# Thread: help needed to finish exercises in a book

1. ## help needed to finish exercises in a book

As a refresher i set myself a task of completing all the misc exercises in the bostock and chandler A-level book (red ). I have ploughed through what must be in region of 500 past A-level questions (from 1970s-1980s) and out of them i am unable to give a solution to the follwing.

If anyone could be so kind to help me finish off this book i would be grateful.

It is not necessarily the case they are hard problems.
They may well have been met when i'd already done 30-40 questions from the chapter and i ran out of mental power or im just not able to do them

1) a curve is
x=a(5cosp+cos5p)
y=a(5sinp-sin5p)

Find equation of normal at point with parameter p and find the point in
1st quadrant of which normal is also normal to curve at another point.

2) If Sn=a^r(1+a+a^2+.....a^r) by considering (1-a)Sn show that

(1-a)Sn=(1-a^(2n+2))/(1-a^2) - a^(n+1)[1+a^(n+1)]/(1-a)

3) In the binomial expansion of (p+q)^n write down the term containing p^r. If p=1/6, q=5/6 and n=30, find the value of r for which this term is greatest in value.

book answer: 5. is there quick way apart from working the values out upto 5 then seeing they go down at 6?

4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.

2. so the term is

$(nCr)(p^r)(q^{n-r})=(30Cr)((\frac{1}{6})^r(\frac{5}{6})^{30-r}$

$=(30Cr)((\frac{1}{5})^r(\frac{5}{6}^{30})$

we want to find for which term, the next is smaller

$(30Cr)((\frac{1}{5})^r(\frac{5}{6})^{30}>(30Cr+1)( (\frac{1}{5})^{r+1}(\frac{5}{6})^{30}$

$(30Cr)>(30Cr+1)(\frac{1}{5})$

$\frac{30!}{r!(30-r)!}>\frac{30!}{(r+1)!(30-r-1)!}(\frac{1}{5})$

$\frac{1}{30-r}>\frac{1}{r+1}(\frac{1}{5})$

$r>4.1$

but r integer

$5\leq r$

3. Hello jiboom

Welcome to Math Help Forum!
Originally Posted by jiboom
As a refresher i set myself a task of completing all the misc exercises in the bostock and chandler A-level book (red ). I have ploughed through what must be in region of 500 past A-level questions (from 1970s-1980s) and out of them i am unable to give a solution to the follwing.

If anyone could be so kind to help me finish off this book i would be grateful.

It is not necessarily the case they are hard problems.
They may well have been met when i'd already done 30-40 questions from the chapter and i ran out of mental power or im just not able to do them

1) a curve is
x=a(5cosp+cos5p)
y=a(5sinp-sin5p)

Find equation of normal at point with parameter p and find the point in
1st quadrant of which normal is also normal to curve at another point.

2) If Sn=a^r(1+a+a^2+.....a^r) by considering (1-a)Sn show that

(1-a)Sn=(1-a^(2n+2))/(1-a^2) - a^(n+1)[1+a^(n+1)]/(1-a)

3) In the binomial expansion of (p+q)^n write down the term containing p^r. If p=1/6, q=5/6 and n=30, find the value of r for which this term is greatest in value.

book answer: 5. is there quick way apart from working the values out upto 5 then seeing they go down at 6?

4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.
You will find it easier to get help if you post questions separately, each in the appropriate sub-forum - see Rule 14.

But here's a start for the first one.

$\frac{dx}{dp}=-5a(\sin p +\sin 5p)$

$\frac{dy}{dp}=5a(\cos p -\cos 5p)$

So the gradient of the normal $= -\frac{dx}{dy}=\frac{\sin p +\sin 5p}{\cos p -\cos 5p}$

So its equation is
$y-a(5\sin p -\sin 5p) = \frac{\sin p +\sin 5p}{\cos p -\cos 5p}\Big(x -a(5\cos p + \cos 5p)\Big)$
Can you take it from here?

4. posted questions in different sections.
Thanks for replies so far,

regarding 3)
are we certain that once the term goes down it does not increase again?

Any ideas for number 2?

5. Hello jiboom
Originally Posted by jiboom
posted questions in different sections.
Thanks for replies so far,

regarding 3)
are we certain that once the term goes down it does not increase again?

Any ideas for number 2?
For number 2, I assume you mean
$S_n=a^n(1+a + a^2+...+a^n)$ ( $n$ not $r$)
in which case I should think you'd start like this:
$aS_n = a^n(a+a^2+...+a^n+a^{n+1})$

$\Rightarrow (1-a)S_n=a^n(1 - a^{n+1})$
As far as number 3 is concerned, it is well know that the binomial distribution has a single maximum peak, so, yes, I am certain that once the term goes down it does not increase again. But I realise that's not a proof.

