2022TrialQ15 (b) (i)ABCDABCDABCD is a parallelogram. NNN is on ADADAD so that . is on the diagonal . Letting and , express in terms of and .
2022TrialQ15 (b) (ii)ABCDABCDABCD is a parallelogram. NNN is on ADADAD so that AN→=13AD→\overrightarrow{AN}=\frac13\overrightarrow{AD}AN=31AD. MMM is on the diagonal ACACAC. By letting AM→=λAC→\overrightarrow{AM}=\lambda\overrightarrow{AC}AM=λAC and MB→=δNB→\overrightarrow{MB}=\delta\overrightarrow{NB}MB=δNB, prove that AM→=14AC→.\overrightarrow{AM}=\frac14\overrightarrow{AC}.AM=413 marksShort answer
2022TrialQ15 (a) (i)Show that a2+9b2≥6aba^2+9b^2\ge6aba2+9b2≥6ab, where aaa and bbb are real numbers.1 markShort answer
2022TrialQ15 (a) (iii)Hence, if a>b>c>0a>b>c>0a>b>c>0, show that a2+5b2+9c2>9bc.a^2+5b^2+9c^2>9bc.a2+5b2+9c2>9bc.2 marksShort answer
2022HSCQ15 (a) (ii)Hence, or otherwise, show that the point PPP cannot be lifted to a position 2h3\frac{2h}{3}32h metres above the floor.2 marksShort answer
2022HSCQ15 (b)The diagrams show two positions of a single piston in the cylinder chamber of a motorcycle. The piston moves vertically, in simple harmonic motion, between a maximum height of 0.170.170.17 metres and a minimum height of 0.050.050.05 metres. The mass of the piston is 0.8 kg0.8\,\mathrm{kg}0.8kg. The piston completes 404040 cycles per second. What is the resultant force on the piston, in newtons, that produces the maximum acceleration of the piston? Give your answer correct to the nearest newton.3 marksShort answer
2022HSCQ15 (c)Using the substitution x=tan2θx=\tan^2\thetax=tan2θ, evaluate ∫01sin−1x1+x dx.\int_0^1\sin^{-1}\sqrt{\frac{x}{1+x}}\,dx.∫01sin−11+xxdx.4 marksShort answer
2022HSCQ15 (d)The complex number zzz satisfies ∣z−4z∣=2.\left|z-\frac4z\right|=2.z−z4=2. Using the triangle inequality, or otherwise, show that ∣z∣≤5+1.|z|\le\sqrt5+1.∣z∣≤5+1.3 marksShort answer
2022TrialQ15 (b)Consider In=∫01xn1−x dx.I_n=\int_0^1\frac{x^n}{\sqrt{1-x}}\,dx.In=∫011−xxndx. Use integration by parts to show that In=2n2n+1In−1.I_n=\frac{2n}{2n+1}I_{n-1}.In=2n+12nI4 marksShort answer
2022TrialQ15 (c) (i)The line has vector equation r⃗=λ[131].\vec r=\lambda\begin{bmatrix}1\\3\\1\end{bmatrix}.r=λ131. For P(x1,y1,z1)P(x_1,y_1,z_1)P(x1,y1,z1), the point QQQ on the line is such that PQPQPQ is a minimum. Using vector projection methods, show that OQ→=x1+3y1+z111[131].\overrightarrow{OQ}=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}.OQ=2 marksShort answer
2022TrialQ15 (c) (ii)It is given that c⃗=[140]+t[−115]\vec c=\begin{bmatrix}1\\4\\0\end{bmatrix}+t\begin{bmatrix}-1\\1\\5\end{bmatrix}c=140+t−115 does not intersect r⃗\vec rr. Point TTT is on c⃗\vec cc and SSS is a point on r⃗\vec rr such that TSTSTS is a minimum. By first expressing c⃗\vec cc in parametric form, find an expression for TS→\overrightarrow{TS}TS in terms of ttt.3 marksShort answer
2022TrialQ15 (a) (ii)Hence, or otherwise, show that a2+5b2+9c2≥3(ab+bc+ac).a^2+5b^2+9c^2\ge3(ab+bc+ac).a2+5b2+9c2≥3(ab+bc+ac).2 marksShort answer
2022TrialQ15 (a) (i)Show by finding it that the only point of intersection between (xyz)=(1105)+λ(2−31)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}xyz=1105λ2−31 and the sphere (x−5)2+(y+2)2+(z−3)2=38(x-5)^2+(y+2)^2+(z-3)^2=38(x−5)2+(y+2)2+( is (718)\begin{pmatrix}7\\1\\8\end{pmatrix}718.3 marksShort answer
