Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.
Noticing the denominator of y, we find that there is an asymptote at x = -1.
And because of the exponential properties of e^-x, y tends to zero as x→+∞
y tends to −∞ as x→−∞ since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.
While we're nitpicking - taking the 2nd derivative to deduce the nature of the turning points is a lot more work than simply sketching x(2-x^2) to argue whether the derivative is increasing or decreasing.
STEP III Q10 Consider the positions of the dogs as time goes on. They will form a square of decreasing side length that rotates about the centre. If we take the position of one of the dogs to be (x,y) from the centre, then we see by symmetry that the position of one the next dogs along is (−y,x). We can also see that the tangent at any moment goes through the position of the next dog along, so: dxdy=x−(−y)y−x=1+xyxy−1 let: u=xy⇒dxdy=dxdux+u[br]∴dxdux+u=1+uu−1[br]⇒dxdux=1+u−u2−1[br]⇒dudxx1=−u2−11+u[br]⇒x1dx=−u2−11+udu let θ be the angle between the x axis and the straight line to the point (x,y). Since u=xy and, by trigonometry, xy=Tanθ⇒u=Tanθ. So, substituting into the RHS and integrating we get: lnx+c1=−ln(secθ)−θ+c2. By trigonometry, again, x=rcosθ where r is the distance from the origin to the point (x,y)⇒ln(rcosθ)+ln(secθ)=−θ+c⇒ln(r)=−θ+c⇒r=e−θec ec is constant so r=e−θλ for some constant λ as required. To generalise to n-dogs held at the vertices of an n-gon it must be noted that the next dog along will have the coordinates of the first dog bur rotated n2π about the centre of the field i.e: (cos(n2π)sin(n2π)−sin(n2π)cos(n2π))(xy)=(xcosn2π−ysinn2πxsinn2π+ycosn2π) So, like the previous part,: dxdy=x−xcosn2π+ysinn2πy−xsinn2π−ycosn2π=1−cosn2π+yx−1sinn2πyx−1−sinn2π−yx−1cosn2π let: u=xy⇒dxdy=dxdux+u[br]∴dxdux=1−cosn2π+usinn2π−sinn2π−u2sinn2π[br]⇒dudxx1=−cosec(n2π)u2+11−cosn2π+usinn2π let: u=Tanθ⇒dθdu=sec2θ[br]⇒dudxx1=−cosec(n2π)sec2θ1−cosn2π+tanθsinn2π[br]⇒x1dx=(−cosec(n2π)+cotn2π−Tanθ)dθ Integrating both sides we get: lnx+c=(−cosec(n2π)+cotn2π)θ−lnsecθ+c And now, just like in the previous part: ln(rcosθ)+ln(secθ)+c=(−cosec(n2π)+cotn2π)θ+c[br]⇒r=e−θ(cosec(n2π)−cotn2π)ec=e−θ(sinn2π1−cosn2π)λ=e−θTan(nπ)λ for some constant λ. Letting n=4 we can check and see that it gives the same answer as the first part. EDIT: Just realised that the final result simplified to the much nicer form that it is currently in.
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
∣z−i∣=2 is the circle on the complex plane with radius 2 and centre (0,i)
so x^2 + (y - 1)^2 = 2
and therefore in paramatrised form
x=2cosθ
y=1+2sinθ
I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.
(i)
w=z−iz+i
wz−wi=z+i
wz−z=i+wi
z(w−1)=i(w+1)
z=w−1i(w+1)
z−i=w−1i(w+1)−i=w−1i(w+1)−i(w−1)
=w−11+i
∣z−i∣=w−11+i
2=∣w−1∣∣1+i∣
∣w−1∣=22
Okay so its a circle of radius 22 and the centre is (1,0)
(ii)
If z = x + iy and is real then y must = 0
so
w=u+iv=x+i(y−1)x+i(y+1)=x−ix+i
(u+iv)(x−i)=x+i
ux−ui+vxi+v=x+i
ux+v+i(−u+vx−1)=x
-u + vx - 1 must = 0 (this is important bit on)
so
ux+v=x
x(1−u)=v
x=1−uv
Now from before
−u+vx−1=0
vx=u+1
x=vu+1
so
1−uv=vu+1
v2=(u+1)(1−u)
u2+v2=1
So in the W plane, when z is real, you get a circle centre(0,0) with radius 1
(iii)
if z = x + iy and z is imaginary then x = 0
u+vi=x+i(y−1)x+i(y+1)
u+vi=i(y−1)i(y+1)
(u+vi)(iy−i)=yi+i
uyi−ui−vy+v=yi+i
yi−uyi=−ui−vy+v−i
y(i−ui)=i(−1−u)−vy+v
−vy+v=0
y=i(1−u)i(−1−u)
y=1−u−1−u
1=1−u−1−u
−1−u=1−u
0=2
Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.
