hsn.uk.net Higher Mathematics HSN24400 Course Revision Notes This docum ent was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to us. For m ore details about the copyright on these notes, please see http://creativecom m ons.org/licenses/by-nc-sa/2.0/ Higher Mathematics Course Revision Notes Contents Unit 1 – Mathematics 1 Straight Lines Functions and Graphs Differentiation Sequences 1 2 5 6 Unit 2 – Mathematics 2 Polynomials and Quadratics Integration Trigonometry Circles 7 8 9 12 Unit 3 – Mathematics 3 Vectors Further Calculus Exponentials and Logarithms The Wave Function hsn.uk.net 12 15 16 18 - ii - HSN24400 Higher Mathematics Course Revision Notes Straight Lines Distance Formula ( x2 − x1 ) Distance = 2 + ( y2 − y1 ) between points ( x1, y1 ) and ( x2 , y2 ) 2 Gradients m= y2 − y1 between the points ( x1, y1 ) and ( x2 , y2 ) where x1 ≠ x2 x 2 − x1 Positive gradients, negative gradients, zero gradients, undefined gradients eg y = 4 eg x = 2 Lines with the sam e gradient are parallel eg The line parallel to 2 y + 3 x = 5 has gradient m = − 32 since 2 y + 3 x = 5 2 y = −3 x + 5 y = − 32 x + 25 (m ust be in the form y = mx + c ) Perpendicular lines have gradients such that m × mperp. = −1 eg if m = 23 then mperp. = − 23 m = tan y positive direction x is the angle that the line m akes with the positive direction of the x-axis Equation of a Straight Line The line passing through ( a, b ) with gradient m has equation: y − b = m( x − a ) Medians B A M hsn.uk.net x1 + x 2 y1 + y2 , 2 2 BM is not usually perpendicular to AC, so m1 × m2 = −1 cannot be used To work out the gradient of BM, use the gradient form ula M is the m idpoint of AC, ie M = C Page 1 HSN24400 Higher Mathematics Course Revision Notes Altitudes B A D is not usually the m idpoint of AC C D BD is perpendicular to AC, so m1 × m2 = −1 can be used to work out the gradient of BD Perpendicular Bisectors D B A C B C D CD passes through m idpoint of AC CD is perpendicular to AB, so m1 × m2 = −1 can be used to find the gradient of CD Perpendicular bisectors do not necessarily have to appear within a triangle – they can occur with straight lines A Functions and Graphs Composite Functions Example If f ( x ) = x 2 − 2 and g ( x ) = 1x , find a form ula for (a) h ( x ) = f ( g ( x ) ) (b) k ( x ) = g ( f ( x ) ) and state a suitable dom ain for each. (a) h ( x ) = f ( g ( x ) ) = f ( 1x ) (b) k ( x ) = g ( f ( x ) ) = g( x 2 − 2) = ( 1x ) − 2 2 = 1 −2 x2 = 1 x −2 2 Dom ain: { x : x ∈ , x ≠ ± 2 } Dom ain: { x : x ∈ , x ≠ 0} You will probably only be asked for a dom ain if the function involved a fraction or an even root. Rem em ber that in a fraction the denom inator cannot be zero and any num ber being square rooted cannot be negative eg f ( x ) = x + 1 could have dom ain: { x : x ∈ , x ≥ −1} hsn.uk.net Page 2 HSN24400 Higher Mathematics Course Revision Notes Graphs of Inverses To draw the graph of an inverse function, reflect the graph of the function in the line y = x y=x