Cambridge Prelim MX1 Textbook Marathon/Q&A (8 Viewers)

davidgoes4wce · Feb 27, 2016

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

Drongoski said:
$(x-2)^3$ ??

$The (x-2) term cancels with the term inside the square bracket as$

$\frac{(x-2)^4}{x-2}= (x-2)^3$

AfroNerd · Feb 27, 2016

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

Solve:
a) 12x^3 - 4x^2 -3x + 1 = 0 if the sum of two of its roots is 0
b) 6x^4 + 5x^3 -24x^2 -15x + 18 = 0 if the sum of two of its roots is 0

davidgoes4wce · Feb 27, 2016

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

$CUBIC EQUATION- SUM AND PRODUCT OF ROOTS$

$The general cubic equation ax^3+bx^2+cx+d=0$

$\alpha + \beta +\gamma =\frac{-b}{a} \ \textcircled{1}$

$\alpha \beta + \alpha \gamma + \beta \gamma =\frac{c}{a} \ \textcircled{2}$

$\alpha \beta \gamma =\frac{-d}{a}$

$$ Roots for this question are : $ \alpha, -\alpha \ and \ \beta$

$$ Using Formula $ \textcircled{1}, \ \alpha+(\alpha)+\beta=-\frac{-4}{12} = - \frac{-1}{3}=\frac{1}{3}$

$$ Using Formula $ \textcircled{2}. \ \alpha(-\alpha)+\alpha \ \beta - \alpha \ \beta=\frac{-1}{4}$

$-\alpha^{2}=\frac{-1}{4} . \therefore \alpha =\pm\frac{1}{2}$

$\therefore $ the 3 roots are x $=\frac{1}{2}, \frac{-1}{2} $ and $ \frac{1}{3}$

davidgoes4wce · Feb 27, 2016

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

AfroNerd said:
Solve:
a) 12x^3 - 4x^2 -3x + 1 = 0 if the sum of two of its roots is 0
b) 6x^4 + 5x^3 -24x^2 -15x + 18 = 0 if the sum of two of its roots is 0

$With the Quartic Equation its a bit different- SUM AND PRODUCT OF ROOTS$

$The general quartic equation is ax^4+bx^3+cx^2+dx+e=0$

$\alpha + \beta +\gamma +\delta =\frac{-b}{a}$

$\alpha \beta+\alpha \gamma+\alpha \delta +\beta \gamma + \beta \delta +\gamma \delta =\frac{c}{a}$

$\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta +\beta \gamma \delta =\frac{-d}{a}$

$\alpha \beta \gamma \delta = \frac{e}{a}$

$$ Roots are:$ \alpha, -\alpha, \beta, \gamma$

$\alpha+(-\alpha)+\beta+\gamma=\beta+\gamma=\frac{-5}{6} \ \textcircled{1}$

$\beta+\gamma=\frac{-5}{6} \textcircled{1}$

$-\alpha^{2}+\alpha \beta +\alpha \gamma-\alpha \beta- \alpha \gamma +\beta \gamma=-4 \textcircled{2}$

$$ By simplifying, -\alpha^{2}+\beta \gamma =-4 \textcircled{2}$

$-\alpha^{2} \beta - \alpha^{2} \gamma +\alpha \beta \gamma - \alpha \beta \gamma =\frac{5}{2}$

$$ By simplifying, $ -\alpha^{2} \beta - \alpha^{2} \gamma =-\alpha^{2}(\beta+\gamma)= \frac{5}{2} \textcircled{3}$

$-\alpha^{2} \beta \gamma =3 \textcircled{4}$

$$ By rearranging $ \textcircled{3}, $ we have $ -\alpha^{2} (\beta+\gamma)=\frac{5}{2}$

$$ Given from $ \textcircled{1}, \beta + \gamma =\frac{-5}{6} , $ we substitute this into $ \textcircled{3}$

$-\alpha^{2} (\frac{-5}{6})=\frac{5}{2}$

$\alpha=\sqrt{3}, -\sqrt{3}$

$$ Substitute either $ \alpha =\sqrt{3} $ or $ -\sqrt{3} $ into $ \textcircled{3} , -(\sqrt{3})^2 \beta \gamma=3$

$\beta \gamma =-1 . $ By rearranging, $ \beta=\frac{-1}{\gamma}. $ Substitute back into $ \textcircled{1}$

