2022TrialQ12 (a) (i)Find AAA and BBB such that 15x2−2x−15=Ax+3+Bx−5.\frac{15}{x^2-2x-15}=\frac A{x+3}+\frac B{x-5}.x2−2x−1515=x+3A+x−5B.2 marksShort answer
2022TrialQ12 (a) (ii)Hence, or otherwise, find ∫x2−2xx2−2x−15 dx.\int\frac{x^2-2x}{x^2-2x-15}\,dx.∫x2−2x−15x2−2xdx.2 marksShort answer
2022TrialQ12 (b)Evaluate ∫0π/4sec4xtan2x dx.\int_0^{\pi/4}\sec^4x\tan^2x\,dx.∫0π/4sec4xtan2xdx.3 marksShort answer
2022TrialQ12 (d) (i)Using the substitution x=secθx=\sec\thetax=secθ or otherwise show by integrating that ∫1x2−1 dx=ln(x+x2−1).\int\frac{1}{\sqrt{x^2-1}}\,dx=\ln\left(x+\sqrt{x^2-1}\right).∫x2−11dx=ln(x+x2−1).2 marksShort answer
2022TrialQ12 (c) (ii)Hence find ∫−7x−18(x−1)(x2+4) dx.\int\frac{-7x-18}{(x-1)(x^2+4)}\,dx.∫(x−1)(x2+4)−7x−18dx.2 marksShort answer
2022HSCQ12 (e)Given the complex number z=eiθz=e^{i\theta}z=eiθ, show that w=zˉ2−1zˉ2+1w=\frac{\bar z^2-1}{\bar z^2+1}w=zˉ2+1zˉ2−1 is purely imaginary.3 marksShort answer
2022HSCQ12 (d)Using partial fractions, evaluate ∫2n4+x(1−x)(4+x2) dx,\int_2^n\frac{4+x}{(1-x)(4+x^2)}\,dx,∫2n(1−x)(4+x2)4+xdx, giving your answer in the form 12ln(f(n)8(n−1)2),\frac12\ln\left(\frac{f(n)}{8(n-1)^2}\right),21ln(8(n−1) where f(n)f(n)f(n) is a function of nnn.4 marksShort answer
2022HSCQ12 (c) (i)A particle with mass 1 kg1\,\mathrm{kg}1kg is moving along the xxx-axis. Initially, the particle is at the origin and has speed u m s−1u\,\mathrm{m\,s^{-1}}ums−1 to the right. The particle experiences a resistive force of magnitude v+3v2v+3v^2v+3v2 newtons, where v m s−1v\,\mathrm{m\,s^{-1}}vms−1 is the speed of the particle after ttt seconds. The particle is never at rest. Show that dvdx=−(1+3v).\frac{dv}{dx}=-(1+3v).dxdv=−(1+3v).1 markShort answer
2022HSCQ13 (c) (i)Consider the equation z5+1=0z^5+1=0z5+1=0, where zzz is a complex number. Solve the equation z5+1=0z^5+1=0z5+1=0 by finding the fifth roots of −1-1−1.2 marksShort answer
2022HSCQ13 (a)Prove that for all integers nnn with n≥3n\ge3n≥3, if 2n−12^n-12n−1 is prime, then nnn cannot be even.3 marksShort answer
2022HSCQ13 (b)The numbers ana_nan, for integers n≥1n\ge1n≥1, are defined as a1=2,a2=2+2,a3=2+2+2,a_1=\sqrt2,\quad a_2=\sqrt{2+\sqrt2},\quad a_3=\sqrt{2+\sqrt{2+\sqrt2}},a1=2,a2= and so on. These numbers satisfy the relation an+12=2+ana_{n+1}^2=2+a_nan+12=2+an, for n≥1n\ge1. (Do NOT prove this.) Use mathematical induction to prove that an=2cosπ2n+1,a_n=2\cos\frac{\pi}{2^{n+1}},an=2cos2n+1 for all integers .4 marksShort answer
2022HSCQ13 (c) (ii)Show that if zzz is a solution of z5+1=0z^5+1=0z5+1=0 and z≠−1z\ne-1z=−1, then u=z+1zu=z+\frac1zu=z+z1 is a solution of u2−u−1=0u^2-u-1=0u2−u−1=0.2 marksShort answer
2022HSCQ13 (c) (iii)Hence, find the exact value of cos3π5\cos\frac{3\pi}{5}cos53π.3 marksShort answer
2022TrialQ13 (b) (ii)Prove by contraposition that if n2n^2n2 is not divisible by 181818, then nnn is not divisible by 666.2 marksShort answer
2022TrialQ13 (a)Prove that if aaa is divisible by 555, then a2−5aa^2-5aa2−5a is divisible by 505050. (You may assume that the product of two consecutive numbers is even.)4 marksShort answer
2022TrialQ13 (c) (i)Prove for all real aaa and bbb that a2+b2≥2ab.a^2+b^2\ge 2ab.a2+b2≥2ab.1 markShort answer
2022TrialQ13 (c) (ii)Hence prove for all a>0a>0a>0 that a+1a≥2.a+\frac1a\ge2.a+a1≥2.1 markShort answer
2022TrialQ13 (c) (iii)Hence prove for all a1,a2>0a_1,a_2>0a1,a2>0 that (a1+a2)(1a1+1a2)≥4.(a_1+a_2)\left(\frac1{a_1}+\frac1{a_2}\right)\ge4.(a1+a2)(a11 markShort answer
2022TrialQ13 (c) (iv)Hence prove by induction for all a1,a2,a3,…,an>0a_1,a_2,a_3,\ldots,a_n>0a1,a2,a3,…,an>0 that (a1+a2+⋯+an)(1a1+1a2+⋯+1an)≥n2.(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)\ge n^2.(a1+a2+ (Note: this has already been proven for n=1n=1n=1 and n=2n=2n=2.)3 marksShort answer