y y = g( x ) y = g−1 ( x ) x O Exponential and Logarithmic Functions Exponential y Logarithmic y y = ax, a >1 y = loga x y y = ax, 0 < a <1 ( a ,1) (1, a ) 1 x O O (1, a ) 1 x 1 x O Trigonometric Functions y = sin x y y 1 1 O –1 x 180° y = tan x y = cos x y x O –1 180° 360° x O 360° 180° 360° Period = 360° Period = 360° Period = 180° Am plitude = 1 Am plitude = 1 Am plitude is undefined Graph Transformations The next page shows the effect of transform ations on the two graphs shown below. y y ( 2, 2 ) y = g ( x ) 1 –1 0 hsn.uk.net 3 O –1 x Page 3 x 180° 360° HSN24400 Higher Mathematics Function Course Revision Notes Effect on f ( x ) Effect f ( x ) + a Shifts the graph a up the y-axis y Effect on sinx ° y = g( x ) + 1 ( 2, 3 ) • 2 • ( 3,1) • ( −1,1 ) 0 f ( x + a ) Shifts the graph −a along the x-axis x y = g ( x + 1) y • 0 1 0 x 360° 180° y = sin ( x − 90 ) ° y 1 (1, 2 ) –2 y = sin x ° + 1 y 2 x x 0 –1 360° 180° − f (x) y Reflects the graph in the x-axis 0 –1 3 • f (−x ) Reflects the graph in the y-axis y ( −2, 2 ) Scales the graph vertically Stretches if k >1 Com presses if k <1 f ( kx ) Scales the graph horizontally Com presses if k >1 Stretches if k <1 hsn.uk.net x 1 x ( 2, − 2 ) 0 –1 y = g( −x ) y = sin ( − x ° ) 180° 360° y • 1 0 –3 kf ( x ) y = − sin x ° y y = − g( x ) x 1 −360° y ( 2, 4 ) y = 2 g ( x ) −180° 0 x –1 y = 12 sin x ° y • 1 2 –1 0 3 y (1, 2 ) x x 0 180° y = g(2x ) • − 12 0 − 12 y 360° y = sin 2 x ° 1 3 2 Page 4 x 0 –1 x 180° 360° HSN24400 Higher Mathematics Course Revision Notes Differentiation Differentiating If f ( x ) = ax n then f ′ ( x ) = anx n −1 Before you differentiate, all brackets should be m ultiplied out, and there should be no fractions with an x term in the denom inator (bottom line), for exam ple: 3 1 − 12 −2 1 −2 = 3x = x 1 = x x2 x 3x 2 3 Equations of Tangents Tangents are straight lines, therefore to find the equation of a tangent, you need a point on the line and its gradient to substitute into y − b = m ( x − a ) You will always be given one coordinate of the point which the tangent touches Find the other coordinate by solving the equation of the curve Find the gradient by differentiating then substituting in the x-coordinate of the point Example Find the equation of the tangent to the graph of y = x 3 at the point where x = 9 . y − b = m(x − a) 1 At x = 9, m = 32 × 9 2 y = x3 y = x3 = 93 = x2 = 32 9 = 33 = 27 dy 3 12 = x dx 2 = 32 × 3 3 y − 27 = 92 ( x − 9 ) 2 y − 54 = 9 x − 81 = 92 ( 9, 27 ) 2 y = 9 x − 27 dy =0 dx You m ust justify the nature of turning points or points of inflection Stationary points occur at points where hsn.uk.net Page 5 HSN24400 Higher Mathematics Course Revision Notes Graphs of Derived Functions Quadratic Cubic y Quartic y y 0 + – + 0 – x + + x + + + – + – Linear – y + x + 0 y – – x 0 y 0 0 – x x Quadratic Cubic Optimisation These types of questions are usually practical problem s which involve m axim um or m inim um areas or volum es Rem em ber you m ust show