IBDP Maths AA: Topic: SL 2.1: equation of a straight line: IB style Questions SL Paper 1

Ans:

(a)

(i)

f'(x)= \(-kx^{-2}\)

\(f'(p)= -kp^{-2} (= – \frac{k}{p^{2}})\)

(ii)

attempt to use point and gradient to find equation of L₁

eg y- \(\frac{k}{p}= – kp^{-2}(x-p), \frac{k}{p}= -\frac{k}{p^{2}}(p)\)+b

correct working leading to answer

eg \(p^{2}y-kp = -kx+kp, y -\frac{k}{p} =-\frac{k}{p^{2}}x+ \frac{k}{p}y=-\frac{k}{p^{2}}x+\frac{2k}{p}\)

\(kx+p^{2}\)y-2pk=0

(b)

METHOD 1

– area of a triangle

recognizing x = 0 at B

correct working to find y -coordinate of B

\(p^{2}y-2pk=0\)

y -coordinate of B at  y = \(\frac{2k}{p}\)(may be seen in area formula)

correct substitution to find area of triangle

eg \(\frac{1}{2}(2p)(\frac{2k}{p}),p \times  (\frac{2k}{p})\)

area of triangle AOB = 2k

METHOD 2 –

integration  recognizing to integrate L₁ between 0 and 2p

eg \(\int_{0}^{2p}L_{1}dx, \int_{0}^{2p} – \frac{k}{p^{2}}x+ \frac{2k}{p}\)

correct integration of both terms

eg \(-\frac{kx^{2}}{2p^{2}}+\frac{2kx}{p},-\frac{k}{2p^{2}}x^{2}+\frac{2k}{p}x+c\),

substituting limits into their integrated function and subtracting (in either order)

eg \(-\frac{k(2p^{2})}{2p^{2}}+\frac{2k(2p)}{p}-(0),-\frac{4kp^{2}}{2p^{2}}+\frac{4kp}{p}\)

correct working

eg -2k+4k

area of triangle AOB = 2K

(c)

recognizing use of transformation

eg area of triangle AOB = area of triangle DEF, g(x)= \(\frac{k}{x-4}+3\)

gradient of \(L_{2}\)= gradient of \(L_{1} ,D(4,3)\),2p+4, one correct shift

correct working

eg area of tringle DEF = 2k, CD = 3, DF= 2P, CG= 2P, E(4,\(\frac{2k}{p}+3)\)

area of rectangle CDFG = 2k

valid approach

eg \(\frac{ED\times DF}{2}\)=\(CD\times DF 2P.3= 2K ,ED =CD \), \(\int_{4}^{2P+4}L_{2}\)dx= 4k

correct working

eg ED = 6, E(4,9),K= 3p, gradient = \(\frac{3-(\frac{2k}{p}+3)}{(2p+4)-4},\frac{-6}{\frac{2k}{3}},-\frac{9}{k}\)

eg \(\frac{-6}{2p},\frac{9-3}{4-(2p+4)},-\frac{3p}{p^2},\frac{3-(\frac{2(3p)}{p}+3)}{(2p+4)-4}, – \frac{9}{3p}\)

gradient of

\(L_{2}is – \frac{3}{p}(=-3p^{-1})\)