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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread I'm using the rule \left(\frac{1}{v}\right)' = - \frac{v'}{v^2} \left(\frac{1}{\ln y}\right)' = - \frac{ ( \ln y )' }{ (\ln y)^2 } = - \frac{\frac{1}{y}}{(\ln y)^2} = - \frac{1}{y( \ln y)^2}

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread dx/dy = -1/(y(lny)^2) so dy/dx = -y(lny)^2 = -e^(1/x)(1/x)^2

Re: MX2 2016 Integration Marathon $ Integrating by parts, $ u = \ln (1+ \sin ^2 x ) . dv = cosec ^2 x dx du = \frac{2 \sin x \cos x}{1 + \sin ^2 x} dx . v = - \cot x $I$ = - \cot x \ln (1+ \sin ^2 x ) + 2 \int \frac{ \cos ^2 x}{1+ \sin ^2 x} dx = - \cot x \ln (1+ \sin ^2 x ) + 2...

Harder 3U (combinatorics and inequalities)

Re: MX2 2016 Integration Marathon There should be absolute values so we need to integrate from -2 to -1 of x(x+1) adding it with integral from -1 to 0 of -x (x+1) and finally adding with the integral of x (x+1) from 0 to 1

Re: HSC 2016 3U Marathon $ Question $ $ If $ \log _{e}s $ and $ \log_{e}r $ are roots to the equation $ 4x^2 - 2x + 5 = 0 $ Find $ \log_{r}s + \log_{s}r

Wish my school offered accelerated

Re: MX2 2016 Integration Marathon Alternative method:

Re: MX2 2016 Integration Marathon = \int \frac{ \ln 2 + \ln x }{x( \ln 3 + \ln x)} dx $ let u $ = \ln x du = \frac{dx}{x} = \int \frac{ \ln 2 + u }{ \ln 3 + u} du = \int \frac{ \ln 3 - \ln 3 + \ln 2 + u}{ \ln 3 + u} du = \int 1 du + \ln \frac{2}{3} \int \frac{1}{ \ln 3 + u} du...

Re: HSC 2016 4U Marathon Draw up the vectors. Remember that anti-clockwise angles are positive. Drawing the left hand side semicircle and the vectors, the angle between them will be -pi/2 not pi/2 so you can only accept the right and side.

Re: HSC 2016 4U Marathon The locus for the original question would be the major arc of a circle with centre to the left of the y axis with the angle between vectors z+i and z being 5pi/4

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $ Use the given condition that the x-values always differ by 2a. I.e. $ 2ap - 2aq = 2a \Rightarrow p - q = 1 $ Squaring the x-coordinate, $ \frac{x^2}{a^2} = (p+q)^2 = p^2 + q^2 + 2pq = (p-q)^2 + 4pq = 1 + 4pq \boxed{1}...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread So we should be multiplying the side joining z2 and z1 by +/- i instead, to find the third vertex. The side joining z2 and z1 is vector z2 - z1

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We should be multiplying the side joining z2 and z1 by +/- i instead, to find the side joining z3 and z1. The side joining z2 and z1 is vector z2 - z1, with the rotated side being z3 - z2 So (z2-z1)i = z3 - z1, expand and...

(n + 2) x^ 2+ 3x - 5 = 0 $ has roots $: \alpha , \beta $ But given that one root is three times the other, then $ \alpha = 3 \beta $roots$: \beta , 3 \beta $ Using the relationships between coefficients of a polynomial and its roots, $ $ Sum of roots $ = 4 \beta = \frac{-b}{a}...

Let y =tan−1(ax/(1-bx)) - tan-1((x-b)/a) and then take the tangent of both sides tany = tan( tan−1(ax/(1-bx)) - tan-1((x-b)/a)) Expanding the RHS using the formula: tan (a-b) = (tana - tanb)/(1+tana tanb) to show tany is independent of x and so y is independent of x

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread (a) $ Recall that $ 1 + \cos \alpha = 2 \cos ^2 \frac{\alpha}{2} $ and $ \sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} z = 1 + \cos E + i \sin E z = 2 \cos ^2 \frac{E}{2} + 2 i \sin \frac{E}{2} \cos...

It says that the distance from P to K (D_PK) is twice the distance from P to line x = -1 (I'll call it D_PX) A written expression for this will be: D_{PK} = 2D_{PX} $ When it asks for distance from a point to a straight line, it refers to the perpendicular distance. Hence, the distance $...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread use the formula: cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2) and sin(a) + sin(b) = 2cos((a+b)/2)sin((a-b)/2) when finding z+w

It sure seems like people are easily aggravated when they see someone ask for help on an online forum