STEP 2005 Solutions Thread Watch

Zacken
Badges: 22
Rep:
?
#1
Report Thread starter 10 years ago
#1
You can let someone else take control of the OP whenever.

STEP I:
1: Solution by nuodai
2: Solution by Unbounded
3: Solution by Unbounded
4: Solution by Unbounded
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by nuodai
10: Solution by Sk1lz
11: Solution by nuodai
12: Solution by darkness9999
13: Solution by brianeverit
14: Solution Farhan.Hanif93

STEP II:
1: Solution by SimonM
2: Solution by Daniel Freedman
3: Solution by SimonM
4: Solution by Glutamic Acid
5: Solution by SimonM
6: Solution by Daniel Freedman
7: Solution by SimonM
8: Solution by sonofdot
9: Solution by Farhan.Hanif93
10: Solution by Glutamic Acid
11: Solution by Farhan.Hanif93
12: Solution by nuodai
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by DeanK22
2: Solution by Adje
3: Solution by SimonM
4: Solution by SimonM
5: Solution by Daniel Freedman
6: Solution by sonofdot
7: Solution by sonofdot
8: Solution by SimonM
9: Solution by sonofdot
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by sweeneyrod
13: Solution by brianeverit
14: Solution by sonofdot

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
3
reply
Unbounded
Badges: 12
Rep:
?
#2
Report 10 years ago
#2
STEP I, Question 3, 2005
(i)
 \dfrac{x}{x-a} + \dfrac{x}{x-b} = 1

 \iff x(x-b) + x(x-a) = (x-a)(x-b)

 \iff x^2 -bx +x^2 -ax = x^2 -bx -ax +ab

 \iff x^2 -ab = 0

 x = \pm \sqrt{ab}

So there are two distinct real roots if, and only if, a and b are both positive or both negative. Q.E.D.
(ii)

 \dfrac{x}{x-a} + \dfrac{x}{x-b} = 1+c

 \iff x(x-b)+x(x-a) = (1+c)(x^2-(a+b)x+ab)

 \iff x^2 -ab = cx^2 -c(a+b)x +abc

 \iff x^2(1-c) -x(ca+cb) -ab(1+c) = 0

There is exactly one real solution when the discriminant of this quadratic in x is equal to zero.

 \therefore c^2(a+b)^2 +4ab(1-c)(1+c) = 0

 \iff c^2 (a+b)^2 - 4abc^2 + 4ab = 0

 \iff c^2(4ab-(a+b)^2) = 4ab

 \iff c^2 = \dfrac{4ab}{4ab -(a+b)^2}

 = 1 + \dfrac{(a+b)^2}{4ab-a^2-2ab-b^2} = 1 - \left(\dfrac{a+b}{a-b} \right)^2 \ \ \ \square

Noting that  c^2 \geq 0

But also that  \left(\dfrac{a+b}{a-b} \right)^2 \geq 0 \iff 1- \left(\dfrac{a+b}{a-b} \right)^2 \leq 1

 \iff c^2 \leq 1

Let us simply prove that c is non-zero:


Suppose that c = 0, then the equation for (ii) becomes exactly the same as (i). Hence we require a and b to both be positive or negative. Then the condition of exactly one root, which is  c^2 = \frac{-4ab}{(a-b)^2} no longer holds, since the LHS is zero whereas the RHS is negative - contradiction. Hence c is non-zero.

And finally, we have proven that  0 < c^2 \leq 1 . QED
2
reply
Unbounded
Badges: 12
Rep:
?
#3
Report 10 years ago
#3
Question 4, STEP I, 2005
(a)
 \cos \theta = \frac{3}{5}, \frac{3\pi}{2} < \theta < 2\pi \implies \sin \theta = -\frac{4}{5}

 \therefore \sin 2\theta = 2\sin \theta \cos \theta = 2 \times -\frac{4}{5} \times \frac{3}{5} = -\frac{24}{25} \ \ \ \square

 \cos 3\theta = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta

 = \cos \theta(2\cos^2 \theta -1) - \frac{-24}{25} \times \frac{-4}{5}

 = \frac{3}{5}(\frac{18}{25} -1) - \frac{96}{125}

 = \boxed{-\frac{117}{125}}
(b)
 \tan 3\theta = \dfrac{\tan 2\theta + \tan \theta}{1-\tan 2\theta\tan \theta}

 = \dfrac{\frac{2\tan \theta}{1-\tan^2 \theta} + \tan \theta}{1-\frac{2tan^2 \theta}{1-\tan^2 \theta}}

  = \dfrac{2\tan \theta +\tan \theta - \tan^3 \theta}{1 - \tan^2 \theta -2\tan^2 \theta}

 = \dfrac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} \ \ \ \square

 \dfrac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} = \dfrac{11}{2}

 \iff 6\tan \theta -2\tan^3 \theta = 11 - 33\tan^2 \theta

 \iff 2\tan^3 \theta -33\tan^2 \theta -6\tan \theta + 11 = 0

 \iff (2\tan \theta -1)(\tan^2 \theta -16\tan \theta -11) = 0

 \implies \tan \theta = \frac{1}{2}

or  \tan \theta = \dfrac{16 \pm \sqrt{256 +44}}{2}

 = 8 \pm \sqrt{75}

Given the restriction on theta, we need  \tan \theta > 1

 \therefore \boxed{\tan \theta = 8 + \sqrt{75}}
4
reply
SimonM
Badges: 18
Rep:
?
#4
Report 10 years ago
#4
I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions
0
reply
sonofdot
Badges: 8
Rep:
?
#5
Report 10 years ago
#5
STEP III 2005 Question 7

This is one of my favourites too

We have \displaystyle\int \frac{1}{u f(u)} \, du = F(u) +c

Making a substitution u=x^m \Rightarrow du = mx^{m-1} dx

\displaystyle\therefore \int \frac{mx^{m-1}}{x^m f(x^m)} \, dx = F(x^m) + c \iff \int \frac{m}{x f(x^m)} \, dx = F(x^m) + c

