Mathematics Ext HSC 4 Solutions Suite 43, 4 Elizabeth St, Surry Hills NSW info@keystoneeducation.com.au keystoneeducation.com.au
Mathematics Extension : HSC 4 Solutions Contents Multiple Choice... 3 Question... 3 Question... 3 Question 3... 4 Question 4... 4 Question 5... 5 Question 6... 6 Question 7... 7 Question 8... 7 Question 9... 8 Question... 9 Question... Question...3 Question 3...7 Question 4... Question 5...5 Question 6...3 Keystone Education 5
Mathematics Extension : HSC 4 Solutions Multiple Choice. (D). (A) 3. (B) 4. (C) 5. (C) 6. (D) 7. (B) 8. (B) 9. (A). (D) Question (D) We can multiply both sides by xx(xx + ) to get: 5xx xx + = aa(xx + ) + xx(bbbb + cc) 5xx xx + = (aa + bb)xx + cccc + aa We can then match up coefficients to show that aa = and cc =. We then know that aa + bb = 5, so bb = 4. Question (A) From the conjugate root theorem, we know that zz = + ii is another root. Hence, the quadratic polynomial would have αα = + ii and ββ = ii as roots. We can easily see that: αα + ββ = 4 αααα = ii = 5 So we can construct a quadratic polynomial with sum of roots 4 and product of roots 5 from here. zz 4zz + 5 Note: You could also expand zz ( + ii) zz ( ii) but this would take longer. Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions Question 3 (B) Dividing through by 9 6 = 44 we get: xx 6 + yy 9 = 5 44 From here, aa = 6 and bb = 9. Using the relation bb = aa ( ee ): 9 = 6( ee ) 9 6 = ee ee = 7 6 ee = 7 4 Question 4 (C) When you take the conjugate of a complex number in modulus-argument form, the argument changes sign so: zz = cos ππ 3 + ii sin ππ 3 We can then use de Moivre s theorem: (zz ) = cos ππ 3 + ii sin ππ 3 Keystone Education 5 4
Mathematics Extension : HSC 4 Solutions Question 5 (C) You first take a look at the graph of yy = xx xx: We are trying to find yy = xx xx or yy = ± xx xx. We cannot square root negative numbers, and we take the positive and negative square root of everything else: Note: You might also notice that this is a hyperbola shifted to the right by one unit. You could have rearranged yy = xx xx to obtain (xx ) yy =. Keystone Education 5 5
Mathematics Extension : HSC 4 Solutions Question 6 (D) Since the answers all have dddd, it indicates to us that taking slices perpendicular to the axis of rotation (slices) are the way to go. We can show that δδδδ = ππ( xx) δδδδ 4 VV = lim ππ( xx) δδδδ δδδδ 4 yy= 4 = ππ( xx) dddd 4 4 = ππ yy 4 8 dddd Since the volume is symmetrical about the xx-axis we can write: 4 ππ yy 8 dddd Keystone Education 5 6
Mathematics Extension : HSC 4 Solutions Question 7 (B) The best way to do this is to multiply top and bottom by + sin xx + sin xx sin xx + sin xx dddd + sin xx = sin xx dddd + sin xx = cos xx dddd = sec xx + sec xx tan xx dddd = tan xx + sec xx + CC Note: You could also try to use t-formulae, but you would need to convert it back from half angles. Question 8 (B) It is clear that uu = zz + ww cannot be true, because when we add vectors, they add tip to tail in a parallelogram. So it must be true that uu = zz ww and we can see it is true. When you multiply two complex numbers, their arguments are added, so this shows that uu = zzzz. Keystone Education 5 7
