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STEP Maths I, II, III 1989 solutions

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Reply 1
I am no expert at STEP, so don't hesitate to correct me, it is more than likely made that I have made a mistake. :redface:

STEP I - Question 8

Using good old de Moivre's theorem;

cos(4θ)=(cos(θ)+isin(θ))4=(cos4(θ)+4icos3(θ)sin(θ)6cos2(θ)sin2(θ)4isin3(θ)cos(θ)+sin4(θ))=cos4(θ)6cos2(θ)sin2(θ)=8cos4(θ)8cos2(θ)+1\cos(4\theta) = \Re( \cos(\theta) + i\sin(\theta))^{4} = \Re(\cos^{4}(\theta) + 4i\cos^{3}(\theta)\sin(\theta) - 6\cos^{2}(\theta)\sin^{2}(\theta) - 4i\sin^{3}(\theta)\cos(\theta) + \sin^{4}(\theta) ) = cos^{4}(\theta) - 6\cos^{2}(\theta)\sin^{2}(\theta) = 8\cos^{4}(\theta) - 8\cos^{2}(\theta) + 1


(using cos2(θ)+sin2(θ)1 \cos^{2}(\theta) + \sin^{2}(\theta) \equiv 1

and similary;

[br]cos(6θ)=(cos(θ)+isin(θ))6=(cos6(θ)+6icos5(θ)sin(θ)15cos4(θ)sin2(θ)20isin3(θ)cos3(θ)+15sin4(θ)cos2(θ)+6icos(θ)sin5(θ)sin6(θ))=32cos6(θ)48cos4(θ)+18cos2(θ)1[br]\cos(6\theta) = \Re( \cos(\theta) + i\sin(\theta))^{6} = \Re(\cos^{6}(\theta) + 6i\cos^{5}(\theta)\sin(\theta) - 15\cos^{4}(\theta)\sin^{2}(\theta) - 20i\sin^{3}(\theta)\cos^{3}(\theta) + 15\sin^{4}(\theta)\cos^{2}(\theta) + 6i\cos(\theta)\sin^{5}(\theta) - \sin^{6}(\theta) ) = 32\cos^{6}(\theta) - 48\cos^{4}(\theta) +18\cos^{2}(\theta) - 1

Now, consider: 12cos(6θ)12cos(4θ) \frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta)

12cos(6θ)12cos(4θ)=16c628c4+13c21 \frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta) = 16c^{6} - 28c^{4} + 13c^{2} -1

Where c is cos(θ) \cos(\theta) .

Now let x=cos(θ)x = cos(\theta),

so:

16x628x4+13x21=16c628c4+13c21=12cos(6θ)12cos(4θ)=0[br] 16x^{6} - 28x^{4} + 13x^{2} -1 = 16c^{6} - 28c^{4} + 13c^{2} -1 = \frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta) = 0[br]

cos(6θ)=cos(4θ)\Rightarrow \cos(6\theta) = \cos(4\theta)

6θ=2nπ±4θ \Rightarrow 6\theta = 2n\pi \pm 4\theta

Which gives;

θ=nπ5 \theta = \frac{n\pi}{5} (n=1,2,3,4 n= 1,2,3,4 ).
These values of θ\theta give distinct values of cos(θ)\cos(\theta)

or

θ=nπ \theta = n\pi (n=0,1 n= 0,1 )


Hence the roots of the equation are:

x=cos(0),cos(π),cos(π5),cos(2π5),cos(3π5) x = \cos(0), \cos(\pi), \cos(\frac{\pi}{5}), \cos(\frac{2\pi}{5}), \cos(\frac{3\pi}{5}) and cos(4π5) \cos(\frac{4\pi}{5}) .
Reply 2
STEP III - Question 10

Lets assume that the result is true for n = k;

r=1kr(r+1)(r+2)(r+3)(r+4)=16k(k+1)(k+2)(k+3)(k+4)(k+5)\displaystyle\sum_{r=1}^k r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}k(k+1)(k+2)(k+3)(k+4)(k+5) (*)

[br]r=1k+1r(r+1)(r+2)(r+3)(r+4)=16k(k+1)(k+2)(k+3)(k+4)(k+5)+(k+1)(k+2)(k+3)(k+4)(k+5)[br]\Rightarrow \displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}k(k+1)(k+2)(k+3)(k+4)(k+5) + (k+1)(k+2)(k+3)(k+4)(k+5)

r=1k+1r(r+1)(r+2)(r+3)(r+4)=16(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)\Rightarrow \displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)

This is the same as (*) except k+1 replaces k. Hence if the result is true for n=k, its true for n= k+1.