(i)
\displaystyle\int \frac{1}{x^n - x} \, dx = \int\frac{1}{x(x^{n-1}-1)} \, dx

Consider:
\begin{array}{rl}

\displaystyle\int \frac{1}{u(u-1)} \, du

& \displaystyle = \int \frac{1}{u-1} - \frac{1}{u} \, du \\ \br \\

& \displaystyle = \ln |u-1| - \ln |u| + c \\ \br \\

& \displaystyle = \ln \left| \frac{u-1}{u} \right| + c \end{array}

By above result:

\displaystyle\int \frac{1}{x^n - x} \, dx = \frac{1}{n-1} \ln \left| \frac{x^{n-1} - 1}{x^{n-1}} \right| + c = \boxed{\ln \left| \frac{\sqrt[n-1]{x^{n-1}-1}}{x} \right| + c}

(The integral is undefined when n=1)
(ii)
\displaystyle\int \frac{1}{\sqrt{x^n + x^2}} \, dx = \int \frac{1}{x \sqrt{x^{n-2} + 1}} \, dx

Consider \displaystyle\int \frac{1}{u \sqrt{1+u}} \, du

Making the substitution z^2 = u+1 \Rightarrow du = 2z dz gives:

\begin{array}{rl}

\displaystyle\int \frac{2z}{z(z^2 - 1)} \, dz

& \displaystyle = 2 \int \frac{1}{z^2 - 1} \, dz \\ \br \\

& \displaystyle =  \int \frac{1}{z - 1} - \frac{1}{z+1} \, dz \\ \br \\

& \displaystyle =  \ln |z-1| - \ln |z+1| +c \\ \br \\

& \displaystyle =  \ln \left| \frac{z-1}{z+1} \right|+c \\ \br \\

& \displaystyle =  \ln \left| \frac{\sqrt{u+1}-1}{\sqrt{u+1}+1} \right| +c \end{array}

Hence, by the above result:

\displaystyle\int \frac{1}{\sqrt{x^n + x^2}} \, dx = \boxed{\frac{1}{n-2} \ln \left| \frac{\sqrt{x^{n-2} + 1} - 1}{\sqrt{x^{n-2} + 1} +1} \right| + c}

Special case: when n=2, the integral becomes:

\displaystyle\int \frac{1}{x\sqrt2} \, dx = \frac{\sqrt2}{2} \ln |x| +c
1
reply
Daniel Freedman
Badges: 6
Rep:
?
#6
Report 10 years ago
#6
STEP II 2005, Question 6

Spoiler:
Show


 \\ (1-x)^{-1} = 1 + x + x^2 + ... + x^r + ... \\

\\ (1-x)^{-2} = 1 + 2x + 3x^2 + ... + rx^{r-1} + ... \\

\\ (1-x)^{-3} = 1 + 3x + 6x^2 + ... + \frac{1}{2}r(r-1)x^{r-2} + ...

i)

 \\ \displaystyle (1-x)^{-2} = \sum_{r=1}^{\infty} rx^{r-1} \\ 

\\ \implies x(1-x)^{-2} = \sum_{r=1}^{\infty} rx^r

Let  x = \frac{1}{2}

 \\ \displaystyle \implies \sum_{r=1}^{\infty} r2^{-r} = \frac{1}{2}\left(1-\frac{1}{2}\right)^{-2} = 2

 \\ \displaystyle (1-x)^{-3} = \sum_{r=1}^{\infty} \frac{1}{2}r(r-1)x^{r-2} \\

\\ \implies 2x^2(1-x)^{-3} = \sum_{r=1}^{\infty} r(r-1)x^r \\

\\ \implies 2x^2(1-x)^{-3} + \sum_{r=1}^{\infty} rx^r = \sum_{r=1}^{\infty} r^2 x^r

Let  x = \frac{1}{2}

 \\ \displaystyle \implies \sum_{r=1}^{\infty} r^2 2^{-r} = 2 \left(\frac{1}{2}\right)^2 \left(1 - \frac{1}{2}\right)^{-3} + 2 \\

\\ \implies \sum_{r=1}^{\infty} r^2 2^{-r} = 6

ii)

 \\ \displaystyle (1-x)^{-\frac{1}{2}} = 1 + \left( -\frac{1}{2} \right)(-x) + \frac{\left( -\frac{1}{2} \right)\left( -\frac{1}{2} - 1 \right) (-x)^2}{2!} + \frac{\left( -\frac{1}{2} \right)\left( -\frac{1}{2} - 1 \right)\left( -\frac{1}{2} - 2 \right) (-x)^3}{3!}  + ... \\

\\ = \sum_{n=0}^{\infty} \frac{\left( -\frac{1}{2} \right)\left( -\frac{1}{2} - 1 \right)\left( -\frac{1}{2} - 2 \right) ... \left(-\frac{1}{2} - (n-1) \right) (-x)^n}{n!} \\ \\

\\ = \sum_{n=0}^{\infty} \frac{(\frac{1}{2})^n \left(1)(1+2)(1+4)...(2n-1) x^n}{n!}  \\ \\

\\ = \sum_{n=0}^{\infty} \frac{1.3.5...(2n-1)x^n}{2^n n!} \\ \\

\\ = \sum_{n=0}^{\infty} \frac{ (2n)! x^n}{ (2n)(2n-2)(2n-4)...4.2.1 .2^n n!} \\ \\

\\ = \sum_{n=0}^{\infty} \frac{ (2n)! x^n}{ 2^n (n)(n-1)(n-2)...2.1. 2^n n!} \\ \\

\\ = \sum_{n=0}^{\infty} \frac{(2n)! x^n}{(n!)^2 2^{2n}}

as required.