$\frac{-1}{\gamma}+\gamma=\frac{-5}{6}$

$\frac{-1}{\gamma}+\frac{\gamma^{2}}{\gamma}=\frac{-5\gamma}{6}$

$$ Equate denominator terms, $ \frac{-6+6\gamma^{2}}{6\gamma}=\frac{-5\gamma}{6}$

$6\gamma^{2}+5\gamma-6=0$

$$ Solve like a quadratic ,$ \gamma=\frac{2}{3} $ or $ \gamma=\frac{-3}{2}$

$$ Hence, the x roots are: $ \frac{-3}{2}, \frac{2}{3}, \sqrt{3} $ and $ -\sqrt{3}$

drsabz101 · Feb 27, 2016

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

Can explain the ans for part B.

drsabz101 · Feb 27, 2016

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

Also how do I solve this questions?

InteGrand · Feb 27, 2016

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

drsabz101 said:
Also how do I solve this questions?
View attachment 32953

For part (a), you can use parametric differentiation. For part (b), do it as they say to do it.

drsabz101 · Feb 27, 2016

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

ohh okay thankyou

drsabz101 · Feb 27, 2016

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

This is my working out so far to part B. I don't know where I went wrong

Question:

My working out:

davidgoes4wce · Feb 27, 2016

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

drsabz101 said:
Can explain the ans for part B.

View attachment 32952

$Q7 (b), Given \frac{dy}{dx}= \frac{\alpha-\beta}{(x-\beta)^2}$

$When \frac{dy}{dx}>0,$ this represents a positive Gradient. When $ \frac{dy}{dx}<0 , $ this represents a negative Gradient. The denominator is positive, since it is squared, so the sign of y' is $ \alpha-\beta$

$$ In effectively, we could treat the derivative sign as $ \frac{dy}{dx}= \alpha-\beta$

$If $ \alpha-\beta>0, \alpha>\beta, $ represents a positive gradient.$

$If $ \alpha-\beta<0, \alpha<\beta, $ represents a negative gradient.$

$$ Q7 (c), when the derivative is equal to 0, the curve has a gradient of 0 , $ \frac{dy}{dx}=0. \ Hence, 0=\alpha-\beta , \alpha=\beta$

drsabz101 · Feb 27, 2016

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

thankyou!

davidgoes4wce · Feb 27, 2016

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

drsabz101 said:
Also how do I solve this questions?
View attachment 32953

$$ Part 8 (a), Use parametric differentiation, $ \frac{dy}{dx}=\frac{\frac{dy}{dy}}{\frac{dx}{dt}}$

$$ Use quotient rule ,$ \frac{dx}{dt}=\frac{(t+1)(1)-(t)(1)}{(t+1)^2}=\frac{1}{(t+1)^2}$

$$ Determine the derivative of $ \frac{dy}{dt} = \frac{(t-1)(1)-(t)(1)}{(t-1)^2}= \frac{-1}{(t-1)^2}$

$\frac{dy}{dx}=\frac{-1}{(t-1)^2} \div \frac{1}{(t+1)^2}$

$\frac{dy}{dx}= \frac{-1}{(t-1)^2} \times \frac{(t+1)^2}{1}=\frac{-(t+1)^2}{(t-1)^2}$

$$ At t=2 $, x=\frac{2}{3}, y=2, \frac{dy}{dx}=\frac{-(2+1)^2}{(2-1)^2}=-9$

$$ Use the Perpendicular Rule in order to determine the gradient of the normal m_{1} \times m_{2}=-1$

$m_{1} \times m_{2}=-1 \implies -9 \times m_{2}= -1, m_{2} = \frac{1}{9}$

$$ Use the Point Gradient Formula , $ y-y_{1}=m(x-x_{1})$

$y-2=\frac{1}{9}(x-\frac{2}{3})$

$By rearranging, into standard form , 3x-27y+52=0$

$Part 8 (b) , for both $ x=\frac{t}{t+1} $ and $ y=\frac{t}{t-1} $ the key here is to get the term 't' as the subject, by cross multiplying the denominator for both fractions$

$x(t+1)=t \implies xt+x=t \implies xt-t=-x \implies t(x-1)=-x \implies t=\frac{-x}{x-1} \textcircled{1}$

$$ Doing the same for the y-term, y(t-1)=t $ \implies yt-y=t \implies yt-t=y \implies t(y-1)=y \implies t=\frac{y}{y-1} \textcircled{2}$

$$ Equate equations $ \textcircled{1} \ and \ \textcircled{2}$

$\frac{-x}{x-1}=\frac{y}{y-1}$

$-x(y-1)=y(x-1)$

$-xy+x=xy-y$

$x=2xy-y \implies x=y(2x-1) \implies y=\frac{x}{2x-1}$

$$ Use the quotient rule to determine the derivative, $ \frac{dy}{dx}= \frac{(2x-1)(1)-(x)(2)}{(2x-1)^2} =\frac{-1}{(2x-1)^2}$

