Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

appleibeats · Sep 10, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Not entirely sure what you mean.

InteGrand · Sep 10, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Not entirely sure what you mean.

$The area as a function of $G$ is $A=\frac{72G}{(2+G)^2}$, from part (a), where $0\leq G \leq 2\pi$, as $G$ is an angle of a circle sector. Have you learnt how to find maxima and minima of such functions?$

appleibeats · Sep 13, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Two circles intersect at point P and Q. The tangent to one circle at P meets the other at A and the tangent to the second circle at P meets the first circle at B. Show that triangles PQA and BQP are similar.

rand_althor · Sep 13, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Two circles intersect at point P and Q. The tangent to one circle at P meets the other at A and the tangent to the second circle at P meets the first circle at B. Show that triangles PQA and BQP are similar.

$Let $\angle{DPA}=\alpha$. $\angle{BPE}=\alpha$, as $\angle{DPA}$ and $\angle{BPE}$ are vertically opposite angles. $\angle{PQA}=\angle{DPA}=\alpha$, and $\angle{BQP}=\angle{BPE}=\alpha$, as per the alternate segment theorem. Now let $\angle{APQ}=\beta$. $\angle{APE}=\angle{APQ}+\angle{QPB}+\angle{BPE}= \alpha+\beta+\angle{QPB}=180^{\circ}$, as angles on a straight line add up to $180^{\circ}$. Rearranging: $\angle{QPB}=180^{\circ}-\alpha-\beta$. Now, $\angle{QPB}+\angle{PBQ}+\angle{BQP}=180^{\circ}$, as the angle sum of a triangle is equal to $180^{\circ}$. Rearranging and substituting values: $\angle{PBQ}=180^{\circ}-\angle{QPB}-\angle{BQP}=180^{\circ}-(180^{\circ}-\alpha-\beta)-\alpha=180^{\circ}-180^{\circ}+\alpha+\beta-\alpha=\beta$

$So in $\Delta$s PQA and BQP, we have $\angle{PQA}=\angle{BQP}=\alpha$ and $\angle{APQ}=\angle{PBQ}=\beta$. Therefore the two triangles are equiangular.$

appleibeats · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Not sure if it is just me, but above you diagram of the circles there are two squares. I think they are meant to be images?? Am I the only one seeing this??

Using a mac.....

calamebe · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Not sure if it is just me, but above you diagram of the circles there are two squares. I think they are meant to be images?? Am I the only one seeing this??

Using a mac.....

You're not the only one.

InteGrand · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Not sure if it is just me, but above you diagram of the circles there are two squares. I think they are meant to be images?? Am I the only one seeing this??

Using a mac.....

I think those squares are LaTeX lines, since the other LaTeX stuff on this page is like that too. LaTeX hasn't been working here lately. It doesn't matter though, since you can see the proof from the diagram, it's basically just using the alternate segment theorem. Also, it usually helps to join the common chord for these types of questions, as that diagram did (the chord PQ).

appleibeats · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

PQR is an isosceles triangle with PQ = PR . A line through R, perpendicular to PR meets the bisector of angle QPR at S. Show that P,Q,R and S are concyclic.

rand_althor · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
PQR is an isosceles triangle with PQ = PR . A line through R, perpendicular to PR meets the bisector of angle QPR at S. Show that P,Q,R and S are concyclic.

Construct a line through Q, perpendicular to QP. Note that this line also passes through S. Let angle QPR = 2alpha. Since PS bisects angle QPR, angle QPS=angle SPR=alpha. Now, find angle RQP and angle PRQ, using the fact that triangle QPR is iscosceles and the angle sum of a triangle is 180 degrees. Then find angle SQR and angle QRS, using the fact that angles on a straight line add up to 180 degrees. After doing that find angle RSQ, using the fact that the angle sum of a triangle is 180 degrees. Now, you should have angle QPR=2alpha and RSQ=180-2alpha. Since these angles are supplementary, QPRS is a cyclic quadrilateral, and hence Q, P, R, S are concyclic.

appleibeats · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Solve the equation 2cos2E - 7sinE = 4 for 0 < or equal to E < or equal to 360

kawaiipotato · Sep 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$2\cos2E - 7\sinE = 4 $ for $ 0 \leq E \leq 360$

$2(1-2\sin^{2}E) - 7sinE = 4 ( $ since $ \cos2E = 1-2sin^{2}E$

$-4\sin^{2}E - 7sinE -2 = 0$

$4\sin^{2}E + 7sinE + 2 = 0$

$\delta = 49 - 4*2*4$

$= 17 ( $ what a weird number, maybe I made some mistake somewhere $ )$

$\therefore sinx = \frac{-7 \pm \sqrt{17}}{8}$

$\therefore sinx = \frac{-7 + \sqrt{17}}{8} or sinx = \frac{-7 - \sqrt{17}}{8}$

$$ But $ |sinx| \leq 1$

$Must exclude second solution$

$\therefore sinx = \frac{-7 + \sqrt{17}}{8} = $blah$$

$x = sin^{-1}(blah) + k360 $where k is an integer $$

$or x = 180 - sin^{-1}(blah) + k360$

$Sub in k values (k=0,1,2,3,...) and accept values of x within the restricted domain$

appleibeats · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Show that 4x^5 + 32x^3 + 66x / x^4 + 8x^2 + 16 = 4x + 2x/ (x^2 + 4)^2

I did this.