Let  x = \frac{1}{3}

 \\ \displaystyle \sum_{n=0}^{\infty} \frac{ (2n)!}{(n!)^2 2^{2n} 3^n} = \left(1- \frac{1}{3}\right)^{-\frac{1}{2}} = \frac{\sqrt{6}}{2}

 \\ \displaystyle (1-x)^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \frac{(2n)! x^n}{(n!)^2 2^{2n}} \\ \\

\\ \implies \frac{1}{2}(1-x)^{-\frac{3}{2}} = \sum_{n=0}^{\infty} \frac{ n(2n)! x^{n-1}}{(n!)^2 2^{2n}} \\ \\

\\ \implies \frac{1}{2}x(1-x)^{-\frac{3}{2}} = \sum_{n=0}^{\infty} \frac{n(2n)! x^n}{(n!)^2 2^{2n}}

Let  x = \frac{1}{3}

 \\ \displaystyle \therefore \sum_{n=0}^{\infty} \frac{n(2n)!}{(n!)^2 2^{2n}3^n} = \frac{1}{6} \left(\frac{2}{3}\right)^{-\frac{3}{2}} = \frac{\sqrt{6}}{8}
1
reply
Daniel Freedman
Badges: 6
Rep:
?
#7
Report 10 years ago
#7
(Original post by sonofdot)
.
What happens when n = 2 in part ii)?
0
reply
Daniel Freedman
Badges: 6
Rep:
?
#8
Report 10 years ago
#8
STEP II 2005, Question 2

Spoiler:
Show


a) i)

 \\ f(12) = 12 \left(1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) = 4 \\

\\ f(180) = 180 \left(1 - \frac{1}{2} \right) \left(1 - \frac{1}{3} \right) \left( 1 - \frac{1}{5} \right) = 48

ii)

 \\ f(N) = N \left( 1 - \frac{1}{p_1} \right) \left( 1 - \frac{1}{p_2} \right) \left( 1 - \frac{1}{p_3} \right) ...  \left( 1 - \frac{1}{p_k} \right) \\

\\ = N \left( \frac{p_1 - 1}{p_1} \right)\left( \frac{p_2 - 1}{p_2} \right)\left( \frac{p_3 - 1}{p_3} \right) ... \left( \frac{p_k - 1}{p_k} \right)

But  N = p_1^{\alpha_1} p_2^{\alpha_2} p_3^{\alpha_3} ... p_k^{\alpha_k} where  \alpha_i \geq 1 . Therefore

 f(N) =  p_1^{\alpha_1-1} p_2^{\alpha_2-1} p_3^{\alpha_3-1} ... p_k^{\alpha_k-1}(p_1-1)(p_2-1)(p_3-1)...(p_k-1)

which is a product of integers and is therefore an integer.

b)

i) Statement is false

Take m = n = 2  \implies f(m)f(n) = 1 but  f(mn) = 2

ii) Statement is true

 \\ f(pq) = pq \left(\frac{p-1}{p}\right)\left(\frac{q-1}{q}\right)  = (p-1)(q-1) \\

\\ f(p)f(q) = p \left(\frac{p-1}{p}\right) q \left(\frac{q-1}{q}\right) = (p-1)(q-1)

iii) Statement is false

Let p = 3, q = 4

f(3) = 2, f(4) = 2, f(12) = 4

c)

 f(p^m) = p^m \left(\frac{p-1}{p}\right) = p^{m-1}(p-1)

 p^{m-1}(p-1) = 146410 = 2 \times 5 \times 14641 = 2 \times 5 \times 11^4 = 10 \times 11^4

Therefore p = 11, m = 5.
0
reply
sonofdot
Badges: 8
Rep:
?
#9
Report 10 years ago
#9
(Original post by Daniel Freedman)
What happens when n = 2 in part ii)?
Good point, edited
0
reply
darkness9999
Badges: 9
Rep:
?
#10
Report 10 years ago
#10
STEP I: Question 12
Part 1
Spoiler:
Show

Group 1:


 \lambda\times p + (1- \lambda)\times q



\lambda p + (1- \lambda) q \times\frac{2}{5}=\lambda p





\Rightarrow\lambda p + (1- \lambda) q \times\2=5\times\lambda p





\Rightarrow \lambda\times(3p+2q)=2q





\Rightarrow \boxed{\lambda =\frac{2q}{3p+2q}}

Part 2
Spoiler:
Show

Group 1:


 \lambda\times p + (1- \lambda)\times q
Group 2:


 \lambda\times(1-p) + (1- \lambda)\times(1-q)
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\implies \lambda p=\frac{2}{5}\times({\lambda p +(1-\lambda)q -\frac{1}{4}[\lambda p +(1-\lambda)q] + \frac{1}{4}[\lambda\times(1-p) + (1- \lambda)\times(1-q)])



\Rightarrow 10\lambda p=4\lambda p + 4(1-\lambda)q - \lambda p -(1-\lambda)q +\lambda(1-p) + (1-\lambda)(1-q)



\Rightarrow 10\lambda p= 4\lambda p + 4q - 4\lambda q - \lambda p - q + \lambda q + \lambda - \lambda p + 1 - q - \lambda - \lambda q



\Rightarrow 10\lambda p = 2\lambda p + 2q - 2\lambda q + 1



\Rightarrow \boxed{\lambda= \frac{4q + 2}{5+6p + 4q}}

1
reply
sonofdot
Badges: 8
Rep:
?
#11
Report 10 years ago
#11
STEP III 2005 Question 14

:eek: that was a long one...

Spoiler:
Show
V has the density function \displaystyle f(x) = \frac{C k^{a+1} x^a}{(x+k)^{2a+2}} for 0 \leq x < \infty

\displaystyle\int_0^{\infty} f(x) \, dx = 1 \br



\iff \displaystyle C k^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2}} \, dx =1 \br 



\iff \displaystyle C k^{a+1} \left( \frac{a! a!}{(2a+1)! k^{a+1}} \right) = 1 \br



\iff \displaystyle \boxed{C = \frac{(2a+1)!}{a! a!}}
Spoiler:
Show
Now we have \displaystyle\int_0^v \frac{x^a}{(x+k)^{2a+2}} \, dx

Making a substitution x = \frac{k^2}{u}} \Rightarrow dx = -\frac{k^2}{u^2} \, du gives:

\begin{array}{rl}

\displaystyle\int_0^v \frac{x^a}{(x+k)^{2a+2}} \, dx

& \displaystyle = \int_{\infty}^{\frac{k^2}{v}} \frac{(\frac{k^2}{u})^a}{(\frac{  k^2}{u} + k)^{2a+2}}} \times - \frac{k^2}{u^2} \, du \\ \br \\