6. for 3 we showed that the next term is smaller than the previous for all values larger than 5 so if this was not correct for all r>=5 then proof by induction would have a serious problem.

Hello jiboomFor number 2, I assume you mean
$S_n=a^n(1+a + a^2+...+a^n)$ ( $n$ not $r$)
in which case I should think you'd start like this:
$aS_n = a^n(a+a^2+...+a^n+a^{n+1})$

$\Rightarrow (1-a)S_n=a^n(1 - a^{n+1})$
As far as number 3 is concerned, it is well know that the binomial distribution has a single maximum peak, so, yes, I am certain that once the term goes down it does not increase again. But I realise that's not a proof.

sorry, ive mistyped the question. it should read

Sn=sum_{r=0 to N} a^r(1+a+a^2+...a^r)

8. Hello jiboom
Originally Posted by jiboom
sorry, ive mistyped the question. it should read

Sn=sum_{r=0 to N} a^r(1+a+a^2+...a^r)
Please confirm the question, which you have now written as:
Given $S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r)$, prove that $(1-a)S_n= \frac{1-a^{2n+2}}{1-a^2}-\frac{a^{n+1}(1+a^{n+1})}{1-a}$
The problem is that I don't think this is correct. Checking it when $n=0$ does not give the correct answer.

9. Hello jiboom

This is what question 3 should be, I think:

$S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r)$

$\Rightarrow aS_n = \sum_{r=0}^na^r(a+a^2+a^3+...+a^{r+1})$

$\Rightarrow (1-a)S_n = \sum_{r=0}^na^r(1-a^{r+1})$

$=\sum_{r=0}^na^r-\sum_{r=0}^na^{2r+1}$
Each of these is a GP. I'll leave you to work out what the common ratios, etc, are. The result, then, is:

$(1-a)S_n = \frac{1-a^{n+1}}{1-a}-\frac{a(1-a^{2n+2})}{1-a^2}$

Hello jiboomPlease confirm the question, which you have now written as:
Given $S_n = \sum_{r=0}^na^r(1+a+a^2+...+a^r)$, prove that $(1-a)S_n= \frac{1-a^{2n+2}}{1-a^2}-\frac{a^{n+1}(1+a^{n+1})}{1-a}$
The problem is that I don't think this is correct. Checking it when $n=0$ does not give the correct answer.

what you have posted is the question from the book. i agree with you, the question looks wrong. It seemed strange to me that we get 2 gps which we can sum but one term matches and the other is a country mile away. that is why i posted it to see if i was misunderstanding Sn. the chapter on series has more mistakes in 41 questions than the other 16 chapters !

11. posted

1) a curve is
x=a(5cosp+cos5p)
y=a(5sinp-sin5p)

Find equation of normal at point with parameter p and find the point in
1st quadrant of which normal is also normal to curve at another point.

4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

5) use vector geometry to prove the internal bisectors of the angles of a triangle are concurrent.

in other areas but getting no rplies. Anyone in this section wanna suggest something?

@grandad: i can get that far with 1). i simplify a bit further using cos A+cos B etc, but the final part alludes me.

12. Originally Posted by jiboom
posted

1) a curve is
x=a(5cosp+cos5p)
y=a(5sinp-sin5p)

Find equation of normal at point with parameter p and find the point in
1st quadrant of which normal is also normal to curve at another point.

4) given y=[(1+x)/(2+x)]^1/2 find value of dy/dx at x=2. Deduce the increase in value of y when x increases in value from 2 to 2+e with e small

i am getting closer to finishing,5 is done. Iust these 2 to go. Anyone able to help? I especially need a help with 4. i dont get 30 but 48 (i think or something like that). If anyone else agrees i can put that down as another error in book.

13. Hello jiboom

$y = \left(\frac{1+x}{2+x}\right)^{\frac{1}{2}}$

$\Rightarrow y^2 = \frac{1+x}{2+x}$

$\Rightarrow 2y\frac{dy}{dx}=\frac{(2+x)1-(1+x)1}{(2+x)^2}$
$=\frac{1}{(2+x)^2}$
When $x = 2, y = \left(\frac34\right)^{\frac12}$ and so

$2\left(\frac34\right)^{\frac12}\frac{dy}{dx}= \frac{1}{4^2}=\frac{1}{16}$

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt3}{48}\approx\frac{\delta y}{\delta x}$

So when $\delta x = e, \delta y \approx \frac{\sqrt3e}{48}$

14. thanks for the confirmation. I maybe wondered if they wanted something else but surely the increase in y is the delta y you have.

