2022TrialQ11 (c)In the Argand diagram below, OA→=z1\overrightarrow{OA}=z_1OA=, , and . Express in terms of and .
2022TrialQ11 (d) (ii)Using the points defined in part (d), find ∣CB→∣|\overrightarrow{CB}|∣CB∣.2 marksShort answer
2022HSCQ11 (f)Using the substitution t=tanx2t=\tan\frac{x}{2}t=tan2x, find ∫dx1+cosx−sinx.\int\frac{dx}{1+\cos x-\sin x}.∫1+cosx−sinxdx.3 marksShort answer
2022HSCQ11 (e)Let ℓ1\ell_1ℓ1 be the line with equation (xy)=(−17)+λ(32),λ∈R.\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-1\\7\end{pmatrix}+\lambda\begin{pmatrix}3\\2\end{pmatrix},\quad \lambda\in\mathbb R.(xy)=(−17)+λ( The line ℓ2\ell_2ℓ2 passes through the point A(−6,5)A(-6,5)A(−6,5) and is parallel to ℓ1\ell_1ℓ1. Find the equation of the line in the form .2 marksShort answer
2022HSCQ11 (c) (ii)Hence, find the exact value of (−3+i)10(-\sqrt3+i)^{10}(−3+i)10, giving your answer in the form x+iyx+iyx+iy.2 marksExtended response
2022HSCQ11 (c) (i)Write the complex number −3+i-\sqrt3+i−3+i in exponential form.2 marksShort answer
2022HSCQ11 (b)Evaluate ∫sin3(2x)cos(2x) dx\displaystyle\int \sin^3(2x)\cos(2x)\,dx∫sin3(2x)cos(2x)dx.2 marksShort answer
2022TrialQ11 (a) (i)If z=−3+iz=-\sqrt3+iz=−3+i and w=1+iw=1+iw=1+i, find Arg(zw)\operatorname{Arg}(zw)Arg(zw) in terms of π\piπ.2 marksShort answer
2022TrialQ11 (a) (ii)If z=−3+iz=-\sqrt3+iz=−3+i and w=1+iw=1+iw=1+i, write zw\frac zwwz in the form x+iyx+iyx+iy, where xxx and yyy are real numbers.2 marksShort answer
2022TrialQ11 (a) (i)Find −1+i31+i\frac{-1+i\sqrt3}{1+i}1+i−1+i3 in Cartesian form.2 marksShort answer
2022HSCQ11 (a)Express 3−i2+i\dfrac{3-i}{2+i}2+i3−i in the form x+iyx+iyx+iy, where xxx and yyy are real numbers.2 marksShort answer
2022TrialQ11 (b) (i)By letting (x+iy)2=7+24i(x+iy)^2=7+24i(x+iy)2=7+24i, where x,yx,yx,y are real, find 7+24i\sqrt{7+24i}7+24i in Cartesian form.3 marksShort answer
2022HSCQ11 (d)A triangle is formed in three-dimensional space with vertices A(1,−1,2)A(1,-1,2)A(1,−1,2), B(0,2,−1)B(0,2,-1)B(0,2,−1) and C(2,1,1)C(2,1,1)C(2,1,1). Find the size of ∠ABC\angle ABC∠ABC, giving your answer to the nearest degree.3 marksShort answer
2022TrialQ11 (a) (ii)Express −1+i3-1+i\sqrt3−1+i3 and 1+i1+i1+i in modulus/argument form.2 marksShort answer
2022TrialQ11 (c)Find the vector equation for the sphere x2+4x+y2−6y+z2=12x^2+4x+y^2-6y+z^2=12x2+4x+y2−6y+z2=12.2 marksShort answer
2022TrialQ11 (a) (iii)Hence find the exact value of tan5π12\tan\frac{5\pi}{12}tan125π.2 marksShort answer
2022TrialQ11 (d)Using De Moivre’s Theorem, find the 3 cube roots of −42+4i2-4\sqrt2+4i\sqrt2−42+4i2. Leave your answers in modulus/argument form.2 marksShort answer
2022TrialQ11 (b) (ii)Hence solve the equation z2−3iz−(4+6i)=0z^2-3iz-(4+6i)=0z2−3iz−(4+6i)=0.2 marksShort answer
2022TrialQ11 (d) (i)Point AAA has position vector 8i+13j−2k8\mathbf i+13\mathbf j-2\mathbf k8i+13j−2k, point BBB has position vector 10i+14j−4k10\mathbf i+14\mathbf j-4\mathbf k10i+14j−4k, and point CCC has position vector 9i+9j+6k9\mathbf i+9\mathbf j+6\mathbf k9i+9j+6k. Find AB→\overrightarrow{AB}AB.1 markShort answer
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 (c)Use proof by contrapositive to prove that: If a2−2a+7a^2-2a+7a2−2a+7 is even, then aaa is odd.3 marksShort answer
2022TrialQ12 (c) (i)Find AAA, BBB and CCC so that Ax−1+Bx+Cx2+4=−7x−18(x−1)(x2+4).\frac{A}{x-1}+\frac{Bx+C}{x^2+4}=\frac{-7x-18}{(x-1)(x^2+4)}.x−1A+x2+4Bx+C3 marksShort answer
2022TrialQ12 (b)Find ∫12+cosx dx\displaystyle\int\frac{1}{2+\cos x}\,dx∫2+cosx1dx.3 marksShort answer