& \displaystyle = \int^{\infty}_{\frac{k^2}{v}} \frac{k^{2a+2}}{u^{a+2} (\frac{k(u+k)}{u})^{2a+2}}} \, du \\ \br \\

& \displaystyle = \int^{\infty}_{\frac{k^2}{v}} \frac{k^{2a+2}u^{2a+2}}{u^{a+2} k^{2a+2} (u+k)^{2a+2}} \, du \\ \br \\

& \displaystyle = \int^{\infty}_{\frac{k^2}{v}} \frac{u^a}{(u+k)^{2a+2}} \, du \end{array}

The median value of V, m_v is such that \mathrm{P} (V<m_v) = \mathrm{P} (V>m_v). Clearly from above result:

\displaystyle m_v = \frac{k^2}{m_v} \iff m_v^2 = k^2 \iff \boxed{m_v = k}
(since all values of v are >0)
Spoiler:
Show
\begin{array}{rl}

\mathrm{E} (V) & \displaystyle = \int_0^{\infty} xf(x) \, dx \\ \br \\

& \displaystyle = C k^{a+1} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}}  \, dx \\ \br \\

& \displaystyle = C k^{a+1} \left( \frac{(a+1)!(a-1)!}{(2a+1)! k^a} \right) \\ \br \\

& \displaystyle = k \left( \frac{(a+1)!(a-1)!}{a! a!} \right) \\ \br \\

& \displaystyle = \boxed{\left( \frac{a+1}{a} \right) k} \end{array}
Spoiler:
Show
\begin{array}{rl}

\mathrm{P} (T<t) & = \mathrm{P} (\frac{s}{V} < t}) \\ \br \\ & = \mathrm{P} (V > \frac{s}{t}) \\ \br \\

& \displaystyle = \int_{\frac{s}{t}}^{\infty} f(x) \, dx \\ \br \\

& \displaystyle = \int_{\frac{s}{t}}^{\infty} \frac{C k^{a+1} x^a}{(x+k)^{2a+2}} \, dx \end{array}

Making the substitution x = \frac{s}{u} \Rightarrow dx = -\frac{s}{u^2} \, du gives:

\begin{array}{rl}

\displaystyle\int_{\fracst}^{\in  fty} \frac{C k^{a+1} x^a}{(x+k)^{2a+2}} \, dx

& \displaystyle = \int_t^0 \frac{C k^{a+1} (\frac{s}{u})^a}{(\frac{s}{u} + k )^{2a+2}} \times - \frac{s}{u^2} \, du \\ \br \\

& \displaystyle = \int^t_0 \frac{C k^{a+1} s^{a+1}}{u^{a+2} (\frac{s+uk}{u})^{2a+2}} \, du \\ \br \\

& \displaystyle = \int^t_0 \frac{C (sk)^{a+1} u^a}{(s+uk)^{2a+2}} \, du \end{array}

Hnece the density function of T, g(x), is given by:

\boxed{\displaystyle g(x) = \frac{C (sk)^{a+1} x^a}{(s+kx)^{2a+2}}}
Spoiler:
Show
Making a substitution in P(T<t) of x = \frac{s^2}{k^2 z} \Rightarrow dz = -\frac{s^2}{k^2 z^2} \, dz gives:

\begin{array}{rl}

\displaystyle\int_0^t \frac{C (sk)^{a+1} x^a}{(s+kx)^{2a+2}} \, dx

& \displaystyle = \int_{\infty}^{\frac{s^2}{k^2 t}} \frac{C (sk)^{a+1} (\frac{s^2}{k^2 z})^a}{(s+k(\frac{s^2}{k^2 z}))^{2a+2}} \times - \frac{s^2}{k^2 z^2} \, dz \\ \br \\

& \displaystyle = \int^{\infty}_{\frac{s^2}{k^2 t}} \frac{C (sk)^{a+1} s^{2a+2}}{k^{2a+2} z^{a+2} (\frac{skz +s^2}{kz})^{2a+2}} \, dz \\ \br \\

& \displaystyle = \int^{\infty}_{\frac{s^2}{k^2 t}} \frac{C (sk)^{a+1} z^a}{(s+kz)^{2a+2}} \, dz \\ \br \\

& \displaystyle = \mathrm{P} \left( T &gt; \frac{s^2}{k^2 t} \right) \end{array}

Clearly, the median value of t, m_t is such that:
\displaystyle m_t = \frac{s^2}{k^2 m_t} \iff m_t = \frac{s}{k}
(since all values of t are >0)

Product of median time and median speed:

\displaystyle m_t m_v = \frac{s}{k} \times k = s

\begin{array}{rl}

\mathrm{E} (T)

& \displaystyle = \int_0^{\infty} x g(x) \, dx \\ \br \\

& \displaystyle = \int_0^{\infty} \frac{C (sk)^{a+1} x^{a+1}}{(s+kx)^{2a+2}} \, dx \\ \br \\

& \displaystyle = \frac{C s^{a+1}}{k^{a+1}} \int_0^{\infty} \frac{x^{a+1}}{(x+\frac{s}{k})^{  2a+2}} \, dx \\ \br \\

& \displaystyle = \frac{C s^{a+1}}{k^{a+1}} \left( \frac{(a+1)! (a-1)! k^a}{(2a+1)! s^a} \right) \\ \br \\

& \displaystyle = \frac{(a+1)s}{ak} \end{array}

Product of expected time and expected speed:

\displaystyle \mathrm{E} (T) \mathrm{E} (V) = \frac{(a+1)s}{ak} \times \frac{(a+1)k}{a} = \left( \frac{a+1}{a} \right)^2 s &gt; s
0
reply
Unbounded
Badges: 12
Rep:
?
#12
Report 10 years ago
#12
Question 2, STEP I, 2005
First Part
 y^2 = 4x \implies \dfrac{dy}{dx} = \frac{2}{y}

At P, the gradient of the tangent is hence 1/p and similarly at Q, the gradient of the tangent is 1/q

The equation of the tangent at P is:

 y - 2p = \frac{1}{p} (x-p^2) or  y = \frac{x}{p} +p

Similarly the equation of the tangent at Q is  y = \frac{x}{q} +q

Finding the point of intersection of these two tangents, to find the point R:

 \frac{x}{p} + p = \frac{x}{q} + q \iff qx + p^2q = px + q^2p

 \iff x(q-p) = pq(q-p) \implies \boxed{x = pq}

 y = \frac{x}{q} + q = \boxed{p+q}

Therefore  R = (pq, p+q) \ \ \ \square
Second Part
The line PQ will have a gradient of  \frac{2p-2q}{p^2-q^2} = \frac{2}{p+q}

So the equation of the line PQ is  y - 2p = \frac{2}{p+q} (x-p^2)

As the point (1,0) is on this line, we can find a condition linking p and q.