$At x=\frac{2}{3} , \frac{dy}{dx} = \frac{-1}{(2(\frac{2}{3})-1)^2}= -9$

$m_{1} \times m_{2}=-1 \implies -9 \times m_{2}=-1 . \therefore m_{2}=\frac{1}{9}$

davidgoes4wce · Feb 27, 2016

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

drsabz101 said:
This is my working out so far to part B. I don't know where I went wrong

Question:
View attachment 32954

My working out:
View attachment 32955

$Exercise 7G, Q10 (b)$

$Your writing in the last part is a bit small, so I couldn't see it too clearly. Because I have nothing else to do on a Saturday night, I decided to redo the whole question again.$

$\frac{dy}{dx}= \frac{(x+1)(1)-(x)(1)}{(x+1)^2} = \frac{1}{(x+1)^2}$

$$ At x=a, $ y=\frac{a}{(a+1)}$

$$ Use the Point Gradient Formula, y-y_{1}=m(x-x_{1})$

$y-\frac{a}{a+1} =\frac{1}{(a+1)^2} (x-a)$

$y-\frac{a}{a+1}= \frac{x}{(a+1)^2}-\frac{a}{(a+1)^2}$

$\frac{y(a+1)^2-a(a+1)}{(a+1)^2} = \frac{x-a}{(a+1)^2}$

$Concentrate on Numerator terms, since the denominators are common$

$y(a+1)^2-a(a+1)=x-a$

$y(a+1)^2-a^2-a+a=x$

$y(a+1)^2-a^2=x$

$0=x-(a+1)^2y+a^2$

davidgoes4wce · Feb 27, 2016

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

drsabz101 said:
This is my working out so far to part B. I don't know where I went wrong

Question:
View attachment 32954

My working out:
View attachment 32955

$Exercise 7G, Q10 (a)$

$$$\lim_{x \to \infty} \frac{x}{x+1}=1$

$$$\lim_{x \to -\infty} \frac{x}{x+1}=1$

$\therefore $ Horizontal asymptote at y=1 $$

$Vertical Asymptote by looking at the denominator term and equating that to zero.$

$x+1=0 \implies $ x=-1 is a vertical asymptote $$

$Domain:{{x \in \mathbb{R}, x\neq{-1}}$

$Range:{{y \in \mathbb{R}, y\neq{1}}$

drsabz101 · Feb 28, 2016

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

I realised for y=a/(a+1)^2, how is that, shouldn't it be y=1/(a+1)^2

davidgoes4wce · Feb 28, 2016

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

drsabz101 said:
I realised for y=a/(a+1)^2, how is that, shouldn't it be y=1/(a+1)^2

$Its a coordinate, if you substitute x=a, we want to find the y-coordinate of y at that Point.$

$$ Hence, the coordinate or point at which the tangent goes through that particular point is $ (x,y)=(a, \frac{a}{a+1})$

davidgoes4wce · Feb 28, 2016

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

$$ Given that we know the point $ x_{1}=a, y_{1}=(\frac{a}{a+1}), m= \frac{1}{(x+1)^2} , $ we have enough in order to determine out Linear equation $ y-y_{1}=m(x-x_{1})$

Nailgun · Feb 28, 2016

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

davidgoes4wce said:
$Its a coordinate, if you substitute x=a, we want to find the y-coordinate of y at that Point.$

$$ Hence, the coordinate or point at which the tangent goes through that particular point is $ (x,y)=(a, \frac{a}{a+1})$

...I think she meant the mistake that you just edited out ahahah

EDIT:

davidgoes4wce said:
$Exercise 7G, Q10 (b)$
$$ At x=a, $ y=\frac{a}{(a+1)^2}$

You didn't carry it over to your working though so it made no difference
but like, don't just speed past it like it never happened ahaha

davidgoes4wce · Feb 28, 2016

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

Nailgun said:
...I think she meant the mistake that you just edited out ahahah

EDIT:

You didn't carry it over to your working though so it made no difference
but like, don't just speed past it like it never happened ahaha

Yep , it was a typo on my behalf.

drsabz101 · Feb 28, 2016

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

What steps do I take to solve this?

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