When n=1,

LHS of (*) = 1x2x3x4x5 = 120
RHS of (*) = (1/6)x1x2x3x4x5x6 = 120.

So (*) is true for n=1.
Hence, by induction;

r=1nr(r+1)(r+2)(r+3)(r+4)=16n(n+1)(n+2)(n+3)(n+4)(n+5)\displaystyle\sum_{r=1}^n r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5) n1 \forall n \geq 1 .


Since;

r5<r(r+1)(r+2)(r+3)(r+4) r^{5} < r(r+1)(r+2)(r+3)(r+4)

r=1nr5<r=1nr(r+1)(r+2)(r+3)(r+4) \Rightarrow \displaystyle\sum_{r=1}^n r^{5} < \displaystyle\sum_{r=1}^n r(r+1)(r+2)(r+3)(r+4)

r=1nr5<16n(n+1)(n+2)(n+3)(n+4)(n+5) \Rightarrow \displaystyle\sum_{r=1}^n r^{5} < \frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5).

Using (*);

r=0n1r(r1)(r2)(r3)(r4)=16n(n1)(n2)(n3)(n4)(n5)\displaystyle\sum_{r=0}^{n-1} r(r-1)(r-2)(r-3)(r-4) = \frac{1}{6}n(n-1)(n-2)(n-3)(n-4)(n-5)


r(r1)(r2)(r3)(r4)<r5 r(r-1)(r-2)(r-3)(r-4) < r^{5}

r=0n1r5>r=0n1r(r1)(r2)(r3)(r4) \Rightarrow \displaystyle\sum_{r=0}^{n-1} r^{5} > \displaystyle\sum_{r=0}^{n-1} r(r-1)(r-2)(r-3)(r-4)

r=0n1r5>16(n5)(n4)(n3)(n2)(n1)n \Rightarrow \displaystyle\sum_{r=0}^{n-1} r^{5} > \frac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n


In this case, f(x)=x5f(x) = x^{5}

From the previous parts, we know that:

r=0n1a6n6r5>a66n6(n5)(n4)(n3)(n2)(n1)n\displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} > \frac{a^{6}}{6n^{6}}(n-5)(n-4)(n-3)(n-2)(n-1)n

and clearly, also:

r=0n1a6n6r5<a66n6n(n+1)(n+2)(n+3)(n+4)(n+5) \displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} < \frac{a^{6}}{6n^{6}}n(n+1)(n+2)(n+3)(n+4)(n+5)

as nn\longrightarrow \infty ,

a66n6(n5)(n4)(n3)(n2)(n1)n \frac{a^{6}}{6n^{6}}(n-5)(n-4)(n-3)(n-2)(n-1)n and a66n6n(n+1)(n+2)(n+3)(n+4)(n+5)a66\frac{a^{6}}{6n^{6}}n(n+1)(n+2)(n+3)(n+4)(n+5) \longrightarrow \frac{a^{6}}{6}

So,

limnr=0n1a6n6r5=a66\lim_{n\to \infty} \displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} = \frac{a^{6}}{6}

Using a similar arguement, we can show that:

limnr=1na6n6r5=a66\lim_{n\to \infty} \displaystyle\sum_{r=1}^{n} \frac{a^{6}}{n^{6}}r^{5} = \frac{a^{6}}{6}

We have shown that the limits exist and are equal to a66\frac{a^{6}}{6}.

Hence;


0ax5dx=a66\displaystyle\int^a_0 x^{5} \, \mathrm{d}x = \frac{a^{6}}{6}
Reply 3
STEP I, Q2

The For x>0x > 0, find xlnxdx\displaystyle\int x\ln x dx

Using integration by parts (uvdx=uvuvdx\displaystyle\int u'v dx = uv - \int uv' dx ):

xlnxdx=12x2lnx12x21xdx\displaystyle\int x\ln x dx = \frac{1}{2}x^2 \cdot \ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} dx

=12(x2lnxxdx)\displaystyle = \frac{1}{2}(x^2 \cdot \ln x - \int x dx)

=12(x2lnx12x2)+C\displaystyle = \frac{1}{2}(x^2 \cdot \ln x - \frac{1}{2} x^2) + C