that a m axim um or m inim um exists Sequences Linear Recurrence Relations A linear recurrence relation is in the form un+1 = aun + b . Also be aware that this m ay be written as un = aun−1 + b If −1 < a < 1 then a lim it l = b exists. You m ust state this whenever you use the lim it 1− a form ula hsn.uk.net Page 6 HSN24400 Higher Mathematics Course Revision Notes Polynomials and Quadratics Polynomials The degree of a polynom ial is the value of the highest power, eg 3 x 4 + 3 has degree 4 Synthetic division (nested form ) can be used to factorise polynom ials Example 4 x 3 − 7 x 2 + 11 Find . x +2 –2 4 –7 0 11 4 –8 –15 30 30 –60 –49 Rem em ber to put in 0 if there is no term 4 x 3 − 7 x 2 + 11 = 4 x 2 − 15 x + 30 rem ainder − 49 x +2 ie 4 x 3 − 7 x 2 + 11 = ( x + 2 ) ( 4 x 2 − 15 x + 30 ) − 49 If the divisor is a factor then the rem ainder is zero If the rem ainder is zero then the divisor is a factor Completing the Square The x 2 term m ust have a coefficient of one. If it does not, you m ust take out a com m on factor from the x 2 and x term , but not the constant In the form y = a ( x + p ) + q the turning point of the graph is ( − p, q ) 2 Example Write 3 x 2 − 12 x + 7 in the form a(x + p)2 + q . 3 x 2 − 12 x + 7 = 3( x 2 − 4x ) + 7 = 3 ( x 2 − 4 x + ( −2 )2 − ( −2 )2 ) + 7 = 3 ( ( x − 2 )2 − 4 ) + 7 = 3 ( x − 2 )2 − 12 + 7 = 3 ( x − 2 )2 − 5 Note that in this exam ple, the graph is ∪ -shaped since the x 2 coefficient is positive; and the turning point is ( 2, − 5 ) . hsn.uk.net Page 7 HSN24400 Higher Mathematics Course Revision Notes The Discriminant The discrim inant is part of the quadratic form ula and can be used to indicate how m any roots a quadratic has. For the quadratic ax 2 + bx + c : If b 2 − 4 ac > 0 , the roots are real and unequal (distinct) 2 roots If b 2 − 4 ac = 0 , the roots are real and equal (ie repeated roots) 1 root If b 2 − 4 ac < 0 , the roots are not real; they do not exist no roots The discrim inant can also be used to calculate the num ber of intersections between a line and a curve. To use it, you m ust first equate them and set equal to zero, before using the discrim inant Rem em ber if b 2 − 4 ac = 0 , the line is a tangent Integration Integrating ax n dx = ax n +1 +c n +1 As with differentiation, all brackets m ust be m ultiplied out, and there m ust be no fractions with an x term in the denom inator Examples 1. Find dx 8 x5 dx 8 = . x5 2. 1 8 = x x5 − 58 x 2 + 5x 7 dx = x −2 ( x 2 + 5 x 7 ) dx 2 x = x 0 + 5 x 5 dx dx dx = 1 + 5 x 5 dx 3 x8 = 3 +c = x + 56 x 6 + c 8 3 = 83 x 8 + c = 38 8 x 3 + c hsn.uk.net x 2 + 5x 7 dx x2 Page 8 HSN24400 Higher Mathematics Course Revision Notes The Area under a Curve If F ( x ) is the integral of f ( x ) , then y b a f ( x ) dx = F (b ) − F ( a ) y = f (x) a x b Rem em ber that areas split by the x-axis m ust be calculated separately and any negative signs ignored; these just show that the area is under the axis. The Area between two Curves The area between