 2p = \frac{2}{p+q}(p^2-1)

 \iff p^2 + pq = p^2 - 1 \iff pq = -1

We proceed to find the coordinates of the point S; the equation of the normal at P is:

 y - 2p = p(p^2-x) or  y = p^3-px + 2p

Similarly the equation of the normal at Q is  y = q^3 -qx + 2q

Solving them to find S:

 q^3 -qx +2q = p^3 -px +2p

 \iff q^3 - p^3 +2(q-p) = x(q-p)

and dividing by q-p yields  q^2 + p^2 + qp + 2 = x

 \implies \boxed{x = p^2 + q^2 + 1}

And finding the y-coordinate  y = p^3 -p(p^2+q^2+1)+2p = p-pq^2 = \boxed{p+q}

Therefore  S = (p^2+q^2+1,p+q) \ \ \ \square
Final Part
Neat proof by Hashashin:
Longer and messier proof by myself spoilerised
The gradient of PR is \frac{1}{p} and the gradient of SQ is -q, and as pq = -1 it's clear that p = -\frac{1}{q} and therefore PR and SQ are parallel.

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.
Spoiler:
Show
We can show that the lines RP and PS are perpendicular:

RP has gradient  \dfrac{2p-(p+q)}{p^2-pq} = \dfrac{p-q}{p^2+1}

PS has gradient  \dfrac{2p-(p+q)}{p^2-(p^2+q^2+1)} = \dfrac{q-p}{q^2+1}

To show that they are perpendicular, the product of their gradients must be -1:

 \dfrac{q-p}{q^2+1} \times \dfrac{p-q}{p^2+1} = \dfrac{-(p-q)^2}{(q^2+1)(p^2+1)}

 = \dfrac{-(p-q)^2}{(pq)^2 + p^2 + q^2 + 1} = -\dfrac{p^2+q^2-2pq}{p^2+q^2+2} = -1 \ \ \ \square

By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
1
reply
Unbounded
Badges: 12
Rep:
?
#13
Report 10 years ago
#13
Question 7, STEP I, 2005
(i)
 P_1 = \displaystyle\prod_{r=1}^n \dfrac{r+1}{r} = \dfrac{2}{1} \times \dfrac{3}{2} \times \dfrac{4}{3} \times \cdots \times \dfrac{n+1}{n}

After cancelling, we are left with  \boxed{P_1 = n+1}

Note that  \displaystyle\prod_{r=2}^n \dfrac{r+1}{r} = \dfrac{n+1}{2}
(ii)
Note that  \displaystyle\prod_{r=2}^n \dfrac{r-1}{r} = \dfrac{1}{n}

 P_2 = \displaystyle\prod_{r=2}^n \dfrac{r^2-1}{r^2} = \displaystyle\prod_{r=2}^n \dfrac{(r+1)(r-1)}{r^2} = \left( \displaystyle\prod_{r=2}^n \dfrac{r+1}{r} \right) \times \left( \displaystyle\prod_{r=2}^n \dfrac{r-1}{r} \right)

And using the results we've just found

 \boxed{P_2 = \dfrac{n+1}{2n}}
(iii)
 P_3 = \displaystyle\prod_{r=1}^n \left( \cos \frac{2\pi}{n} + \sin \left(\frac{2\pi}{n} \right) \cdot \cot \left(\frac{(2r-1)\pi}{n}\right) \right) =  \displaystyle\prod_{r=1}^n \left( \cos \frac{2\pi}{n} + \sin \left(\frac{2\pi}{n} \right) \cdot \dfrac{\cos \left(\frac{(2r-1)\pi}{n}\right)}{\sin \left(\frac{(2r-1)\pi}{n}\right)} \right)

 = \displaystyle\prod_{r=1}^n \left( \dfrac{\cos \left(\frac{2\pi}{n}\right) \sin \left(\frac{(2r-1)\pi}{n}\right) + \sin \left(\frac{2\pi}{n}\right) \cos \left(\frac{(2r-1)\pi}{n}\right)}{\sin \left(\frac{(2r-1)\pi}{n}\right)} \right)

 = \displaystyle\prod_{r=1}^n \dfrac{\sin \left(\frac{(2r+1)\pi}{n}\right)  }{\sin \left(\frac{(2r-1)\pi}{n}\right)}

Again, expanding this, and watching the terms drop away, we end up with

 P_3 = \dfrac{\sin \left(\frac{(2n+1)\pi}{n}\right)  }{\sin \left(\frac{\pi}{n}\right)}  = \dfrac{\sin \left( 2\pi + \frac{\pi}{n} \right)}{\sin \left( \frac{\pi}{n} \right)}  = \dfrac{ \sin \left( \frac{\pi }{n} \right) }{\sin \left( \frac{\pi}{n} \right) } = \boxed{1}

And this is justified as n is even, so all the denominators of each term in the product are not zero, as  \frac{2r-1}{n} = \frac{\mathrm{odd}}{\mathrm{even  }} so  \frac{(2r-1)\pi}{n} is not a multiple of  \pi , meaning that  \sin \frac{(2r-1)\pi}{n} \not= 0 \ \forall r

When I originally looked at this part of the question, I just thought WTF, but it turned out to be quite nice =)
0
reply
Glutamic Acid
Badges: 14
#14
Report 10 years ago
#14
II/4:

1st part

\tan^{-1} \left( \dfrac{1}{a + b} \right) + \tan^{-1}\left(\dfrac{1}{a + c} \right) = \tan^{-1} \left( \dfrac{1}{a} \right) (*)

\Leftrightarrow \tan \left[ \tan^{-1} \left( \dfrac{1}{a + b} \right) + \tan^{-1}\left(\dfrac{1}{a + c} \right) \right] = \tan \tan^{-1} \left( \dfrac{1}{a} \right)

\Leftrightarrow \dfrac{\frac{1}{a + b} + \frac{1}{a + c}}{1 - \frac{1}{a + b} \frac{1}{a +c}} = \dfrac{1}{a}

\Leftrightarrow \dfrac{2a + b + c}{a^2 + ba + ca + bc - 1} = \dfrac{1}{a}

Substituting a^2 = bc - 1, LHS = \dfrac{2a + b + c}{2a^2 + ba + ca} = \dfrac{1}{a}, as required. Note, since a, b and c are positive, 1/(a+b), 1/(a+c) and 1/a will all lie between 0 and +infinity, so "implies and is implied by" implication signs can be used.


