2022TrialQ13 (c) (ii)The equation of line l2l_2l2 is x+3k=4−y6=z−1.\frac{x+3}{k}=\frac{4-y}{6}=z-1.kx+3=64−y=z−1. Find the value of kkk for which l1l_1l1 and l2l_2l2 intersect.3 marksShort answer
2022TrialQ13 (d)The points XXX, YYY and ZZZ bisect ABABAB, BCBCBC and CACACA respectively. Show that AY→+BZ→+CX→=0⃗.\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX}=\vec0.AY+BZ3 marksShort answer
2022HSCQ14 (b) (iv)Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn=∫01xne−xdx, where nnn is a non-negative integer. Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all n≥0n\ge0n≥0, Jn=n!−n!e∑r=0n1r!.J_n=n!-\frac{n!}{e}\sum_{r=0}^{n}\frac1{r!}.Jn=n!−en!2 marksShort answer
2022HSCQ14 (b) (iii)Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn=∫01xne−xdx, where nnn is a non-negative integer. Show that Jn=nJn−1−1e,n≥1.J_n=nJ_{n-1}-\frac1e,\qquad n\ge1.Jn=nJn−1−2 marksShort answer
2022HSCQ14 (b) (ii)Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn=∫01xne−xdx, where nnn is a non-negative integer. Show that Jn≤1n+1.J_n\le\frac1{n+1}.Jn≤n+11.2 marksShort answer
2022TrialQ14 (d) (i)Use a suitable substitution to show that ∫0af(x) dx=∫0af(a−x) dx.\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx.∫0af(x)dx=∫0af(a−x)dx.1 markShort answer
2022TrialQ14 (c) (i)On the Argand diagram, zzz and zwzwzw are represented by AAA and BBB. Given z=reiθz=re^{i\theta}z=reiθ and w=eiπ/3w=e^{i\pi/3}w=eiπ/3, where r>0r>0r>0, explain why OABOABOAB is an equilateral triangle.2 marksShort answer
2022TrialQ14 (b)Simplify fully 11+ω+11+ω2,\frac1{1+\omega}+\frac1{1+\omega^2},1+ω1+1+ω21, given that ω\omegaω is a non-real cube root of unity.2 marksShort answer
2022TrialQ14 (a) (i)A particle is moving so that v2=n2(a2−x2)v^2=n^2(a^2-x^2)v2=n2(a2−x2). Show that the particle is moving in simple harmonic motion.1 markShort answer
2022TrialQ14 (a) (ii)If the particle is moving at 15 m/s\sqrt{15}\,\mathrm{m/s}15m/s when it is 2 m2\,\mathrm m2m to the right of the centre of motion and at 23 m/s2\sqrt3\,\mathrm{m/s}23m/s when it is 4 m4\,\mathrm m4m to the right of the centre of motion, find the period and amplitude of the motion.3 marksShort answer
2022TrialQ14 (b) (i)A particle is moving so that x=2sin3t−23cos3tx=2\sin3t-2\sqrt3\cos3tx=2sin3t−23cos3t. By expressing the motion in the form Asin(3t−α)A\sin(3t-\alpha)Asin(3t−α), find the period and amplitude.3 marksShort answer
2022TrialQ14 (b) (ii)A particle is moving so that x=2sin3t−23cos3tx=2\sin3t-2\sqrt3\cos3tx=2sin3t−23cos3t. Find the initial position and velocity of the particle.2 marksShort answer
2022TrialQ14 (c) (i)A 20 kg20\,\mathrm{kg}20kg projectile is launched upwards from the ground at 500 m/s500\,\mathrm{m/s}500m/s. It experiences a gravitational force of 200 N200\,\mathrm N200N and air resistance, opposite to its motion, of v218 N\frac{v^2}{18}\,\mathrm N18v2N. Letting the ground be x=0x=0x=0, show that the acceleration is x¨=−10−v2360.\ddot{x}=-10-\frac{v^2}{360}.x¨=−10−360v21 markShort answer
2022TrialQ14 (c) (ii)Show that x=180ln(2536003600+v2)x=180\ln\left(\frac{253600}{3600+v^2}\right)x=180ln(3600+v2253600) and find the maximum height the projectile reaches.3 marksShort answer
2022TrialQ14 (c) (iii)Show that the time taken for the projectile to reach velocity v m/sv\,\mathrm{m/s}vm/s is t=6tan−1(253)−6tan−1(v60),t=6\tan^{-1}\left(\frac{25}{3}\right)-6\tan^{-1}\left(\frac{v}{60}\right),t=6tan−1(325)−6tan−1(60v), and find the time taken for the projectile to reach its maximum height.2 marksShort answer
2022TrialQ14 (a)If x≤1x\le1x≤1 and y≥1y\ge1y≥1, show that x+y≥1+xyx+y\ge1+xyx+y≥1+xy.2 marksShort answer
2022HSCQ14 (b) (i)Let Jn=∫01xne−x dx,J_n=\int_0^1x^ne^{-x}\,dx,Jn=∫01xne−xdx, where nnn is a non-negative integer. Show that J0=1−1e.J_0=1-\frac1e.J0=1−e1.1 markShort answer
2022HSCQ14 (a) (iv)The point PPP is defined by AP→=−6AB→−8AC→.\overrightarrow{AP}=-6\overrightarrow{AB}-8\overrightarrow{AC}.AP=−6AB−8AC. Does PPP lie on the line ALALAL? Justify your answer.2 marksShort answer
2022HSCQ14 (a) (iii)Using part (ii), or otherwise, determine the position of LLL by showing that BL→=47BC→.\overrightarrow{BL}=\frac47\overrightarrow{BC}.BL=74BC.2 marksShort answer
2022HSCQ14 (b) (v)Let Jn=∫01xne−x dxJ_n=\int_0^1x^ne^{-x}\,dxJn=∫01xne−xdx. Using parts (ii) and (iv), prove that e=limn→∞∑r=0n1r!.e=\lim_{n\to\infty}\sum_{r=0}^{n}\frac1{r!}.e=limn→∞∑r=0n1 markShort answer
2022HSCQ14 (a) (i)The two non-parallel vectors u⃗\vec uu and v⃗\vec vv satisfy λu⃗+μv⃗=0⃗\lambda\vec u+\mu\vec v=\vec0λu+μv for some real numbers λ\lambdaλ and μ\muμ. Show that λ=μ=0\lambda=\mu=0λ=μ=0.2 marksShort answer
2022TrialQ14 (e)A recurrence relation is defined by un+1=3un−14un−1,u1=1.u_{n+1}=\frac{3u_n-1}{4u_n-1},\qquad u_1=1.un+1=4un−13un−1,u1=1. Use mathematical induction to prove that un=n2n−1for n≥1.u_n=\frac n{2n-1}\quad\text{for }n\ge1.un=2n−1nfor n≥3 marksShort answer
2022TrialQ14 (d) (ii)Hence, or otherwise, evaluate ∫0π/4sinxcosx+sinx dx.\int_0^{\pi/4}\frac{\sin x}{\cos x+\sin x}\,dx.∫0π/4cosx+sinxsinxdx.3 marksShort answer
2022TrialQ15 (a) (ii)Show by finding it that the point of intersection between the lines (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 (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} is (718)\begin{pmatrix}7\\1\\8\end{pmatrix}718.3 marksShort answer