⇒∣2i∣=2∣w−1∣ i.e. ∣w−1∣=1 so it is a circle, centre w=1 radius 1 (ii)z real ⇒Im((u−1)i−v(u+1)i−v)= Im((u−1)2+v2((u+1)i−v)((u−1)i−v))=−(u−1)2+v22uv=0 hence, 2uv=0 i.e. It is the two axes (iii)z imaginary ⇒Re(w−=1w+1)i=0 i.e. (u−1)2+v2v2−(u2−1)=0⇒v2−u2+1=0 i.e. A hyperbola
h(x)=xlnx⇒h′(x)=x21−lnx=0 when lnx=1⇒x=e for small h,h′(1−h)>0 while h′(1+h)<0 so a maximum point at (e,e−1 x=1⇒h(x)=0 h(x)→0 as x→∞ h(x)→−∞ as x→0
nm=mn⇒mlnn=nlnm⇒nlnn=mlnm Clearly, m and n must lie on either side of the turning point so the only possible values for the smaller of m and n are 1 and 2 there is obviously no other integer with 1m=m1 but 24=42 so only pair of integers is 2 and 4
g(x)= f(x)+f(x)1⇒ g′(x)= f′(x)−(f(x))2f′(x) and g′′(x)= f′′(x)−(f(x))4(f(x))2f′′(x)−f′(x).2f(x)f′(x) i.e. g′′(x)= f′′(x)−(f(x))2f′′(x)+(f(x))32(f′(x))2
Unparseable latex formula:
\text{f}(x)=4+ \cos2x+2 \sin x \Rightarrow \text{f}'(x)=2\cosx-2 \sin2x \text{ and f}''(x)=-2 \sin x-4\cos2x
so g′(x)=2cosx−2sinx−(4+cos2x+2sinx)22cosx−2sin2x
Unparseable latex formula:
\text{g}'(x)=0 \text{ when }}2\cos x-2\sin 2x=0 \text{ or }(4+\cos 2x+2\sin x)^2=1
4+cos2x+2sin2x>1 for all x so solutions are given by 2cosx−2sin2x=0 ⇒cosx−2sinxcosx=0⇒cosx=0 or sinx=21 so x=6π,2π,65π or 23π g(x) is a continuous function since 4+cos2x+2sinx cannot be zero so we need only consider one of the turning points x=2π⇒ f(x)=5, f′(x)=0 and f′′(x)=2⇒ g′′(x)>0 so a minimum at x=2π
\text{so stationary points are maxima at } \left( \dfrac{\pi}{6}, \dfrac{125}{22} \right) \text{ and }\left( \dfrac{5\pi}{6}, \dfrac{125}{22} \right) \text{ and minima at }\left( \dfrac{\pi}{2}, \dfrac{26}{5} \right) \text{ and } \left( \dfrac{3\piu}{2}, 2\right)
f(x)=ax2+bx+c⇒ f′(x)=2ax+b f(1)(x+21)+ f(−1)(x−21)−2 f(0)x=(a+b+c)(x+21)+(a−b+c)(x−21)−2cx =2ax+2cx+b−2cx=2ax+b= f′(x) Since f′(x) is linear it must attain it’s maximum value at either x=1 or x=−1 f′(−1)=−21f(1)−23f(−1)−2f(0) and f′(1)=23f(1)+21f(−1)−2f(0)
Unparseable latex formula:
\text{f}(-1),\tewxt{ f}(0) \text{ and f}(1) \text{ are all between }\pm{1} \Rightarrow |\text{f}'(x)| \leq4
Consider 2x2−1 then −1≤ f(x0)≤1 i.e. f(x)∣≤1 for all x with ∣x∣≤1 and f′(x)=4x lies between ±4 so this function satisfies the requirements
i) ∫1ex2logxdx Recall that ∫logxdx=x(log(x)−1)+c, now we can use this to integrate by parts. I=∫x2logxdx=x21x(log(x)−1)−∫−2x3x(log(x)−1)dx −I=x1(log(x)−1)+2x1=xlogx+1 Putting in the limits gives −I=e2−1= So I=1−e2
ii) ∫sin(x)1+sin(x)cos(x)dx Let u2=1+sin(x)⇒2udxdu=cos(x)⇔dx=cos(x)2udu So the integral becomes
Going back to x we have log∣u+1u−1∣+c=log∣1+sinx+11+sinx−1∣+c
Alternative solution for part (b)
∫sinx1+sinxcosxdx so putting sinx=tan2θ so cosxdθdx=2tanxsec2x integral becomes ∫tan2θsecθ2tanθsec2θdθ=∫tanθ2secθdθ=∫2cscθdθ=−ln∣cscθ+cotθ∣+c tanθ=x⇒cotθ=x1,sinθ=1+xx so cscθ=x1+x Hence ∫sinx1+sinxcosxdx=−lnAx1+1+x=lnA1+1+xx