1)a) dy/dx=5(6x-5)(3x^2-5x)^4
1)b) f'(0)=1 (lolx at the slight misprint in the question stem!)
2) Negative hyperbolic graph: x-int at (1,0), y-int at (0,-2)
Vertical asymptote x=-1, horizontal asymptote y=2
3) x={-2pi/9, 2pi/9} over [-pi/2,pi/2]
4)a) Show that k=pi/2 by integration such that A=1
4)b) Pr (X<1/4 | x<1/2) = [2-sqrt(2)]/2
5) C= 0.5log(6)
6)a) Mode=3 (0.4 > {0.3,0.2,0.1})
6)b) Probability=0.3*
{*Derived from Binomial Distribution, where n=2, r=2, p=Pr(X=x):
Pr(X=2)=(2 C 2)[Pr(X=x)]^2[1-Pr(X=x)]^0 = [Pr(X=x)]^2
=> Probability = 0.1^2+0.2^2+0.3^2+0.4^2 = 0.3}
7) Markov Chain: Pr(CCD) + Pr(CDC) + Pr(DCC) = 0.336
8)a) dom f' = R/{1,2}
8)b) Modulus graph...stationary points (local maxima) (-1,4), (2/3 , 17/27)
Non-inclusive endpoint at (1,0), flip the given curve such that y>=0
9)a) y=[4000sqrt(3)]/[3(x)^2]
9)b) Show that A=[4000sqrt(3)]/x + [(sqrt(3)x^2)/2]
9)c) Minimum** Surface Area at x=cuberoot(4000) =10cuberoot(4) cm (looking for tangible evidence that VCAA overlooks root simplification...)
{**The graph for A(x) is an oblique curve whose only stationary point is a local minimum at x=10cuberoot(4), hence the minimum surface area occurs at x=10cuberoot(4)}
10)a) Inverse f: (-1,infinity)->R, inverse f(x) = 0.5loge(x+1)
10)b) Graph of y=x, x>1 with non-inclusive endpoint at (-1,-1)
10)c) (-2x)/(2x+1)*** => a=-2,b=2,c=1
{***We need to evaluate f[-inverse f(2x)] => -inverse f(2x) = -0.5loge(2x+1) = 0.5loge[(2x+1)^(-1)] = 0.5loge[1/(2x+1)]
f[-inverse f(2x)] = e^2[0.5loge[1/(2x+1)]]-1 = 1/(2x+1) - 1 = [1-2x-1]/[2x-1] = (-2x)/(2x-1), QED}
From another forum.
Apparently there was a misprint in the first question.
Pretty good. I struggled a bit with that question, but realised that the height wasn't xcm, so I got it out in the end. That last question (f(^-1(x))) was a shocker.