Mathematics Extension : HSC 4 Solutions Question 9 (A) The way we do this is to test each one. For (A), xx = and vv = can be verified straight away. To find acceleration, we use the fact that xx = vv dddd dddd. dddd = cos(xx ) dddd xx = vv cos(xx ) This can be used to verify that aa = 4 when xx = and vv = The other options can be done similarly to show that they do not describe the motion of the particle, and you would need xx = dd dddd vv for (C). Keystone Education 5 8
Mathematics Extension : HSC 4 Solutions Question (D) aa ff(xx aa)dddd aa + ff(aa xx)dddd The second integral can be simplified using the definitely integral property. aa ff(xx) dddd For the first integral, we will use the substitution uu = xx aa, so dddd = dddd. ff(uu)dddd aa = ff(xx)dddd aa So if we add them back together: ff(xx)dddd aa aa + ff(xx) aa dddd = ff(xx) dddd aa Keystone Education 5 9
Mathematics Extension : HSC 4 Solutions Question (a) (i) zz + ww = ( ii) + (3 + ii) = ii = cos ππ 4 + ii sin ππ 4 (ii) zz = ii 3 ii = 6+ii 6ii+ii = 8 4ii = 4 ii ww 3+ii 3 ii 3 + 5 5 (b) Integrate by parts (uu = 3xx, vv = cos ππππ) (3xx ) dd dddd sin(ππππ) dddd ππ = (3xx ) sin(ππππ) ππ = ππ 3 cos ππππ ππ ππ 3 (sin ππ xx)dddd ππ = ππ 3 ππ ππ = ππ 6 ππ (c) Keystone Education 5
Mathematics Extension : HSC 4 Solutions (d) The curve approaches yy = xx as xx. There are intercepts at xx xx = or xx = ±. As xx from the right hand side, yy approaches which approaches. A xx similar argument can be made when xx from the left hand side. Keystone Education 5
Mathematics Extension : HSC 4 Solutions (e) Take a point (xx, yy) on the curve and draw a cylindrical shell. You can unfold the cylindrical shell to obtain a rectangular prism with dimensions xx ππππ δδδδ. So: δδδδ = ππππππππππ VV = lim ππππππππππ δδδδ 6 yy= 6 = ππ yy (6 yy)dddd 6 = ππ (6yy yy 3 )dddd = ππ yy 3 yy4 4 = 6ππ uu 3 6 Keystone Education 5
Mathematics Extension : HSC 4 Solutions Question (a) (i) Everything on the right half plane gets reflected onto the left half plane. (ii) Note that there is an open circle at xx = and the original horizontal asymptote has become xx =. Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions (b) (i) Substitute xx = cos θθ into xx 3 3xx = 3. (ii) ( cos θθ) 3 3( cos θθ) = 3 8 cos 3 θθ 6 cos θθ = 3 (4 cos 3 θθ 3 cos θθ) = 3 4 cos 3 θθ 3 cos θθ = 3 cos 3θθ = 3 (using given identity) Hence by solving cos 3θθ = 3 we can solve for θθ and hence solve for xx, since xx = cos θθ. Now we will take θθ ππ since any other value would provide the same value of xx. 3θθ = ππ 6, ππ 6, 3ππ 6 θθ = ππ 8, ππ 8, 3ππ 8 Hence, xx = cos ππ 8 ππ 3ππ, cos, cos 8 8 Keystone Education 5 4
Mathematics Extension : HSC 4 Solutions (c) For xx yy = 5, we can implicitly differentiate: xx yy dddd dddd = dddd dddd = xx yy So the gradient of the tangent at (xx, yy ) is dddd dddd = xx yy. For xxxx = 6: yy + xx dddd dddd = dddd dddd = yy xx So the gradient of the tangent at (xx, yy ) is dddd = yy. dddd xx So multiplying the two gradients together: xx yy yy xx =, which shows that they are perpendicular. Keystone Education 5 5
Mathematics Extension : HSC 4 Solutions (d) (i) II = xx + dddd = [tan xx] = ππ 4 (ii) II nn + II nn = xxnn xx + + xxnn xx + dddd = xxnn (xx + ) xx dddd + = xx nn dddd = nn (iii) II = 3 II = 3 ( II ) = 3 + ππ 4 Keystone Education 5 6