=14x2(2lnx1)+C\displaystyle = \frac{1}{4}x^2(2 \ln x - 1) + C



By approximating the area corresponding to 01xln(1/x)dx\int^1_0 x\ln(1/x) dx by n rectangles of equal width and with their top right hand vertices on the curve y=xln(1/x)y = x\ln(1/x), show that, as nn \to \infty,

12(1+1n)lnn1n2[ln(n!0!)+ln(n!1!)+ln(n!2!)+...ln(n!(n1)!)]14\displaystyle\frac{1}{2}(1 + \frac{1}{n})\ln n - \frac{1}{n^2}[ln(\frac{n!}{0!}) + ln(\frac{n!}{1!}) + ln(\frac{n!}{2!}) + ... ln(\frac{n!}{(n-1)!})] \to \frac{1}{4}



()(*) Firstly, note that

01xln(1/x)dx\displaystyle \int^1_0 x\ln(1/x) dx

=01x(ln(x))dx\displaystyle= \int^1_0 x(-\ln(x)) dx

=01xln(x)dx\displaystyle= -\int^1_0 x\ln(x) dx

=[14x2(2lnx1)]01\displaystyle= -[\frac{1}{4}x^2(2 \ln x - 1)]^1_0 (by the above indefinite integral)

=limt0[14x2(2lnx1)]t1\displaystyle= \lim_{t \to 0} -[\frac{1}{4}x^2(2 \ln x - 1)]^1_t

=limt014(12(ln11)t2(2lnt1))\displaystyle= \lim_{t \to 0} -\frac{1}{4}(1^2(\ln 1 - 1) - t^2(2 \ln t - 1))

=limt014(1+2t2lntt2))\displaystyle= \lim_{t \to 0} \frac{1}{4}(1 + 2t^2\ln t - t^2))

=limt014(1+2t(tlnt)t2))\displaystyle= \lim_{t \to 0} \frac{1}{4}(1 + 2t(t\ln t) - t^2))

=14\displaystyle= \frac{1}{4}



The sum of the areas of our rectangles is going to be

()r=1n1n(rnln(1r/n))\displaystyle (**) \sum^n_{r=1} \frac{1}{n} (\frac{r}{n}\ln(\frac{1}{r/n}))

=r=1n1n(rnln(nr))\displaystyle = \sum^n_{r=1} \frac{1}{n} (\frac{r}{n}\ln(\frac{n}{r}))

=1n2r=1nrln(nr)\displaystyle = \frac{1}{n^2} \sum^n_{r=1} r\ln(\frac{n}{r})

=1n2(1ln(n1)+2ln(n2)+3ln(n3)+...+nln(nn))\displaystyle = \frac{1}{n^2} (1\ln(\frac{n}{1}) + 2\ln(\frac{n}{2}) + 3\ln(\frac{n}{3}) + ... + n\ln(\frac{n}{n}))

=1n2ln((n1)(n2)2(n3)3(nn)n)\displaystyle = \frac{1}{n^2} \ln((\frac{n}{1})(\frac{n}{2})^2(\frac{n}{3})^3 \cdot \cdot \cdot (\frac{n}{n})^n)

=1n2ln(nn2n3nn12233nn)\displaystyle = \frac{1}{n^2} \ln(\frac{n\cdot n^2\cdot n^3\cdot\cdot\cdot n^n}{1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n})

=1n2ln(n1+2+3+4+...+n12233nn)\displaystyle = \frac{1}{n^2} \ln(\frac{n^{1+2+3+4+...+n}}{1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n})

=1n2(ln(n1+2+3+4+...+n)ln(12233nn))\displaystyle = \frac{1}{n^2} (ln(n^{1+2+3+4+...+n})-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

=1n2(ln(n12n(n+1))ln(12233nn))\displaystyle = \frac{1}{n^2} (ln(n^{\frac{1}{2}n(n+1)})-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

=1n2(12n(n+1)ln(n)ln(12233nn))\displaystyle = \frac{1}{n^2} ({\frac{1}{2}n(n+1)}ln(n)-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

=1n212n(n+1)ln(n)1n2ln(12233nn)\displaystyle = \frac{1}{n^2}{\frac{1}{2}n(n+1)}ln(n) - \frac{1}{n^2} \ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n)

=12(1+1n)ln(n)1n2ln(12233nn)\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} \ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n)



Claim: 12233kk=k!0!k!1!k!2!k!k!1\cdot 2^2\cdot 3^3\cdot\cdot\cdot k^k = \frac{k!}{0!}\cdot\frac{k!}{1!}\cdot\frac{k!}{2!} \cdot\cdot\cdot\frac{k!}{k!}

So for k + 1, we must prove that k!0!k!1!k!2!(k+1)!(k+1)!\frac{k!}{0!}\cdot\frac{k!}{1!}\cdot\frac{k!}{2!} \cdot \cdot \cdot \frac{(k+1)!}{(k+1)!} is the equation we arrive at.

