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}.
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 (e)Use the substitution x=2sinθx=2\sin\thetax=2sinθ to find ∫dx(4−x2)3/2.\int\frac{dx}{(4-x^2)^{3/2}}.∫(4−x2)3/2dx.3 marksShort answer
2022TrialQ13 (b)Use the substitution t=tanx2t=\tan\frac x2t=tan2x, or otherwise, to evaluate ∫0π/213sinx−4cosx+5 dx.\int_0^{\pi/2}\frac{1}{3\sin x-4\cos x+5}\,dx.∫0π/23sinx−4cosx+513 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 (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) (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
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
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) (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 (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
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
2022TrialQ16 (a) (iii)Hence find ∫0π/2cos7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2cos7θdθ.2 marksShort answer
2022TrialQ16 (c)Find ∫lnx−1(x+lnx)2 dx.\int\frac{\ln x-1}{(x+\ln x)^2}\,dx.∫(x+lnx)2lnx−1dx.3 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
2022TrialQ16 (b) (ii)If In=∫0π/2cosnθ dθ,I_n=\int_0^{\pi/2}\cos^n\theta\,d\theta,In=∫0π/2cosnθdθ, hence find ∫0π/2cos7θ dθ.\int_0^{\pi/2}\cos^7\theta\,d\theta.∫0π/2cos7θdθ.2 marksShort answer
2021HSCQ2Which expression is equal to ∫x5e7x dx\int x^5e^{7x}\,dx∫x5e7xdx? A. 17x5e7x−57∫x4e7x dx\frac17x^5e^{7x}-\frac57\int x^4e^{7x}\,dx71x5e7x−7 B. 17x5e7x−57∫x5e7x dx\frac17x^5e^{7x}-\frac57\int x^5e^{7x}\,dx71x5e7x− C. 57x4e7x−57∫x4e7x dx\frac57x^4e^{7x}-\frac57\int x^4e^{7x}\,dx75x4e7x− D. 57x4e7x−57∫x5e7x dx\frac57x^4e^{7x}-\frac57\int x^5e^{7x}\,dx75x4e7x−1 markMultiple choice
2021HSCQ11 (f)Express 3x2−5(x−2)(x2+x+1)\frac{3x^2-5}{(x-2)(x^2+x+1)}(x−2)(x2+x+1)3x2−5 as a sum of partial fractions over R\mathbb RR.3 marksShort answer
2021HSCQ13 (c) (ii)Region AAA is bounded by y=1y=1y=1 and x2+y2=1x^2+y^2=1x2+y2=1 between x=0x=0x=0 and x=1x=1x=1. Region BBB is bounded by y=1y=1y=1 and y=lnxy=\ln xy=lnx between x=1x=1x=1 and x=ex=ex=e. Using part (i), or otherwise, show that the ratio of the volumes formed by rotating regions AAA and BBB about the xxx-axis is VA:VB=1:3V_A:V_B=1:3VA:VB=1:3.4 marksShort answer
2021HSCQ15 (c) (i)An object of mass 1 kg1\,\mathrm{kg}1kg is projected vertically upwards with initial velocity u m s−1u\,\mathrm{m\,s^{-1}}ums−1. It experiences air resistance of magnitude kv2kv^2kv2 newtons, where vvv is velocity in m s−1\mathrm{m\,s^{-1}}ms−1 and k>0k>0k>0. The height above its starting point is xxx metres, time is ttt seconds, and acceleration due to gravity is g m s−2g\,\mathrm{m\,s^{-2}}gms−2. Show that the time to reach maximum height is 1gkarctan(ukg) seconds.\frac1{\sqrt{gk}}\arctan\left(u\sqrt{\frac{k}{g}}\right)\text{ seconds}.gk13 marksShort answer
2021HSCQ15 (c) (ii)For the resisted vertical motion described in part (i), find an expression for the maximum height reached by the object, in terms of kkk, ggg and uuu.3 marksShort answer