Mathematics Extension : HSC 4 Solutions Question 3 (a) Let tt = tan xx. So xx = tan tt and dddd = dddd +tt. 3 = = = = 3 tt + tt 4 tt + tt + 5 3 3 3 3 + tt dddd 3(tt) 4( tt ) + 5( + tt ) dddd 3(tt) 4( tt ) + 5( + tt ) dddd 9tt + 6tt + dddd (3tt + ) dddd = 3(3tt + ) 3 = 3 3 + = 3 3 6 Keystone Education 5 7
Mathematics Extension : HSC 4 Solutions (b) You can find the area of the trapezium in terms of xx and yy where yy = xx : You can probably see that it is made of three equilateral triangles we can find the area of one of the equilateral triangles using the area sine rule: yy sin 6 = 3 4 yy Hence the area of the trapezium is 3 3 4 yy. You could have also done this by finding the height of the trapezium through trigonometry and using the area formula for a trapezium. δδδδ = 3 3 4 yy δδδδ VV = lim 3 3 δδδδ 4 yy δδδδ xx= = 3 3 4 yy dddd = 3 3 4 xx4 dddd = 3 3 4 xx5 5 = 4 3 5 uu 3 Keystone Education 5 8
Mathematics Extension : HSC 4 Solutions (c) (i) Substitute aa tt + tt, bb tt into xx tt yy aa bb and show that it equal to. aa(tt + ) tt aa bb(tt ) tt = bb aa (tt + ) 4tt aa bb (tt ) 4tt bb = (tt + ) (tt ) 4tt 4tt = tt4 + tt + tt 4 + tt 4tt = 4tt 4tt = (ii) The gradient of line PPPP is: bbbb + bb tt aaaa aa tt = bb(tt + ) aa(tt ) Hence, line PPPP is a unique line that passes through MM with gradient bb(tt +) aa(tt ). The gradient of the tangent at MM can be found by implicitly differentiating the hyperbola: At aa tt + tt, bb tt : tt xx aa yy dddd dddd bb = dddd dddd = bb xx aa yy aa(tt ddyy bb + ) dddd = tt = bb(tt + ) aa bb(tt ) aa(tt ) tt Hence, the tangent at MM is a line passing through MM with gradient bb(tt +). aa(tt ) Since a line can be defined by a gradient and a point, the line through PP and QQ is the tangent at MM. Keystone Education 5 9
Mathematics Extension : HSC 4 Solutions (iii) Hence OOOO OOOO = aa + bb. OOOO = aa tt + bb tt = tt aa + bb OOOO = aa tt + bb tt = tt aa + bb OOSS = aa ee However, we have bb = aa (ee ) which can be rearranged to show that aa + bb = aa ee which shows that OOOO OOOO = OOSS. (iv) If PP and SS have the same xx coordinate, then Hence MM is now: So the gradient of MMMM is: aa(ee + ) ee bb(ee ) ee aa(ee + ) aaaa ee = bb(ee ) aa(ee ) aaaa = aaaa tt = ee, bb(ee ) ee bb(ee ) = aa(ee + ) aaee = bb aa Which is parallel to one of the asymptotes. Keystone Education 5
Mathematics Extension : HSC 4 Solutions Question 4 (a) (i) Using the multiple root theorem, if we show that PP() = PP () = PP () =, then we show that it is a root of multiplicity three. PP(xx) = xx 5 xx + 5xx 6 PP (xx) = 5xx 4 xx + 5 PP (xx) = xx 3 PP() = + 5 6 = PP () = 5 + 5 = PP () = = (ii) Suppose the other two roots are αα and ββ. Using sum of roots: + + + αα + ββ = αα + ββ = 3 Using product of roots: αααα = 6 αααα = 6 We can form a quadratic with this and solve it: xx + 3xx + 6 = 3 ± 9 4(6) xx = 3 ± 5ii = Keystone Education 5
Mathematics Extension : HSC 4 Solutions (b) (i) Gradient OOOO: bb sin θθ aa cos θθ = bb tan θθ aa To find normal gradient, we implicitly differentiate: xx aa + yy dddd dddd = bb dddd dddd = bb xx aa yy At PP: dddd dddd = bb aa cos θθ aa bb sin θθ = bb cot θθ aa So normal gradient is aa tan θθ. bb To find the acute angle between the two lines, we use the formula: tan φφ = mm mm + mm mm aa = bb tan θθ bb tan θθ aa + tan θθ = aa bb bb tan θθ aa sec θθ = aa bb aaaa sin θθ cos θθ cos θθ = aa bb sin θθ cos θθ (assuming that θθ is acute) aaaa Keystone Education 5