Basis case (let k=1): 11=1!0!1!1!=11^1 = \frac{1!}{0!}\cdot\frac{1!}{1!} = 1

Inductive step: 12233kk(k+1)k+1=(k+1)k+1(k!0!k!1!k!2!k!k!)1\cdot 2^2 \cdot 3^3 \cdot \cdot \cdot k^k\cdot (k+1)^{k+1} = (k+1)^{k+1}(\frac{k!}{0!} \cdot \frac {k!}{1!} \cdot \frac{k!}{2!} \cdot \cdot \cdot \frac{k!}{k!})

=((k+1)k!0!(k+1)k!1!(k+1)k!2!(k+1)k!k!) = ((k+1)\frac{k!}{0!}(k+1) \cdot \frac{k!}{1!}(k+1) \cdot \frac{k!}{2!} \cdot\cdot\cdot (k+1)\frac{k!}{k!})

=((k+1)!0!(k+1)!1!(k+1)!2!(k+1)!k!) = (\frac{(k+1)!}{0!}\cdot\frac{(k+1)!}{1!}\cdot\frac{(k+1)!}{2!}\cdot\cdot\cdot\frac{(k+1)!}{k!})

=((k+1)!0!(k+1)!1!(k+1)!2!(k+1)!k!(k+1)!(k+1)!) = (\frac{(k+1)!}{0!}\cdot\frac{(k+1)!}{1!}\cdot\frac{(k+1)!}{2!}\cdot\cdot\cdot\frac{(k+1)!}{k!}\cdot \frac{(k+1)!}{(k+1)!})



So,

()=12(1+1n)ln(n)1n2ln(n!0!n!1!n!2!n!n!)\displaystyle (**) = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} \ln(\frac{n!}{0!}\cdot\frac{n!}{1!}\cdot\frac{n!}{2!}\cdot\cdot\cdot\frac{n!}{n!})

=12(1+1n)ln(n)1n2(ln(n!0!)+ln(n!1!)+ln(n!2!)+...+ln(n!(n1)!)+ln(n!n!))\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!}) + \ln(\frac{n!}{n!}))

=12(1+1n)ln(n)1n2(ln(n!0!)+ln(n!1!)+ln(n!2!)+...+ln(n!(n1)!))\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!}))


As nn \to \infty, our sum tends to the integral ()(*). So,

12(1+1n)ln(n)1n2(ln(n!0!)+ln(n!1!)+ln(n!2!)+...+ln(n!(n1)!))14\displaystyle \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!})) \to \frac{1}{4}


is this ok? have i missed anything out or not explained each step in enough detail?
Reply 4
Question 4, STEP I

With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points.

Diagram showing five points is attached.
Reply 5
Question 5, STEP I

(1+x)n=(n0)+(n1)x+(n2)x2++(nn)xn\displaystyle (1+x)^{n} = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + \cdots + \binom{n}{n}x^{n}

a)
Let x = 1.

2n=(n0)+(n1)+(n2)++(nn)\displaystyle 2^{n} = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}

Let x = -1

0=(n0)(n1)+(n2)+(nn)\displaystyle 0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + \binom{n}{n}

As n is even. So

(n0)+(n2)++(nn)=(n1)+(n3)++(nn1)=2n1\displaystyle \binom{n}{0} + \binom{n}{2} + \cdots + \binom{n}{n} = \binom{n}{1} + \binom{n}{3} + \cdots + \binom{n}{n-1} = 2^{n-1}

As their sum is 2n\displaystyle 2^{n}

b) Suppose that

(k1)+2(k2)++k(kk)=k2k1\displaystyle \binom{k}{1} + 2\binom{k}{2} + \cdots + k\binom{k}{k} = k2^{k-1}

Then (k0)+2(k1)+3(k2)++(k+1)(kk)=k2k1+2k\displaystyle \binom{k}{0} + 2\binom{k}{1} + 3\binom{k}{2} + \cdots + (k+1)\binom{k}{k} = k2^{k-1} + 2^{k}