the graphs of y = f ( x ) and y = g ( x ) is defined as b a f ( x ) − g ( x ) dx y y = g( x ) If the lim its are not given, f ( x ) and g ( x ) should be equated to find a and b a x y = f (x) b Trigonometry Background Knowledge You should know how to use all of the inform ation below: SOH CAH TOA tan x = sin x cos x sin 2 x + cos2 x = 1 The sine rule: a b c = = sin A sin B sin C b2 + c 2 − a2 The cosine rule: a = b + c − 2bc cos A or cos A = 2bc 2 hsn.uk.net 2 2 Page 9 HSN24400 Higher Mathematics Course Revision Notes The area of a triangle, A = 12 ab sin C CAST diagram s Exact values: 2 45° 30° 2 1 3 60° 45° 1 1 Radians You should know how to convert between radians and degrees: 90° = 2 45° = 4 ÷ 180 × → Radians Degrees 60° = 3 5 × 180 eg 56 = = 150° 6 30° = 6 × 180 ÷ → Degrees Radians 360° = 2 180° = Trigonometric Equations Look at the restrictions on the dom ain, eg 0 ≤ x ° < 360, or 0 ≤ x < Be aware of whether the answer is required in degrees or radians Rem em ber a CAST diagram whenever you are asked to “solve” Examples 1. Solve 3sin 2 x° = 1 where 0 ≤ x ° < 360 . 3sin 2 x ° = 1 3 ( sin x ° ) = 1 2 ( sin x ° )2 = 13 sin x ° = ± 13 x ° = sin x° = 35.3° −1 ( ± 13 ) S A T C x° = 180 − 35.3 x° = 180 + 35.3 x° = 360 − 35.3 = 144.7° = 215.3° = 324.7° Solution set = {35.3°, 144.7°, 215.3°, 324.7°} hsn.uk.net Page 10 HSN24400 Higher Mathematics Course Revision Notes 2. Solve 2 sin 2 x − 1 = 0 , 0 ≤ x < 2 . 2 sin 2 x − 1 = 0 2sin 2 x = 1 sin 2 x = 12 S 2 x = sin −1 ( 12 ) A T C 2 x ° = 30° 2 x ° = 180° − 30° 2 x ° = 150° x ° = 15° x ° = 75° 2 x ° = 360° + 30° 2 x ° = 360° +180° − 30° 2 x ° = 390° 2 x ° = 510° x ° = 195° x ° = 255° 75 75° = 180 15 15° = 180 195 195° = 180 255 255° = 180 3 = 36 = 15 36 39 = 36 51 = 36 = 12 5 = 12 13 = 12 = 17 12 { 5 , 13 , 17 Solutions set = 12 , 12 12 12 } Compound Angle Formulae cos ( A ± B ) = cos A cos B sin A sin B sin ( A ± B ) = sin A cos B ± cos A sin B These are given on the form ula sheet Double Angle Formulae sin 2 A = 2sin A cos A cos 2 A = cos2 A − sin 2 A = 1 − 2 sin 2 A = 2 cos 2 A − 1 These are given on the form ula sheet hsn.uk.net Page 11 HSN24400 Higher Mathematics Course Revision Notes Circles Equations of Circles A circle with centre ( a,b ) and radius r has the equation ( x − a )2 + ( y − b ) = r2 2 Note that if a circle has centre ( 0, 0 ) then the equation is x 2 + y 2 = r2 The equation can also be given in the form x 2 + y 2 + 2 gx + 2 fy + c = 0 where the centre is ( − g,− f ) and the radius r = g2 + f 2 − c You do not have to rem em ber any of these equations, since they are all given in the exam You will have to rem em ber the distance form ula, d = is not given, and is frequently used in circle questions ( x2 − x1 ) 2 + ( y2 − y1 ) , since this 2 Intersection of a Line and a Circle two intersections one intersection (tangency) no intersections Rem em ber, a tangent and a line from the centre of a circle will m eet at right angles, which m eans that m1 × m2 = −1 can be