Mathematics Extension : HSC 4 Solutions (ii) By symmetry, we can assume that an acute θθ will maximise φφ. Let TT = aa bb bb sin θθ cos θθ = aa aaaa aaaa sin θθ dddd dddd = bb aa cos θθ aaaa dd TT ddθθ = bb aa sin θθ aaaa When dddd bb =, aa cos θθ = which means θθ = ππ or θθ = ππ. dddd aaaa 4 Using the second derivative test, when θθ = ππ 4 : dd TT ddθθ = bb aa sin ππ aaaa == bb aa < aaaa Which shows that it is a maximum. Note that the first derivative test can be used as well. φφ is maximised when θθ = ππ 4. (c) (i) The forward force is FF and backward force is KKvv so the net force is: mmxx = FF KKvv When xx =, vv = 3 (terminal velocity). So: = FF KK 3 KK = FF 3 mmxx = FF FF 3 vv = FF vv 3 Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions (ii) We re looking for a relationship between time and velocity, so we have to use xx = dddd dddd. xx = dddd dddd = FF vv mm 3 = FF vv mm 3 3 dddd dddd = mm FF 3 3 vv tt = mm FF 3 3 vv dddd Observe that + = 6 so we can conclude that: 3 vv 3+vv 3 vv 3 5 6 3 = vv 3 vv = 5 3 vv + 3 + vv So tt = 5mm FF 3 vv + 3 + vv dddd = 5mm FF log 3 + vv + CC 3 vv When vv =, tt =, so CC = so: tt = 5mm + vv log 3 FF 3 vv Hence when vv = : tt = 5mm FF log 5 5mm = FF log 5 Keystone Education 5 4
Mathematics Extension : HSC 4 Solutions Question 5 (a) = (aa + bb + cc) = aa + bb + cc + aaaa + aaaa + bbbb Now since < aa bb cc: aa aaaa aa aaaa bb bbbb Hence: aa + bb + cc + aa + aa + bb = 5aa + 3bb + cc (b) (i) Now + ii and ii are conjugate, which means ( + ii) nn and ( ii) nn are conjugates from conjugate laws. So ( + ii) nn + ( ii) nn = Re( + ii) nn ( + ii) nn = cos ππ 4 + ii sin ππ 4 nn = nn cos nnnn 4 + ii sin nnnn 4 (De Moivre' s theorem) ( + ii) nn + ( ii) nn = Re( + ii) nn = nn cos nnnn 4 Keystone Education 5 5
Mathematics Extension : HSC 4 Solutions (ii) If we expand ( + ii) nn when nn is a multiple of 4 using the binomial theorem ( + ii) nn = nn + nn ii + nn ii + nn 3 ii3 + nn 4 ii4 + nn iinn nn = nn + nn ii nn nn 3 ii + nn 4 + + nn nn Re( + ii) nn = nn nn + nn 4 + + nn nn nn nn + nn 4 + + nn nn = nn cos nnnn 4 Now when nn is divisible by 4, we have: cos ππ = cos ππ = cos 3ππ = cos nnnn 4 = ( )nn 4 Dividing both sides by two we get the result: nn nn + nn 4 + + nn nn = ( )nn 4 nn Keystone Education 5 6
Mathematics Extension : HSC 4 Solutions (c) (i) Resolving horizontally and vertically: TT cos φφ = mmvv rr kkvv = mmmm + TT sin φφ We can rearrange to get TT cos φφ = kkvv mmmm and TT sin φφ = mmvv. By dividing, we get: rr TT sin φφ TT cos φφ = kkvv mmmm mmvv /rr We also identify that cos φφ = rr or rr = l cos φφ. l TT sin φφ TT cos φφ = kkvv mmmm mmvv l cos φφ = l cos φφ (kkvv mmmm) mmvv sin φφ cos φφ = l(kkvv mmmm) mmvv = lkkvv lmmmm mmvv = lkk mm lgg vv Keystone Education 5 7
Mathematics Extension : HSC 4 Solutions (ii) We know that: sin φφ cos φφ < lkk mm sin φφ sin φφ < lkk mm sin φφ < lkk mm ( sin φφ) lkk mm sin φφ + sin φφ lkk mm < This is a quadratic equation in sin φφ so we can solve by graphing: + 4 lkk mm lkk/mm < sin φφ < + + 4 lkk mm lkk/mm Taking just the right hand side, and multiplying top and bottom by mm. sin φφ < mm + mm + 4l kk lkk (iii) Suppose ff(φφ) = sin φφ cos φφ = sin φφ sec φφ. ff (φφ) = cos φφ sec φφ + sin φφ sec φφ tan φφ = cos φφ + sin φφ cos 3 φφ = cos 3 φφ (cos φφ + sin φφ) = cos 3 φφ ( + sin φφ) For ππ < φφ < ππ, cos φφ > and hence cos3 φφ >. Since squares are nonnegative, sin φφ, hence + sin φφ >. Hence, ff (φφ) = +sin φφ cos 3 φφ the domain. >, which shows that it is an increasing function in Keystone Education 5 8