Note that (kr)+(kr1)=(k+1r)\displaystyle \binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r} and that (kk)=(k+1k+1)\displaystyle \binom{k}{k} = \binom{k+1}{k+1}

So, upon adding these two expressions, we get

(k+11)+2(k+12)++(k+1)(k+1k+1)=k2k1+k2k1+2k=(k+1)2k\displaystyle \binom{k+1}{1} + 2\binom{k+1}{2} + \cdots + (k+1)\binom{k+1}{k+1} = k2^{k-1} + k2^{k-1} + 2^{k} = (k+1)2^{k}

Also, (11)=1=1×20\displaystyle \binom{1}{1} = 1 = 1 \times 2^{0}

So true by induction.

r=0n(r+(1)r)(nr)=((n1)+2(n2)++n(nn))+((n0)(n1)++(1)n(nn))\displaystyle \displaystyle \sum_{r=0}^{n} \left(r + (-1)^{r}\right) \binom{n}{r} = \left(\binom{n}{1} + 2\binom{n}{2} + \cdots + n \binom{n}{n}\right) + \left(\binom{n}{0} - \binom{n}{1} + \cdots + (-1)^{n}\binom{n}{n}\right)

=n2n1\displaystyle = n2^{n-1}
Reply 6
STEP II Q1

[br]cos(3θ)=cos(2θ+θ)=cos(2θ)cos(θ)sin(2θ)sin(θ)=(2cos2(θ)1)cos(θ)2sin(θ)cos(θ)sin(θ)=2cos3(θ)cos(θ)2(1cos2(θ))cos(θ)=4cos3(θ)3cos(θ).[br][br]\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) = (2cos^{2}(\theta) - 1)\cos(\theta) - 2\sin(\theta)\cos(\theta)sin(\theta) = 2cos^{3}(\theta) - \cos(\theta) - 2(1-\cos^{2}(\theta))\cos(\theta) = 4cos^{3}(\theta) - 3\cos(\theta).[br]

Substituting x= y-a into the equation:

24x372x2+66x19=24(ya)372(ya)2+66(ya)19=0 24x^{3} - 72x^{2} + 66x -19 = 24(y-a)^{3} - 72(y-a)^{2} + 66(y-a) -19 = 0

Notice that the reduced form shown in the question has no z2z^{2} term, so we need to find the value of a which will get rid of the y2y^{2} term.

Expanding out and collectiing terms we get:

24y372(a+1)y2+(72a2+144a+66)y24a372a266a19=0 24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 0

So the value of a to get rid of the y2y^{2} is -1.

So we now have:

24y372(a+1)y2+(72a2+144a+66)y24a372a266a19=24t36y1=0 24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 24t^{3} - 6y -1 = 0

24y36y=1 \Rightarrow 24y^{3} - 6y = 1 (*)

Substituting y=z/b into (*);

24(zb)36(zb)=124(\frac{z}{b})^{3} - 6(\frac{z}{b}) = 1

dividing both sides by 2 and multiplying by bb;

12z3b23z=b212\frac{z^{3}}{b^{2}} - 3z = \frac{b}{2} .

And so to make this equal to;

4z33z=b24z^{3}- 3z = \frac{b}{2}

we make b = ±3\pm \sqrt3.

Lets take b = 3\sqrt3 and let z = cos(θ)\cos(\theta)

so;

4z33z=4cos3(θ)3cos(θ)=cos(3θ)=32 4z^{3}- 3z = 4\cos^{3}(\theta) - 3\cos(\theta) = \cos(3\theta) = \frac{\sqrt3}{2}

3θ=pi6,11π6\Rightarrow 3\theta = \frac{pi}{6}, \frac{11\pi}{6} and 13π6 \frac{13\pi}{6} (these give distinct values of cosθcos\theta, and hence, three distinct solutions.)


we know that; x = y - a = (z/b) - a.

Hence the solutions of the equation are:

x=cos(π18)3+1,cos(11π18)3+1 x = \frac{\cos(\frac{\pi}{18})}{\sqrt3} + 1 , \frac{\cos(\frac{11\pi}{18})}{\sqrt3} +1 and cos(13π18)3+1 \frac{\cos(\frac{13\pi}{18})}{\sqrt3} + 1
Question 11, STEP I.