For the curve question i simplify dy/dx as

dy/dx= 5a(5cos t-cos 5t)/[-5a(sin t+ sin 5t)]
=cos 5t-cos t/[sin 5t+sin t]
= -2sin3tsin2t/[2sin3tcos2t]
=-sin2t/cos2t

so grad of normal is cos(2t)/sin(2t) hence normal is given by

y-a(5 sin t -sin 5t)= cos(2t)/sin(2t)[x-a(5cos t+cos 5t)]

sin(2t)y-cos(2t)x=a[5sin(t)sin(2t)-sin(5t)sin(2t)-cos(5t)cos(2t)-5cos(t)cos(2t)
.......................=a[-5(cos(2t)cos(t)-sin(2t)sin(t))-(cos(5t)cos(2t)+sin(5t)sin(2t)
sin(2t)y-cos(2t)x=a[-5cos(3t)-cos(3t)]=-6acos(3t)

so need slick way of finding the other points in first quadrant which normal is normal at another point.

15. Hello jiboom
Originally Posted by jiboom
thanks for the confirmation. I maybe wondered if they wanted something else but surely the increase in y is the delta y you have.

For the curve question i simplify dy/dx as

dy/dx= 5a(5cos t-cos 5t)/[-5a(sin t+ sin 5t)]
=cos 5t-cos t/[sin 5t+sin t]
= -2sin3tsin2t/[2sin3tcos2t]
=-sin2t/cos2t

so grad of normal is cos(2t)/sin(2t) hence normal is given by

y-a(5 sin t -sin 5t)= cos(2t)/sin(2t)[x-a(5cos t+cos 5t)]

sin(2t)y-cos(2t)x=a[5sin(t)sin(2t)-sin(5t)sin(2t)-cos(5t)cos(2t)-5cos(t)cos(2t)
.......................=a[-5(cos(2t)cos(t)-sin(2t)sin(t))-(cos(5t)cos(2t)+sin(5t)sin(2t)
sin(2t)y-cos(2t)x=a[-5cos(3t)-cos(3t)]=-6acos(3t)

so need slick way of finding the other points in first quadrant which normal is normal at another point.
Ah, I see you spotted the mistake you'd made in the sign, and you've corrected it. This is a very tricky question (unless I'm overlooking something obvious - which is entirely possible!).

Yes, I agree that the equation of the normal at $p$ is:
$y\sin2p=x\cos2p-6a\cos3p$
Therefore the normal at $q$ is:
$y\sin2q=x\cos2q-6a\cos3q$
The normal at $p$ is normal to the curve again at $q$, if these are one and the same line; i.e. if (making $y$ the subject and equating coefficients):
$\frac{\cos2p}{\sin2p}=\frac{\cos2q}{\sin2q}$ (1)
and
$\frac{\cos3p}{\sin2p}=\frac{\cos3q}{\sin2q}$ (2)
Equation (1) gives:
$\tan2q=\tan2p$

$\Rightarrow 2q = 2p + n\pi, n = 1, 2, 3$ (Note: $n =4$ gives $p = q +2\pi$; i.e. $p$ and $q$ represent the same point; and $n > 4$ gives repeated values.)

$\Rightarrow q = p + n\pi/2$
Substitute into (2):
$\frac{\cos3p}{\sin2p}= \frac{\cos (3p+3n\pi/2)}{\sin(2p +n\pi)}$
There's a bit of work to do now with the three different values of $n$, but I think when you simplify everything, you get:

$n=1:$
$\tan3p = -1$

$\Rightarrow p = \pi/4, 7\pi/12, 11\pi/12, ...$
$n=2:$
$\cos 3p = 0$

$\Rightarrow p = \pi/6, \pi/2, 5\pi/6, ...$
$n=3:$
$\tan3p = 1$

$\Rightarrow p = \pi/12, 5\pi/12, 3\pi/4, ...$
And it's then a question of choosing the values of $p$ that give points in the first quadrant. I think these are:
$p = \pi/12, \pi/6, \pi/4, 5\pi/12$
I attach an Excel spreadsheet that shows the curve (with $a = 1$)and a normal (in fact the one where $p = \pi/6$), which you can (if you have Excel) play around with to confirm that these values seem to be correct.