Mathematics Extension : HSC 4 Solutions (iv) As vv increases, decreases, lkk lgg sin φφ increases. So since, = lkk vv mm vv cos φφ mm lgg sin φφ vv, cos φφ increases. Since it is an increasing function, φφ increases. Keystone Education 5 9
Mathematics Extension : HSC 4 Solutions Question 6 (a) (i) Suppose AAAAAA = xx. AAAAAA = xx (alternate segment theorem) DDDDDD = xx (alternate angles and parallel lines) (ii) Suppose AAAAAA = DDDDDD = xx. Under a similar argument to part (i), YYYYYY = RRRRRR. Let this angle be yy. Let XXXXXX = zz. QQQQQQ = zz (vertically opposite) Hence AAAAAA = xx + yy + zz and DDDDDD = xx + yy + zz. Adding angles around the point PP: AAAAAA + (xx + yy + zz) + BBBBBB = 36 Now, AAAAAA = BBBBBB = 9 (angle in semicircle). So from this, we can solve: xx + yy + zz = 9 Hence AAAAAA = AAAAAA + CCCCDD = 9 + xx + yy + zz = 8, which shows that AA, PP and CC are collinear. (iii) Now since APC is collinear, AAAAAA = RRRRRR (vertically opposite). This shows that xx = yy, which shows that AAAAAA = AAAAAA. This shows that AAAAAAAA is cyclic (equal angles subtended at DD and CC on the same side) Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions (b) (i) + xx ( xx + xx 4 xx 6 + + ( ) nn xx nn ) = + xx ( xx ) nn + xx = ( )nn (xx nn ) + xx If nn is even, xxnn is clearly greater than equal to +xx xxnn. Since xx, + xx, so if you divide by a number greater than or equal to, it remains the same, or it becomes smaller. Hence xx nn xxnn + xx xxnn If nn is odd, xxnn +xx is clearly less than or equal to xxnn. A similar argument above can be made to show that Hence xx nn xxnn + xx xxnn xx nn ( )nn xx nn + xx xx nn Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions (ii) If we integrate both sides with respect to xx between and : xx nn dddd + xx ( xx + xx 4 xx 6 + + ( ) nn xx nn )dddd xx nn dddd nn + [tan xx] xx xx3 3 + xx5 5 xx7 7 + + ( )nn xx nn nn nn + ππ 4 3 + 5 7 + + ( )nn nn nn + nn + (iii) If we let nn ππ 4 3 + 5 7 + Which means ππ + + + =, and this is what we want. 4 3 5 7 (iv) The tricky part of this is that using substitution is hard, because we don t have the derivative of ln xx anywhere. You have to add it in: ln xx ( + ln xx) dddd = xx ln xx xx( + ln xx) dddd = xx ln xx dddd xx( + ln xx) Now use integration by parts, where vv = integrate this by substitution to get vv = xx(+ln xx) +ln xx and uu = xx ln xx. We can xx ln xx + ln xx xx ln xx (ln xx + )dddd = + dddd + ln xx + ln xx xx ln xx = + xx + CC + ln xx xx ln xx xx( + ln xx) = + + ln xx + ln xx + CC xx = + ln xx + CC Keystone Education 5 3
Mathematics Extension : HSC 4 Solutions Alternate solution An alternate way to do this: ln xx ln xx + dddd = ( + ln xx) ( + ln xx) dddd = + ln xx dddd ( + ln xx) dddd We can now try to solve the first one using integration by parts, where vv = and uu =. So vv = xx and +ln xx uu = xx(+ln xx) + ln xx dddd = xx + ln xx xx xx( + ln xx) dddd = xx + ln xx + ( + ln xx) dddd If we substitute it back, we have: ln xx ( + ln xx) dddd = xx + ln xx + ( + ln xx) dddd ( + ln xx) dddd xx = + ln xx + CC Keystone Education 5 33