Vertical component = vsinθv \sin \theta
Using s = ut + 1/2at^2.
h=vsinθ1/2gt2-h = v \sin \theta - 1/2gt^2
1/2gt2vsinθh=01/2gt^2 - v \sin \theta - h = 0
Quadratic in t. Ignoring the negative root.
t=vsinθ+v2sin2θ+2hggt = \dfrac{v \sin \theta + \sqrt{v^2 \sin^2 \theta + 2hg}}{g}
R = horizontal component x time.
Horizontal component = vcosθv \cos \theta
R=vcosθ(vsinθ+v2sin2θ+2hggR = v \cos \theta (\dfrac{v \sin \theta + \sqrt{v^2 \sin^2 \theta + 2hg}}{g}
R=v2sinθcosθ+vcosθv2sin2θ+2hggR = \dfrac{v^2 \sin \theta \cos \theta + v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{g}
Multiply both numerator and denominator by 2.
R=2v2sinθcosθ+2vcosθv2sin2θ+2hg2gR = \dfrac{2v^2 \sin \theta \cos \theta + 2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}
Using sin(2θ)=2sinθcosθ\sin (2\theta) = 2 \sin \theta \cos \theta
We can simplify it.
R=v2sin2θ+2vcosθv2sin2θ+2hg2gR = \dfrac{v^2 \sin 2\theta + 2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}
Splitting up Mr. Big Fraction.
R=v2sin2θ2g+2vcosθv2sin2θ+2hg2gR = \dfrac{v^2 \sin 2\theta}{2g} + \dfrac{2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}

Ignore the first fraction for the moment, we'll work on the second one.
2vcosθv2sin2θ(1+2hgv2sin2θ)2g\dfrac{2v \cos \theta \sqrt{v^2 \sin^2 \theta (1 + \dfrac{2hg}{v^2 \sin^2 \theta})}}{2g}

Unparseable latex formula:

\dfrac{2v^2 \cos \theta \sin \theta}{2g} \times \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}


We can use our old friend Mr. Double Angle Formula to simplify it:
v2sin2θ2g1+2hgv2sin2θ\dfrac{v^2 \sin 2\theta}{2g} \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}}
The whole fraction looks like:
Unparseable latex formula:

R = (\dfrac{v^2\sin 2\theta}{2g}) + (\dfrac{v^2 \sin 2\theta}{2g}) \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}


And lo and behold, there's a common factor to both fractions. Let's take it out:
R=v22gsin2θ(1+1+2hgv2sin2θ)R = \dfrac{v^2}{2g}\sin 2\theta(1 + \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}})
And that's what we're looking for.
*I'll finish off the last small bit in a few minutes*
Reply 8
STEP I Question 6

y=f(x)y = f(x)
dy/dx=f(x)dy/dx = f'(x)

The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be

yf(x)=1f(x)(xx)y - f(x) = \frac{-1}{f'(x)}(x' - x) (where x' is the general x coordinate of the normal).

At the point Q, x' = 0 as it cuts the y axis.

yf(x)=xf(x)y - f(x) = \frac{x}{f'(x)}

y=xf(x)+f(x)y = \frac{x}{f'(x)} + f(x)

The distance PQ can be worked out using Pythagoras' theorem.

PQ2=(f(x)(xf(x)+f(x)))2+x2PQ^2 = (f(x) - (\frac{x}{f'(x)} + f(x)))^2 + x^2

=x2f(x)2+x2= \frac{x^2}{f'(x)^2} + x^2

It's given that PQ2=ex2+x2PQ^2 = e^{x^2} + x^2

So

x2f(x)2+x2=ex2+x2\frac{x^2}{f'(x)^2} + x^2 = e^{x^2} + x^2

x2f(x)2=ex2\frac{x^2}{f'(x)^2} = e^{x^2}

x2ex2=f(x)2\frac{x^2}{e^{x^2}} = f'(x)^2

xe0.5x2=f(x)\frac{x}{e^{0.5x^2}} = f'(x)

1dy=xe0.5x2dx\int 1 \mathrm{d}y = \int \frac{x}{e^{0.5x^2}} \mathrm{d}x

y=xe0.5x2dxy = \int xe^{-0.5x^2} \mathrm{d}x

y=e0.5x2+cy = -e^{-0.5x^2} + c (use a substitution of u = x^2 if you can't see why)

2=1+c-2 = -1 + c

c=1c = -1

y=e0.5x21y = -e^{-0.5x^2} - 1