2nd part

Let p + q = "a", s = "b", t = "c" \Rightarrow \tan^{-1} \left( \dfrac{1}{p + q + s} \right) + \tan^{-1}\left(\dfrac{1}{p + q + t}\right) = \tan^{-1} \left( \dfrac{1}{p + q} \right), from (*)

Let p + r = "a", u = b, v = c \Rightarrow \Rightarrow \tan^{-1} \left( \dfrac{1}{p + r + u} \right) + \tan^{-1}\left(\dfrac{1}{p + r + v}\right) = \tan^{-1} \left( \dfrac{1}{p + r} \right), from (*)

Let p = "a", b = "q" and r = "c", so \Rightarrow \tan^{-1} \left( \dfrac{1}{p + q} \right) + \tan^{-1}\left(\dfrac{1}{p + r}\right) = \tan^{-1} \left( \dfrac{1}{p} \right), from (*), and we're done.


3rd part

We can use the result in the second part by choosing p = 7, q =1, s = 5, t = 13, r = 50, u = 25 and v = 130, and everything falls out neatly.
3
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#15
Report 10 years ago
#15
STEP III

Question 1


1st part


If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<->  Sin(A)(1+Sin(x))=CosA)Cos(x)
<-> tan(A) = \frac{Cos(x)}{1+Sin(x)}

Spoiler:
Show

It is relativel easy to show that  \frac{cos(x)}{1+sin(x)} = tan(\frac{\pi}{4} - \frac{x}{2}) if after searching for GA formula / derivation it has not been found, we use the identity sin(x) = cos(pi/2 - x) = cos(2(pi/4 - x/2)) and expand using the double angle formulae and the result follows fairly soon.


We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain;

 Tan(A) = Tan(\frac{\pi}{4} - \frac{x}{2} + n\pi) \; n \in \mathbb{Z}

<->  A = \pi(n+\frac{1}{4}) - \frac{x}{2}

and the result

 A = \frac{(4n+1)\pi}{2} + B follows - although interestingly we do not have  \pm B wip is obviously needed to obtain this.


2nd part


Rcos(x-a) = Rcos(a)cos(x) + Rsin(a)sin(x)

It follws that if

psin(x) + qcos(x) = Rcos(x-a) we have;

p = Rsin(a) and q = Rcos(a)

Adding p^2 + q^2 will reveal that this is equal to R^2 so ;

R^2 = p^2 + q^2

Divding reveals that (p/q) = tan(a)

Consider;

cos(x) + sin(x)

p=q=1 and it follows cos(x) + sin(x) = root(2)cos(x-(pi/4))

consider

sin(x) - cos(x) = root(2)cos(x+(pi/4))

We now note that |cos(x)| <= 1 and it follows that

 |cos(x) \pm sin(x)| \le \sqrt{2}

Using these results we see that for sin(sin(x)) = cos(cos(x)) we must have  sin(x) \pm cos(x) = (4n+1)\frac{\pi}{2}

Noting that  |cos(x) \pm sin(x)| \le \sqrt{2} and that |(4n+1)pi/2| => pi/2 for n in Z we need to determine whether root(2) => pi/2 which even crude approimations will quickly reveal to not be the case so it follows that there is no value of x that cos(cos(x)) = sin(sin(x))



Cheating with the graphs.
Attached files
2
reply
sonofdot
Badges: 8
Rep:
?
#16
Report 10 years ago
#16
STEP III 2005 Question 6

Spoiler:
Show
We have the equation x^3 - 3a^2 x = 2a^3 \cosh T and have to show it is satisfied by 2a \cosh (\frac13 T). Substituting x = 2a \cosh (\frac13 T) gives:

\begin{array}{rl}

x^2 - 3 a^2 x

& \displaystyle = 8a^3 \cosh^3 \left(\frac13 T \right) - 3a^2 \left(2a \cosh \left( \frac13 T \right) \right) \\ \br \\

& \displaystyle = 2a^3 \left( 4 \cosh^3 \left(\frac13 T \right) - 3 \cosh \left( \frac13 T \right) \right) \\ \br \\

& \displaystyle = 2a^3 \cosh T \end{array}

Spoiler:
Show
Now we have x^3 - 3bx = 2c with c^2 \geq b^3 &gt; 0

We can let b=a^2 for some a, since b^3 &gt; 0 \Leftrightarrow b &gt;0

We can also let c = a^3 \cosh T for some T, since cosh T>1 so c^3 \geq b^2 as required.

c = a^3 \cosh T \Leftrightarrow T = \cosh^{-1} \frac{c}{a^3} (if we say T>0)

\begin{array}{rl} T

& \displaystyle = \cosh^{-1} \frac{c}{a^3} \\ \br \\

& \displaystyle = \ln \left( \frac{c}{a^3} + \sqrt{\frac{c^2}{b^3} -1} \right) \\ \br \\

& \displaystyle = \ln \left( \frac{c+ \sqrt{c^2 - b^3}}{b^{\frac32}}  \right) \\ \br \\

& \displaystyle = \ln \left(c+ \sqrt{c^2 - b^3} \right) - \frac32 \ln b \end{array}

From first part, one of the roots is 2a \cosh (\frac13 T)

\begin{array}{rl}

\displaystyle 2a \cosh \left( \frac13 T \right)

