Cambridge Prelim MX1 Textbook Marathon/Q&A (7 Viewers)

Drongoski · May 10, 2016

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

Yea - I did it on 29 May 2013. See if I can locate my solution.

Drongoski · May 10, 2016

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

Found it.

Q17 b(ii)

From b(i):
$$if $ \beta = \frac {1}{2}\alpha $ then $ \alpha = 2\beta \\ \\ \therefore V cos\beta = Ucos 2\beta = U(2cos^2 \beta - 1)\\ \\ \therefore 2Ucos^2 \beta - V cos \beta - U = 0 $ (a quadratic equation in cos B)$ \\ \\ \therefore cos \beta = \frac {-(-V) \pm \sqrt {(-V)^2 - 4 \times 2U \times (-U)}}{2 \times 2U} = \frac {V \pm \sqrt {V^2 + 8U^2}}{4U}\\ \\ $since B is acute, cos B is positive $ \\ \\ \therefore cos \beta = \frac {V+\sqrt {V^2+8U^2}}{4U}\\ \\ \therefore \beta = cos^{-1} \left( \cdots \cdots \right)$

drsabz101 · May 30, 2016

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

how do you solve these questions?

InteGrand · May 30, 2016

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

drsabz101 said:
how do you solve these questions?
View attachment 33223

Essentially, use trig. identities.

$\noindent For example, for (a), get everything in terms of just $\cot ^2 \theta$ or just $\csc^2 \theta$, using the trig. identity $\cot^2 \theta + 1 = \csc^2 \theta$. Say you did everything in terms of $\cot^2 \theta$. Then you can solve for $\cot^2 \theta$ from one equation and sub. it into the other one, thus eliminating $\theta$.$

drsabz101 · Jun 2, 2016

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

Can someone help solve these questions, with working out. I really am not getting these.

davidgoes4wce · Jun 3, 2016

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

$Q6 a \ sin2x=\frac{1}{2} $ in the interval [0,360] $$

$2x=30^\circ+k \ 360^\circ$

$2x=150^\circ+k \ 360^\circ$

$Where k=0,1$

$$ Solutions of x $=15^\circ, 75^\circ, 195^\circ , 255^\circ$

Here is the graph:

drsabz101 · Jun 3, 2016

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

Thankyou! Could u please also show the steps for solving questions 7 c and d

davidgoes4wce · Jun 3, 2016

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

$Question 6 (b)$

$cos 2x=\frac{-1}{\sqrt{2}}$

$k=0,1$

$2x=135^\circ+k \ 360^\circ \ and \ 2x=225^\circ+k \ 360^\circ$

$Let k=0,1$

$x=67.5^\circ, 112.5^\circ, 247.5^\circ, 292.5^\circ$

davidgoes4wce · Jun 3, 2016

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

$Question 6 (c)$

$tan \ 3x=\sqrt{3}$

$3x=60^\circ+180^\circ k$

$Let k=0,1,2,3,4,5$

$x=20^\circ, 80^\circ, 140^\circ, 200^\circ, 260^\circ, 320^\circ$

davidgoes4wce · Jun 3, 2016

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

$Q 6(d) . There are no solutions for sec(3x)=0$

davidgoes4wce · Jun 3, 2016

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

$Q 7 (a)$

$tan \ (\alpha-45^\circ)=\frac{1}{\sqrt{3}}$

$\alpha-45^\circ=30^\circ+k \ 180^\circ$

$k=0,1$

$\alpha=75^\circ , 255^\circ$

davidgoes4wce · Jun 3, 2016

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

$Q 7 (b)$

$sin (\alpha+30)=\frac{-\sqrt{3}}{2}$

$\alpha+30^\circ=240^\circ + k \ 360^\circ$

$\alpha+30^\circ=300^\circ + k \ 360^\circ$

$$ Let k =0, \alpha=210^\circ , \alpha=270^\circ$

davidgoes4wce · Jun 3, 2016

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

$Q 7 (c)$

$cot (\alpha+60^\circ)=1$

$\frac{1}{tan(\alpha+60^\circ)}=1$

$1=tan (\alpha+60^\circ)$

$Drawing the ASTC Diagram, Tan lies in the positive in the 1st and 3rd quadrants$

$\alpha+60^\circ=45^\circ+k \ 180^\circ$

$Let k=1,2$

$\alpha=165^\circ, 345^\circ$

davidgoes4wce · Jun 3, 2016

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

$Q 7 (d)$

${cosec(\alpha-75^\circ)}=-2$

$\frac{1}{sin(\alpha-75^\circ)}=-2$

$\frac{-1}{2}=sin(\alpha-75^\circ)$

$\alpha-75^\circ=210^\circ$

$\alpha-75^\circ=-30^\circ$

$\alpha=285^\circ, 45^\circ$

appleibeats · Jun 4, 2016

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

Screen Shot 2016-06-04 at 12.39.31 pm.png

Unsure how to do part b). I get the answer 13/6 but the answer is n = 7

InteGrand · Jun 4, 2016

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

appleibeats said:
View attachment 33235

Unsure how to do part b). I get the answer 13/6 but the answer is n = 7

$\noindent The answer should be an integer. Note that the coefficients of $x,x^2$ and $x^3$ are $n,\frac{n\left(n-1\right)}{2}$ and $\frac{n(n-1)(n-2)}{3!}$ (i.e. binomial coefficients). Since this is an A.P., we have $\frac{n\left(n-1\right)}{2} = \frac{n+\frac{n(n-1)(n-2)}{3!}}{2}$. Now you just need to solve this equation for $n$ (rearrange it etc. first. You can use the integer roots theorem to test appropriate integer values of $n$.). Make sure to take the positive integer solution.$

appleibeats · Jun 4, 2016

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

Screen Shot 2016-06-04 at 1.21.44 pm.png

Not sure how to begin this question. I tried finding the general form for the binomial expansion. Is that correct to do? But I am having some difficuties collecting x terms.

drsabz101 · Jun 5, 2016

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

how do you solve these questions?

InteGrand · Jun 5, 2016

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

drsabz101 said:
how do you solve these questions?
View attachment 33239

Inspection.

$\noindent (a) $T_n = T_{n-1} + 5$$

$\noindent (b) $T_n = 2T_{n-1}$$

$\noindent (c) $T_n = T_{n-1} -7$$

$\noindent (d) $T_n = -T_{n-1}$$

drsabz101 · Jun 5, 2016

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

okay

Cambridge Prelim MX1 Textbook Marathon/Q&A (7 Viewers)

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Member

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 7)