2023TrialQ11 (b) (i)The nnnth pentagonal number can be represented by the arithmetic series P5(n)=1+4+7+⋯+(3n−2).P_5(n)=1+4+7+\cdots +(3n-2).P5
2023HSCQ11 (a)Solve the quadratic equation z2−3z+4=0z^2-3z+4=0z2−3z+4=0, where zzz is a complex number. Give your answers in Cartesian form.2 marksShort answer
2023TrialQ11 (e) (i)In the diagram below, OABCDOABCDOABCD is a solid figure where ∣OA→∣=∣OB→∣=4|\overrightarrow{OA}|=|\overrightarrow{OB}|=4∣OA∣=∣OB∣=4 units and ∣OD→∣=3|\overrightarrow{OD}|=3∣OD∣=3 units. The edge OD→\overrightarrow{OD}OD is vertical, DC→\overrightarrow{DC}DC is parallel to OB→\overrightarrow{OB}OB and ∣DC→∣=1|\overrightarrow{DC}|=1∣DC∣=1 unit. The base, OABOABOAB, is horizontal and ∠AOB=90∘\angle AOB=90^\circ∠AOB=90∘. Unit vectors i⃗\vec ii, j⃗\vec jj and k⃗\vec kk are parallel to OA→\overrightarrow{OA}OA, OB→\overrightarrow{OB}OB and OD→\overrightarrow{OD}OD respectively. The midpoint of AB‾\overline{AB}AB is MMM and the point NNN on BC‾\overline{BC}BC is such that NC→=2BN→\overrightarrow{NC}=2\overrightarrow{BN}NC=2BN. Express vectors MD→\overrightarrow{MD}MD and ON→\overrightarrow{ON}ON in terms of , and .2 marksShort answer
2023HSCQ11 (e)A particle moves in simple harmonic motion described by x¨=−9(x−4)\ddot x=-9(x-4)x¨=−9(x−4). Find the period and the central point of motion.2 marksShort answer
2023TrialQ11 (c)Given a∈Za\in\mathbb{Z}a∈Z, prove that if 3a2−4a+53a^2-4a+53a2−4a+5 is even, then aaa is odd.2 marksShort answer
2023HSCQ11 (b)Find the angle between the vectors a=i+2j−3k\mathbf a=\mathbf i+2\mathbf j-3\mathbf ka=i+2j−3k and b=−i+4j+2k\mathbf b=-\mathbf i+4\mathbf j+2\mathbf kb=−i+4j+2k, giving your answer to the nearest degree.3 marksShort answer
2023TrialQ11 (b) (ii)A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers. For an rrr-sided regular polygon, where r∈Z+r\in\mathbb{Z}^+r∈Z+, r≥3r\geq 3r≥3, the nnnth polygonal number Pr(n)P_r(n)Pr(n) is given by Pr(n)=(r−2)n2−(r−4)n2P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}Pr(n)=2(r−2) where n∈Z+n\in\mathbb{Z}^+n∈Z+. Hence, the nnnth triangular number can be expressed as P3(n)=n2+n2.P_3(n)=\frac{n^2+n}{2}.P3(n)=2n2+n The nnnth polygonal number, Pr(n)P_r(n)Pr(n), can be represented by the series Pr(n)=∑m=1n(1+(m−1)(r−2)),P_r(n)=\sum_{m=1}^{n}\left(1+(m-1)(r-2)\right),Pr(n)=m=1∑ where r∈Z+r\in\mathbb{Z}^+r∈Z+, r≥3r\ge 3r≥3. Use mathematical induction to prove that Pr(n)=(r−2)n2−(r−4)n2P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}Pr(n)=2(r−2) where n∈Z+n\in\mathbb{Z}^+n∈Z+.3 marksShort answer
2023TrialQ12 (b) (iii)Hence or otherwise, evaluate ∫0π/2x1+cosx+sinx dx.\int_0^{\pi/2}\frac{x}{1+\cos x+\sin x}\,dx.∫0π/21+cosx+sinxxdx.2 marksShort answer
2023TrialQ12 (a) (iii)Let α\alphaα be a real number and suppose that zzz is a complex number such that z+1z=2cosα.z+\frac{1}{z}=2\cos\alpha.z+z1=2cosα. You may assume that zn+1zn=2cos(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1= for all positive integer nnn. Let ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1. Hence or otherwise, find all solutions of 16(cosα)4+8(cosα)3−12(cosα)2−4cosα+1=0.16(\cos\alpha)^4+8(\cos\alpha)^3-12(\cos\alpha)^2-4\cos\alpha+1=0.16(cosα)4+8(cosα)3−2 marksShort answer
2023HSCQ12 (e) (i)The complex number 2+i2+i2+i is a zero of P(z)=z4−3z3+cz2+dz−30P(z)=z^4-3z^3+cz^2+dz-30P(z)=z4−3z3+cz2+dz−30, where c,d∈Rc,d\in\mathbb Rc,d∈R. Explain why 2−i2-i2−i is also a zero of P(z)P(z)P(z).1 markShort answer