& \displaystyle = 2a \cosh \left(\frac13 \ln \left(c+ \sqrt{c^2 - b^3} \right) - \frac12 \ln b \right) \\ \br \\

& \displaystyle = a \left( e^{\frac13 \ln \left(c+ \sqrt{c^2 - b^3} \right) - \frac12 \ln b \right)} + e^{-\frac13 \ln \left(c+ \sqrt{c^2 - b^3} \right) + \frac12 \ln b \right)} \right) \\ \br \\

& \displaystyle = \sqrt{b} \left( \frac{\left(c+ \sqrt{c^2 - b^3} \right)^{\frac13}}{\sqrt{b}} + \frac{\sqrt{b}}{\left(c+ \sqrt{c^2 - b^3} \right)^{\frac13}} \right) \\ \br \\

& \displaystyle = u + \frac{b}{u} \end{array}

Spoiler:
Show
Let \alpha = u + \frac{b}{u}

\begin{array}{rrl} & x^3 - 3bx - 2c

& \equiv (x-\alpha)(x^2 + Ax + B) \\ \br \\

\Leftrightarrow & x^3 - 3bx - 2c

& \equiv x^3 (A-\alpha)x^2 + (B - A\alpha)x - B\alpha \end{array}

By comparing coefficients, A = \alpha = u+\frac{b}{u} and B = \alpha^2 - 3b = u^2 + \frac{b^2}{u^2} -b

So other roots are satisfied by:
\displaystyle x^2 + \left( u + \frac{b}{u} \right) x + u^2 + \frac{b^2}{u^2} - b =0

Using the quadratic formula:

\begin{array}{rl} x

& \displaystyle = \frac12 \left( - \left( u+\frac{b}{u} \right) \pm \sqrt{u^2 + \frac{b^2}{u^2} +2b - 4 \left( u^2 + \frac{b^2}{u^2} -b \right)} \right) \\ \br \\

& \displaystyle = \frac12 \left( - \left( u+\frac{b}{u} \right) \pm \sqrt{-3 \left(u^2 + \frac{b^2}{u^2} - 2b \right)} \right) \\ \br \\

& \displaystyle = \frac12 \left( - \left( u+\frac{b}{u} \right) \pm \left( u - \frac{b}{u} \right) \sqrt3 i \right) \\ \br \\

& \displaystyle = \frac12 \left( -u \pm u\sqrt3 i + \frac{-b \mp b\sqrt3 i}{u} \right) \end{array}

This gives roots x = u \omega + \frac{b}{u} \omega^2 and x = u \omega^2 + \frac{b}{u} \omega, where \omega = \frac{-1 + i\sqrt3}{2} \Rightarrow \omega^2 = \frac{-1 - i\sqrt3}{2}

Spoiler:
Show
We now have x^3 - 6x = 6

From above, if we let b=2 and c=3, the roots are:

\displaystyle x=u + \frac{b}{u} , \, x = u\omega + \frac{b}{u} \omega^2, \, u\omega^2 + \frac{b}{u} \omega

\begin{array}{rl} u

& \displaystyle = \left(c+ \sqrt{c^2 - b^3} \right)^{\frac13} \\ \br \\

& \displaystyle = \left(3+ \sqrt{9 - 8} \right)^{\frac13} \\ \br \\

& \displaystyle = 2^{\frac23} \end{array}

\begin{array}{rl} \displaystyle u + \frac{b}{u}

& \displaystyle = 2^{\frac23} + \frac{2}{2^{\frac23}} \\ \br \\

& \displaystyle = 2^{\frac13} \left( 1 + 2^{\frac13} \right) \end{array}

Therefore the roots are:

\boxed{\sqrt[3]2 \left(1+\sqrt[3]2 \right) , \, \sqrt[3]2 \omega \left( \omega + \sqrt[3]2 \right) , \, \sqrt[3]2 \omega \left(1 + \sqrt[3]2 \omega \right) }

0
reply
Adjective
Badges: 12
Rep:
?
#17
Report 10 years ago
#17
III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation


 \displaystyle \frac{dy}{dx} = -\frac{xy}{x^2 + a^2}

Separating variables and integrating:

 \displaystyle \int \frac{1}{y} \ dy = - \int \frac{x}{x^2 + a^2} \ dx

 \displaystyle \ln y = -\frac{1}{2} \ln (x^2 + a^2) + K = -\frac{1}{2} \ln [B(x^2 + a^2)] with  K = - \frac{1}{2}\ln B

 \displaystyle y = \frac{1}{\sqrt{B}\sqrt{x^2 + a^2}}

 \displaystyle y \sqrt{x^2 + a^2} = c with  c = \frac{1}{\sqrt{B}}

 \displaystyle y^2(x^2 + a^2) = c^2


d/dx and d²/dx² of (x² + y²)


 \displaystyle \frac{d}{dx} (x^2 + y^2) = 2x + 2y\frac{dy}{dx}

 \displaystyle = 2x - \frac{2xy^2}{x^2 + a^2}

 \displaystyle = 2x - \frac{2xc^2}{(x^2 + a^2)^2}

 \displaystyle = 2x(1 - \frac{c^2}{(x^2 + a^2)^2}

 \displaystyle \frac{d^2}{dx^2} (x^2 + y^2) = \frac{d}{dx} \left( 2x(1 - \frac{c^2}{(x^2 + a^2)^2} \right)

 = \displaystyle 2\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right) + \frac{2.2x.2x.c^2}{(x^2 + a^2)^3}

 = \displaystyle 2\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right) + \frac{8x^2c^2}{(x^2 + a^2)^3}


Graph
This is a picture of y^2 = \frac{1}{x^2 + 1}. In general, the graph crosses the y-axis at ± c/a.


Point of minimum distance to origin with 0 < c < a^2


The points of minimum distance to the origin will be when  \sqrt{x^2 + y^2} is a minimum. Which is a minimum when  x^2 + y^2 is a minimum. Which is a minimum when:

 \displaystyle 2x\left( 1 - \frac{c^2}{(x^2 + a^2)^2} \right) = 0

and

 \displaystyle 2\left(1 - \frac{c^2}{(x^2 + a^2)^2}\right) + \frac{8x^2c^2}{(x^2 + a^2)^3} &gt; 0 .

