1(a)⎝⎛−11−3⎠⎞.(b)49.2∘.25x+9y=14.3(a)k=−9.(b)f(x)=3x−29x2+29x3+…4(a)212(cos125π+isin125π).(b)z4=2(cos121π+isin121π) or z4=2(cos−1211π+isin−1211π).5(a)a is parallel to b. a=kb,k∈R,k=0.(b)(i)The set of all possible positions of the point R forms a line passing through the
point P and parallel to the direction vector q.(ii)3x−5y+2z=−5.
The set of all possible positions of the point R forms a plane passing through the
point P and perpendicular to the normal vector q.6k=2. t=−8. Other root=−52+56i.7(a)2x+41cos(4x)+C.(b)41π2+81π.(c)49π.8(a)(i)u30=295.(ii)23345.(b)(i)S∞=20.(ii)Smallest n=18.9(a)y=2x and y=−21x are perpendicular since their gradients
are 2 and −21 respectively and m1m2=2(−21)=−1. y=2x makes an angle tan−1(2) clockwise to the positive x-axis and
y=−21x makes an angle −tan−1(−21) anticlockwise to the positive x-axis.
Hence tan−1(2)−tan−1(−21)=21π.(b)0<k≤4.5.(c)(d)(2π−34ln2) units2.10(a)dtdP=−1003P.(b)P=Ae−1003t. As t→∞,P→0.
Hence over many years the number of sheep will approach 0 and there will be no more sheep.(c)dtdP=−1003P+n(d)P=Ae−1003t+3100n.(e)n=15.11(b)x=4a+a2. tanθ=4a+a22.(c)The optimal point and optimal angle may not correspond to the highest chance of scoring.(c)∠KDA≈2π rad.(e)0.0801≤θ<2π