2023TrialQ12 (c) (i)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Show that In=1−nIn−1I_n=1-nI_{n-1}In=1−nIn−1 for n=1,2,3,….2 marksShort answer
2023TrialQ12 (c) (ii)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Show that Jn=1−(n+2)InJ_n=1-(n+2)I_nJn=1−(n+2)In for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,….2 marksShort answer
2023HSCQ12 (c) (ii)Using the same inclined-plane model, given that the object is initially at rest, find its velocity v(t)\mathbf v(t)v(t) in terms of ggg, θ\thetaθ, ttt and i\mathbf ii.2 marksShort answer
2023TrialQ12 (a) (i)Let α\alphaα be a real number and suppose that zzz is a complex number such that z+1z=2cosα.z+\frac{1}{z}=2\cos\alpha.z+z1=2cosα. You may assume that zn+1zn=2cos(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1= for all positive integer nnn. Let ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1. Show that ω4+ω3−3ω2−2ω=z+1z+z2+1z2+z3+1z3+z4+1z4.\omega^4+\omega^3-3\omega^2-2\omega=z+\frac{1}{z}+z^2+\frac{1}{z^2}+z^3+\frac{1}{z^3}+z^4+\frac{1}{z^4}.ω4+ω3−32 marksShort answer
2023HSCQ12 (b)Prove that for all real numbers xxx and yyy, where x2+y2≠0x^2+y^2\ne0x2+y2=0, (x+y)2x2+y2≤2.\frac{(x+y)^2}{x^2+y^2}\le2.x2+y2(x+y)2 marksShort answer
2023TrialQ12 (a) (ii)Let α\alphaα be a real number and suppose that zzz is a complex number such that z+1z=2cosα.z+\frac{1}{z}=2\cos\alpha.z+z1=2cosα. You may assume that zn+1zn=2cos(nα)z^n+\frac{1}{z^n}=2\cos(n\alpha)zn+zn1= for all positive integer nnn. Let ω=z+1z.\omega=z+\frac{1}{z}.ω=z+z1. Find the ninth roots of unity.2 marksShort answer
2023HSCQ12 (d)Find the cube roots of 2−2i2-2i2−2i. Give your answer in exponential form.3 marksShort answer
2023HSCQ12 (c) (i)An object of mass mmm slides down a smooth inclined plane. The plane makes an angle θ\thetaθ with the horizontal. The normal reaction is R\mathbf RR, gravity is g\mathbf gg with magnitude ggg, and i,j\mathbf i,\mathbf ji,j are unit vectors parallel and perpendicular to the plane. Show that the resultant force is F=−(mgsinθ)i.\mathbf F=-(mg\sin\theta)\mathbf i.F=−(mgsinθ)i.2 marksShort answer
2023TrialQ12 (b) (i)Show that ∫0π/2dx1+cosx+sinx=ln2.\int_0^{\pi/2}\frac{dx}{1+\cos x+\sin x}=\ln 2.∫0π/21+cosx+sinxdx=ln2.3 marksShort answer
2023TrialQ12 (b) (ii)By making the substitution u=a−xu=a-xu=a−x, 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
2023TrialQ12 (c) (iii)Let In=∫e−11(1+logex)n dxI_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dxIn=∫e−11(1+logex)ndx and Jn=∫e−11(logex)(1+logex)n dxJ_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dxJn=∫e for n=0,1,2,3,…n=0,1,2,3,\ldotsn=0,1,2,3,…. Hence find the value of J3J_3J3 in simplest exact form.2 marksShort answer
2023TrialQ13 (d) (ii)Another line, ℓ2\ell_2ℓ2, is defined by the vector equation r⃗=(012)+λ(a1−1),λ,a∈R.\vec r=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}a\\1\\-1\end{pmatrix},\qquad \lambda,a\in\mathbb{R}.r=0 Find the possible values of aaa when the angle between ℓ1\ell_1ℓ1 and ℓ2\ell_2ℓ2 is π4\frac{\pi}{4}.2 marksShort answer
2023TrialQ13 (c) (i)Find real numbers aaa, bbb and ccc such that 10(x+1)(x2+4)≡ax+1+bx+cx2+4.\frac{10}{(x+1)(x^2+4)}\equiv\frac{a}{x+1}+\frac{bx+c}{x^2+4}.(x+1)(x2+4)10≡x+1a3 marksShort answer
2023TrialQ13 (a) (i)Use De Moivre’s Theorem to solve the equation z3=42(1+i).z^3=4\sqrt{2}(1+i).z3=42(1+i).2 marksShort answer