But I got the next part wrong.

Using the substitution u = x^2 + 4 or otherwise, find the Integral of 4x^5 + ....... (above) dx

I let u = x^2 + 4

x^2 = u - 4

x = (u -4) ^ 1/2

du /dx = 2x

du = 2x dx

Integral of 4x^5 ...... dx

= Integral of 4x + 2x/ ...... dx

= Integral 4 (u-4)^1/2 + u^-2 du

= 4( u -4)^3/2 . 2/3 + u^-1 . -1 + C

= 8/3 x^3 - 1/x^2 + 4 + C

Not sure where I have gone wrong.

appleibeats · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$The area as a function of $G$ is $A=\frac{72G}{(2+G)^2}$, from part (a), where $0\leq G \leq 2\pi$, as $G$ is an angle of a circle sector. Have you learnt how to find maxima and minima of such functions?$

No I don't think so.

InteGrand · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
No I don't think so.

Learn that theory first then before attempting that Q.

rand_althor · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
= Integral of 4x + 2x/ ...... dx

= Integral 4 (u-4)^1/2 + u^-2 du

$\begin{align*}\int \frac{4x^5 + 32x^3 + 66x}{x^4 + 8x^2 + 16} \, \mathrm{d}x &= \int 4x+\frac{2x}{(x^2 + 4)^2} \, \mathrm{d}x \\&= \int \left (4x+\frac{2x}{\underset{u}{(\underbrace{x^2 + 4}})^2} \right ) \, \frac{1}{2x}\times \underset{\mathrm{d}u}{\underbrace{2x\times\mathrm{d}x}} \\&= \int 2+\frac{1}{u^2} \,\mathrm{d}u \\&= 2u-\frac{1}{u} + C \\&=2(x^2+4)-\frac{1}{x^2+4}+C \\\end{align*}$

appleibeats · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Solve 2cos^2 x - 3sinxcosx + 2sin^2 x = 0

I tried dividing by cos^2 x

to get

2tan^2 x - 3tanx + 2 = 0

Not sure where to go from here though, as you can't factorise, use completing the square or the quadratic formula.

InteGrand · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Solve 2cos^2 x - 3sinxcosx + 2sin^2 x = 0

I tried dividing by cos^2 x

to get

2tan^2 x - 3tanx + 2 = 0

Not sure where to go from here though, as you can't factorise, use completing the square or the quadratic formula.

$Let $u=\tan x$ and you have yourself a quadratic equation in $u$. You can easily solve for $u$, and then for $x$ (but discriminant here is negative, so no real solutions).$

rand_althor · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Not sure where to go from here though, as you can't factorise, use completing the square or the quadratic formula.

This would mean there are no solutions. Here is another way of reaching the same answer:

$\begin{align*}2\cos^2{x} - 3\sin{x}\cos{x} + 2\sin^2{x}&=0 \\2(\sin^2{x}+\cos^2{x})-3\sin{x}\cos{x}&=0 \\2-3\sin{x}\cos{x}&=0 \\2-\frac{3}{2}\times 2\sin{x}\cos{x}&=0 \\2-\frac{3}{2}\sin{2x}&=0 \\\frac{3}{2}\sin{2x}&=2 \\\sin{2x}&=\frac{4}{3} \\\textup{But }\sin{2x} \textup{ has a range of }&-1\leq y \leq 1\textup{. Therefore there are no solutions.}\end{align*}$

appleibeats · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Solve the equation sec^2 x - 2 tanx = 0 for 0 < or equal to x < or equal to 180

I got this but I think I am very wrong.

1/cos^2 x -2 sinx/cosx = 0

( x cos^2x)

1 - 2sinxcosx = 0

1 - sin2x = 0

sin2x = 1

2x = 90
x = 45

Is that wrong??

braintic · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Solve the equation sec^2 x - 2 tanx = 0 for 0 < or equal to x < or equal to 180

I got this but I think I am very wrong.

1/cos^2 x -2 sinx/cosx = 0

( x cos^2x)

1 - 2sinxcosx = 0

1 - sin2x = 0

sin2x = 1

2x = 90
x = 45

Is that wrong??

Why don't you substitute to check?
Also consider whether you have missed any solutions.

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