used Vectors Basic Facts A vector is a quantity with both m agnitude (size) and direction A vector is nam ed either by using a directed line segm ent (eg AB ) or a bold letter (eg u written u) A vector m ay also be defined in term s of i, j and k, the unit vectors in three perpendicular directions: 1 i= 0 0 0 j= 1 0 0 k= 0 1 a The m agnitude of vector AB = b is defined as AB = a 2 + b 2 hsn.uk.net Page 12 HSN24400 Higher Mathematics Course Revision Notes b1 a1 ± b1 a1 a2 ± b2 = a2 ± b2 a3 b3 a3 ± b3 ka1 a1 k a2 = ka2 where k is a scalar a3 ka3 0 Z ero vector: 0 0 OA is called the position vector of the point A relative to the origin, written a AB = b − a where a and b are the position vectors of A and B If AB = k BC where k is a scalar, then AB is parallel to BC . Since B is com m on to both AB and k BC , then A, B and C are collinear Dividing Vectors in a Ratio The point P can also be worked out from first principles, or U sing the section form ula. If P divides AB in the ratio m : n , then: n m p= a+ b where p is the position vector OP m+n m+n Example P is the point ( −2, 4, − 1) and R is the point ( 8, − 1,19 ) . Point T divides PR in the ratio 2:3. Work out the coordinates of point T. P 2:3 R T U sing the section form ula The ratio is 2:3, so let m = 2 and n = 3 From first principles n m p+ r m+n m+n = 35 p + 25 r t= ( = 15 3 p + 2r = 15 ( 3t − 3 p = 2r − 2t 3t + 2t = 2r + 3 p −6 16 12 + −2 −3 38 16 −6 5t = −2 + 12 38 −3 10 10 5t = 10 35 = 15 10 35 2 = 2 7 hsn.uk.net ) 3 t − p = 2 (r − t ) ) Therefore T is the point ( 2, 2, 7 ) . PT 2 = TR 3 3PT = 2TR 2 t= 2 7 Therefore T is the point ( 2, 2, 7 ) . Page 13 HSN24400 Higher Mathematics Course Revision Notes The Scalar Product The scalar product a.b = a b cos , where is the sm allest angle between a and b Rem em ber that both vectors m ust point away from the angle, eg a b b1 a1 If a = a2 and b = b2 then a.b = a1b1 + a2b2 + a3b3 a3 b3 cos = ab +a b +a b a.b or cos = 1 1 2 2 3 3 ab a b If a and b are perpendicular then a.b = 0 If a.b = 0 then a and b are perpendicular Example 8 4 If u = 0 and v = 0 , calculate the angle between the vectors u + v and u − v . 1 4 Let a = u + v Let b = u − v 8 4 a= 0 + 0 1 4 8 4 b= 0 − 0 1 4 12 a= 0 5 4 b= 0 3 cos = = a.b ab (12 × 4 ) + ( 0 × 0 ) + ( 5 × 3 ) 122 + 02 + 52 4 2 + 02 + 32 63 = 169 25 = cos −1 63 169 25 = 14·3° hsn.uk.net Page 14 HSN24400 Higher Mathematics Course Revision Notes Further Calculus Trigonometry D ifferentiation This is straightforward, since the form ulae are given on the form ula sheet: f (x) f ′( x ) sin ax a cos ax cos ax − a sin ax Integration Again, the form ulae are provided in the paper: f ( x ) dx f (x) sin ax − 1a cos ax + c cos ax 1 sin ax + c a Examples 1. Differentiate x 3 + cos3 x with respect to x. ( ) d x 3 + cos3 x = 3 x 2 − 3sin3 x dx 2. Find 4 x 3 + sin3 x dx . 