So... there is a stationary value of x² + y² when x = 0. When this is the case,

 \displaystyle \frac{d^2}{dx^2} (x^2 + y^2) = 2(1 - \frac{c^2}{a^4}) , which is greater than 0 since c < a² . Subbing x = 0 into  y^2(x^2 + a^2) = c^2 yields  y = \pm \frac{c}{a} .

So our minimum distances from the origin are at  (0, \pm \frac{c}{a}) .


If c > a²


Looking at the other case in which  \displaystyle 2x \left(1 - \frac{c^2}{(x^2 + a^2)^2} \right) = 0

i.e.  \displaystyle 1 - \frac{c^2}{(x^2 + a^2)^2} = 0

leads to  x^2 = \pm c - a^2 .

Taking x² with positive c, when subbed into  \frac{d^2}{dx^2} (x^2 + y^2) this gives:

 \displaystyle 2(1 - \frac{c^2}{c^2}) + \frac{8(c - a^2)c^2}{c^3}

which is greater than 0 since c > a². Taking x² with negative c leads to a minimum too, but also to an imaginary x, which we don't want to bother with.

When  \displaystyle x^2 = c - a^2 :

 \displaystyle y^2 = \frac{c^2}{c - a^2 + a^2} = c \Rightarrow y = \pm \sqrt c .

So our new minima are  ( \pm \sqrt{c -a^2}, \pm \sqrt{c}) .
Attached files
1
reply
Unbounded
Badges: 12
Rep:
?
#18
Report 10 years ago
#18
Question 8, STEP I, 2005
 y^2 = x^kf(x) \implies xy^2 = x^{k+1}f(x)

 \implies 2xy\dfrac{\mathrm{d}y}{\mathrm{d  }x} + y^2 = (k+1)x^kf(x) + x^{k+1}f'(x)

 \iff 2xy\dfrac{\mathrm{d}y}{\mathrm{d  }x} + y^2 = (k+1)y^2 + x^{k+1}\dfrac{\mathrm{d}f}{\math  rm{d}x}

 \iff 2xy\dfrac{\mathrm{d}y}{\mathrm{d  }x} = ky^2 + x^{k+1}\dfrac{\mathrm{d}f}{\math  rm{d}x} \ \ \ \square
(i)
Let the solution be of the form  y^2 = x^k f(x)

 2xy \dfrac{\mathrm{d}y}{\mathrm{d}x} = y^2 + x^2 - 1

We see, and are told, that k = 1

 \therefore x^{k+1}\dfrac{\mathrm{d}f}{\math  rm{d}x} = x^2\dfrac{\mathrm{d}f}{\mathrm{d  }x} = x^2 - 1

 \iff \displaystyle\int \mathrm{d}f = \displaystyle \int \left(1 - \dfrac{1}{x^2}\right) \ \mathrm{d}x

 \iff f(x) = x - \dfrac{1}{x} + C

So the solution is:

 y^2 = x\left(x+\dfrac{1}{x} + C\right)

Subbing y = 2, x = 1 gives C = 2

 \therefore y^2 = x\left(x+\dfrac{1}{x}+2\right) = x^2+ 2x + 1 = (x+1)^2

 \therefore \boxed{y = x+1, y = -x-1} which is a pair of lines perpendicular to each other, which form a cross centred at (-1,0).
(ii)
 2x^2y\dfrac{\mathrm{d}y}{\mathrm  {d}x} = 2\ln x - xy^2

 \implies 2xy\dfrac{\mathrm{d}y}{\mathrm{d  }x} = \dfrac{2\ln x}{x} - y^2

Let the solution be of the form  y^2 = \dfrac{f(x)}{x} ie. k = -1

 \therefore x^{-1+1} \dfrac{\mathrm{d}f}{\mathrm{d}x} = \dfrac{2\ln x}{x}

 \iff \displaystyle\int \mathrm{d}f = 2 \displaystyle\int \dfrac{\ln x}{x} \ \mathrm{d}x

 \iff f(x) = (\ln x)^2 + C

 \therefore y^2 = \dfrac{(\ln x)^2 + C}{x}

Subbing y = 1, x = 1 yields C = 1

 \therefore \boxed{y^2 = \dfrac{(\ln x)^2 + 1}{x}}
0
reply
Hashshashin
Badges: 13
Rep:
?
#19
Report 10 years ago
#19
(Original post by GHOSH-5)
Question 2, STEP I, 2005

Final Part
We can show that the lines RP and PS are perpendicular:

RP has gradient  \dfrac{2p-(p+q)}{p^2-pq} = \dfrac{p-q}{p^2+1}

PS has gradient  \dfrac{2p-(p+q)}{p^2-(p^2+q^2+1)} = \dfrac{q-p}{q^2+1}

To show that they are perpendicular, the product of their gradients must be -1:

 \dfrac{q-p}{q^2+1} \times \dfrac{p-q}{p^2+1} = \dfrac{-(p-q)^2}{(q^2+1)(p^2+1)}

 = \dfrac{-(p-q)^2}{(pq)^2 + p^2 + q^2 + 1} = -\dfrac{p^2+q^2-2pq}{p^2+q^2+2} = -1 \ \ \ \square

By symmetry, the lines RQ and QS must be perpendicular, and by some simple geometry, we can see that QR and RP, and also QS and SP must also be perpendicular, hence we have a quadrilateral with four right-angles, and hence a rectangle. Q.E.D.
For this part, is it not far simpler to note that as the gradient of PR is \frac{1}{p} and the gradient of SQ is -q, and as pq = -1 it's clear that p = -\frac{1}{q} and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
1
reply
Unbounded
Badges: 12
Rep:
?
#20
Report 10 years ago
#20
(Original post by Hashshashin)
For this part, is it not far simpler to note that as the gradient of PR is \frac{1}{p} and the gradient of SQ is -q, and as pq = -1 it's clear that p = -\frac{1}{q} and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
That is a lot neater. I'll edit it :smile:
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (487)
31.12%
The paper was reasonable (603)
38.53%
Not feeling great about that exam... (256)
16.36%
It was TERRIBLE (219)
13.99%

Watched Threads

View All