2022TrialQ15 (a) (iii)Show that the line (xyz)=(15−16)+δ(4−1−1)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}xyz=15−16δ4−1−1 is also tangent to the sphere (x−5)2+(y+2)2+(z−3)2=38(x-5)^2+(y+2)^2+(z-3)^2=38(x−5)2+(y+2)2+( by showing that it is perpendicular to the radius of the sphere at (718)\begin{pmatrix}7\\1\\8\end{pmatrix}718.2 marksShort answer
2022TrialQ16 (a) (ii)Hence show that cos7θ=164(cos7θ+7cos5θ+21cos3θ+35cosθ).\cos^7\theta=\frac1{64}\left(\cos7\theta+7\cos5\theta+21\cos3\theta+35\cos\theta\right).cos7θ=641(cos7θ+7cos5θ+21cos3θ+35cosθ).2 marksShort answer
2022TrialQ16 (a) (i)Prove that cosθ+isinθ−1cosθ+isinθ+1=itanθ2.\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=i\tan\frac\theta2.cosθ+isinθ+1cosθ+isinθ−1=itan2θ.2 marksShort answer
2022TrialQ16 (a) (ii)Find the roots of w5=1w^5=1w5=1. Express your answers in the form cosθ+isinθ\cos\theta+i\sin\thetacosθ+isinθ, where −π<θ≤π-\pi<\theta\le\pi−π<θ≤π.2 marksShort answer
2022TrialQ16 (a) (iii)Using parts (i) and (ii), find the roots of (2+z2−z)5=1.\left(\frac{2+z}{2-z}\right)^5=1.(2−z2+z)5=1.3 marksShort answer
2022HSCQ16 (a)A square in the Argand plane has vertices 5+5i5+5i5+5i, 5−5i5-5i5−5i, −5−5i-5-5i−5−5i and −5+5i-5+5i−5+5i. The complex numbers zA=5+iz_A=5+izA=5+i, zBz_BzB and zCz_CzC lie on the square and form the vertices of an equilateral triangle, as shown in the diagram. Find the exact value of the complex number zBz_BzB.4 marksShort answer
2022HSCQ16 (b)A projectile of mass M kgM\,\mathrm{kg}Mkg is launched vertically upwards from a horizontal plane with initial speed v0 m s−1v_0\,\mathrm{m\,s^{-1}}v0ms−1, where v0<100 m s−1v_0<100\,\mathrm{m\,s^{-1}}v0<100ms−1. The projectile experiences a resistive force of magnitude 0.1Mv0.1Mv0.1Mv newtons, where v m s−1v\,\mathrm{m\,s^{-1}}vms−1 is the speed of the projectile. The acceleration due to gravity is 10 m s−210\,\mathrm{m\,s^{-2}}10ms−2. The projectile lands on the horizontal plane 777 seconds after launch. Find the value of v0v_0v0, correct to 111 decimal place.4 marksShort answer
2022HSCQ16 (c) (i)It is given that for positive numbers x1,x2,x3,…,xnx_1,x_2,x_3,\ldots,x_nx1,x2,x3,…,xn with arithmetic mean AAA, x1x2x3⋯xnAn≤1.\frac{x_1x_2x_3\cdots x_n}{A^n}\le1.Anx1x (Do NOT prove this.) Suppose a rectangular prism has dimensions a,b,ca,b,ca,b,c and surface area SSS. Show that abc≤(S6)3/2.abc\le\left(\frac S6\right)^{3/2}.abc≤(2 marksShort answer
2022HSCQ16 (c) (ii)Using part (i), show that when the rectangular prism with surface area SSS is a cube, it has maximum volume.2 marksShort answer
2022TrialQ16 (a) (i)Using Euler’s theorem, show that if z=eiθ=cosθ+isinθz=e^{i\theta}=\cos\theta+i\sin\thetaz=eiθ=cosθ+isinθ, then zn+z−n=2cosnθ.z^n+z^{-n}=2\cos n\theta.zn+z−n=2cosnθ.1 markShort answer
2022TrialQ16 (a) (iii)Hence find ∫0π/2cos7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2cos7θdθ.2 marksShort answer
2022TrialQ16 (b) (i)If In=∫0π/2cosnθ dθ,I_n=\int_0^{\pi/2}\cos^n\theta\,d\theta,In=∫0π/2cosnθdθ, show that In=n−1nIn−2.I_n=\frac{n-1}{n}I_{n-2}.In=nn−1I2 marksShort answer