4x 4 1 4 x + sin3 x dx = − 3 cos3 x + c 4 = x 4 − 13 cos3 x + c 3 Chain Rule Differentiation n −1 If f ( x ) = ( ax + b )n then f ′ ( x ) = n ( ax + b ) n −1 × a = an ( ax + b ) or n −1 If f ( x ) = ( p ( x ) ) then f ′ ( x ) = n ( p ( x ) ) n × p′ ( x ) “The power m ultiplies to the front, the bracket stays the sam e, the power lowers by one and everything is m ultiplied by the differential of the bracket” hsn.uk.net Page 15 HSN24400 Higher Mathematics Course Revision Notes Examples 1. G iven f ( x ) = 1 + x − sin3 x , find f ′ ( x ) . x2 1 f ( x ) = x −2 + x 2 − sin3 x −1 f ′ ( x ) = −2 x −3 + 12 x 2 − 3cos3 x 2 1 =− 3 + − 3cos3 x x 2 x 2. G iven f ( x ) = ( 3 x 2 + 2 x + 1) , find f ′ ( x ) . 3 f ′ ( x ) = 3 ( 3 x 2 + 2 x + 1) × ( 6 x + 2 ) 2 = 3 ( 6 x + 2 ) ( 3 x 2 + 2 x + 1) 2 3. Differentiate y = cos 2 x = ( cos x )2 with respect to x. dy = 2 ( cos x ) × ( − sin x ) dx = −2cos x sin x n Integration of (ax + b) ( ax + b ) dx = n ( ax + b )n+1 ( n + 1) × a +c Example Find ( 3 x + 5 )4 dx . ( 3 x + 5 ) dx = 4 ( 3 x + 5 )5 5× 3 +c = ( 3 x + 5 )5 15 +c It is possible for any type of ‘further calculus’ to be exam ined in the style of a standard calculus question (eg optim isation, area under a curve, etc) Exponentials and Logarithms An exponential is a function in the form f ( x ) = a x Logarithm s and exponentials are inverses y = a x ⇔ loga y = x On a calculator, log is log10 and ln is loge hsn.uk.net Page 16 HSN24400 Higher Mathematics Course Revision Notes Laws of Logarithms loga x + loga y = loga xy (Squash) loga x − loga y = loga xy (Split) loga x n = n loga x (Fly) Examples 1. Evaluate log2 4 + log2 6 − log2 3 log2 4 + log2 6 − log2 3 4×6 3 = log2 = log2 8 =3 (since 23 = 8) 2. Below is a diagram of part of the graph of y = ke 0.7 x y y = ke 0.7 x 3 0 x =1 x (a) Find the value of k (b) The line with equation x = 1 intersects at R. Find the coordinates of R. (a) At ( 0, 3 ) , y = ke 0.7 x 3 = ke 0.7×0 3 = ke 0 k =3 (b) x = 1 y = 3e 0.7×1 = 6.04 So R is the point (1, 6.04 ) . hsn.uk.net Page 17 HSN24400 Higher Mathematics Course Revision Notes The Wave Function Example 1. Express 6 sin x ° − 2 cos x ° in the form k cos ( x − a ) ° where 0 ≤ a ° < 360 . k cos ( x − a ) ° = k cos x ° cos a ° + k sin x ° sin a ° = k cos a ° cos x ° + k sin a ° sin x ° k cos a ° = − 2 k sin a ° = 6 (− 2 ) k= 2 + 6 2 = 2+6 S A = 8 T C =2 2 k sin a ° k cos a ° 6 =− 2 =− 3 tan a ° = a ° = 180° − tan −1 ( 3 ) = 180° − 60° = 120° Therefore 6 sin x ° − 2 cos x ° = 2 2 cos ( x − 120 ) ° . 2. Express cos x − sin x in the form k sin ( x + ) where 0 ≤ < 2 . k sin ( x + ) = k sin x cos + k cos x sin = k cos sin x + k sin cos x k cos = −1 k sin = 1 k = ( −1)2 + 12 = 2 S A k sin k cos = −1 tan = ° = 180° − tan −1 (1) T C = 180° − 45° = 135° 135 = 180 = 34 ( Therefore cos x − sin x = 2 sin x + 34 ). The m axim um value of an expression in the form k cos ( x ± a ) occurs when cos ( x ± a ) = 1 ; and sin ( x ± a ) = 1 for k sin ( x ± a ) The m inim um value of an expression in the form k cos ( x ± a ) occurs when cos ( x ± a ) = −1 ; and sin ( x ± a ) = −1 for k sin ( x ± a ) hsn.uk.net Page 18 HSN24400