2x – 1)3 2 15(2x – 1)4(3x 2)4 8(2x – 1)3(3x 2)5 (2x – 1)3(3x 2)4[15(2x – 1) 8(3x 2)] (2x – 1)3(3x 2)4[30x – 15 24x 16] (2x – 1)3(3x 2)4(54x 1)dy 4x 8dxFor turning pointdy 0dx4x – 8 0x 2x 2, y 2(4) – 16 3 5d2y 4 0, thus the point (2, 5) is adx 2minimum point.Rate of Change of Related QuantitiesExample: The radius of a circle increases whicha rate of 0.2 cm s-1, find the rate of change of thearea of the circle when the radius is 5 cm.dA 2 rdrdr 0.2 cm s-1dtdA dA dr dt dr dt 2 r 0.2 0.4 rWhen r 5 cm,dA 0.4 5dt 2 cm2 s-1Equation of Tangent and NormalGradient of tangent gradient of curve Maximum and Minimum ValueGiven y 2x2 – 8x 3. Find the coordinates ofthe turning point. Hence, determine if the turningpoint is maximum or minimum.y 2x2 – 8x 3A r2 24(3x 5)7(b)y – 0 1( x – 1)y x – 1.dydxExample: Find the equation of the tangent to thecurve y 3x2 – 5x 2 at the point x 1.Small Changes and Approximation y dy xdxy 3x2 – 5x 2Example: Given y 2x2 – 5x 3, find the smallchange in y when x increases from 2 to 2.01dy 6x – 5dxy 2x2 – 5x 3x 1, y 3 – 5 2 0dy 6–5 1dxEquation of tangent :zefry@sas.edu.mydy 4x – 5dx x 2.02 – 2 0.018

y dy xdx(c) (4x – 5) 0.01Substitute the original value, x 2, y (8 – 5) 0.01 0.03Thus the small increment in y is 0.03.p1 100 1256060p1 125 RM75100CHAPTER 9: INDEX NUMBER1.Price Index, I p1 100p0p1 price at a certain timep0 price in the base year2.Composite index I Iw wI price indexw weightageExample:ItemBookBegShirtShoesPrice index100x125140Weightage62y3The table above shows the price indices and theweightage for four items in the year 2004 basedin the year 2000 as base year.If the price of a beg in the year 2000 and 2004are RM40 and RM44 respectively. Thecomposite index for 2004 is 116. Find(a) the value of x(b) the value of y(c) the price of a shirt in 2004, if the price in2000 was RM60.(a)(b)44 100 110406 100 2 110 125 y 3 140 1166 2 y 3600 220 125 y 420 11611 yx 1240 125y 116(11 y)1240 125y 1276 116y125y – 116y 1276 – 12409y 36y 4zefry@sas.edu.my9

ADDITIONAL MATHEMATICS FORM 4 CHAPTER 1: FUNCTION 1. f : x x 3 x is the object, x 3 is the image f: x x 3 can be written as f(x) x 3. To find the image for 2 means f(2) 2 3 5 Image for 2 is 5. Find the object for 8 means f(x) 8 what is the value of x 3? x 3 8 ; x 5 The object is 5. I

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