Cambridge Prelim MX1 Textbook Marathon/Q&A (9 Viewers)

davidgoes4wce · Feb 14, 2016

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

I should have looked at the 'True North' bearing at point C, its much more clearer after adding up the 180+90 (from south to west) and the 34.6 degrees .

davidgoes4wce · Feb 14, 2016

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

Q7A from 3 Unit Cambridge page 284

Write down the general form of a monic quadratic for which one of the zeroes is x=1. Then find the equation of such a quadratic in which:

(a) The curve passes through the origin.

$$The answer has $ y=(x-\alpha)(x-1)$

$Just wondering if its a typo? should it be y=x(x-1)$

davidgoes4wce · Feb 14, 2016

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

Origin I assume to be going through (0,0)

drsabz101 · Feb 15, 2016

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

Can someone please help solve these two questions , my answers are close but still incorrect:

porcupinetree · Feb 15, 2016

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

drsabz101 said:
Can someone please help solve these two questions , my answers are close but still incorrect:

View attachment 32907

View attachment 32908

Will write up solutions now

porcupinetree · Feb 15, 2016

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

drsabz101 said:
View attachment 32907

Scanner isn't working atm :/

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porcupinetree · Feb 15, 2016

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

drsabz101 said:
View attachment 32908

Sent from my iPhone using Tapatalk

drsabz101 · Feb 15, 2016

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

Help solving these two:

InteGrand · Feb 15, 2016

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

drsabz101 said:
Help solving these two:
View attachment 32909

$\noindent Here's how to do Q.26. Let $\left(u,v\right)$ be an arbitrary point in the plane. Let the cubic be $f(x) = ax^3 + bx^2 + cx +d$, so $f^\prime (x) = 3ax^2 + 2bx + c$. Hence the slope of the tangent to the cubic at $x=t$ ($t\in \mathbb{R}$) is $3at^2 + 2bt + c$. Since the point on the cubic at $x=t$ is $T\left(t,at^3 + bt^2 + ct +d\right)$, the equation of the tangent to the cubic at $T$ is $y-\left(at^3 + bt^2 + ct +d\right) = 3at^2 + 2bt + c \left(x-t\right)$.$

$\noindent Therefore, the point $(u,v)$ will lie on a tangent to the cubic if and only if there exists some $t\in \mathbb{R}$ such that $v-\left(at^3 + bt^2 + ct +d\right) = 3at^2 + 2bt + c \left(u-t\right)$ holds. But this equation is a cubic equation in $t$, and since any cubic equation has at least one real root (this fact follows from the intermediate value theorem), there is indeed a real $t$ that satisfies that equation, thus proving the claim.$

davidgoes4wce · Feb 15, 2016

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

$Q25 (a)$
$\frac{dy}{dx}=2ax+b$
$\frac{dy}{dx}=m$
$m=2ax+b$
$x=\frac{m-b}{2a}$
$$Substituting it into the equation $y=ax^2+bx+c$
$y=a(\frac{m-b}{2a})^2+b(\frac{m-b}{2a})+c$
$$The coordinates of $ (x_{1},y_{1}) \ are: (\frac{m-b}{2a}, a(\frac{m-b}{2a})^2+b(\frac{m-b}{2a})+c)$

$By substituting $ x=0 (y-intercept), m=2ax+b, x_{1}=\frac{m-b}{2a}, y_{1}=a(\frac{m-b}{2a})^2+b(\frac{m-b}{2a})+c) $into the Point Gradient Formula$

$y-y_{1}=m(x-x_{1})$

$y-a(\frac{m-b}{2a})^2-b(\frac{m-b}{2a})-c=(2ax+b)[0-(\frac{m-b}{2a})]$

$By cancelling similar terms we are left with $ y=c+\frac{(m-b)^2}{4a} $as the y-intercept$

davidgoes4wce · Feb 15, 2016

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

Q25 (b)
$The question states that the quadratic goes through the origin$ (0,0), so for y=ax^2+bx+c, $c must be equal to 0$
$$ Given tangent is $ y=-2x-4, gradient=-2, y-intercept=-4 \ \textcircled{1}$
$$ Given tangent is $ y=8x-49, gradient=8, y-intercept=-49 \ \textcircled{2}$
$Combining$ \textcircled{1} $with the expression found in part (a) $ [y=c-\frac{(m-b)^2}{4a}] $but now we have c=0$

$-4=-\frac{(-2-b)^2}{4a}$

$4a=\frac{(-2-b)^2}{4}$

$4a=\frac{[-1(2+b)]^2}{4}$

$4a=\frac{b+2)^2}{4} \textcircled{3}$

$$Combining$ \textcircled{2} $with the expression obtained in part$ (a)$

$-49=\frac{-(8-b)^2}{4a} \textcircled{4}$

$4a=\frac{(8-b)^2}{49}$

$Using \textcircled{3} and \textcircled{4}$

$\frac{(b+2)^2}{4}=\frac{(8-b)^2}{49}$

$b=\frac{2}{9} or b=-6$

$$By substituting $ b=\frac{2}{9} into \textcircled{3} a= \frac{25}{81}$

$$By substituting $ b=-6 into \textcircled{3} a=1$

$$Hence, the quadratic equations obtained are$: y=x^2-6x $and y$=\frac{25x^2}{81}+\frac{2x}{9}$

davidgoes4wce · Feb 15, 2016

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

Q25 (c)
$$Similarly to part(b) $ y=ax^2+bx+c, \frac{dy}{dx}=2ax+b$

$y=-5x-10, gradient=-5, y-intercept=-10 \textcircled{1}$

$y=-3x-7, gradient=-3, y-intercept=-7 \textcircled{2}$

$y=x-7, gradient=1, y-intercept=-7 \textcircled{3}$

$Combining the \textcircled{1} \textcircled{2} and \textcircled{3} into the expression found in part$ (a) y=c-\frac{(m-b)^2}{4a} $but now c is not equal to 0.$

$-10=c-\frac{(-5-b)^2}{4a} \textcircled{4} \implies 4ac=(-5-b)^2-40a$

$-7=c-\frac{(-3-b)^2}{4a} \textcircled{5} \implies 4ac=(-3-b)^2-28a$

$-7=c-\frac{(1-b)^2}{4a} \textcircled{6}\implies 4ac=(1-b)^2-28a$

$Equating \textcircled{4} and \textcircled{5}$

$(-5-b)^2-40a=(-3-b)^2-28a \implies 16=12a-4b \textcircled{7}$

$Equating \textcircled{4} and \textcircled{6}$

$(-5-b)^2-40a=(1-b)^2-28a \implies 24=12a-12b \textcircled{8}$

$We can then use simultaneous equations \textcircled{7} and \textcircled{8} in order to solve for part (a) and part(b)$

$Using simultaneous equations gives a=1 and b=-1. We can then substitute these values into \textcircled{6}, to solve for 'c'$

$4ac=(1-b)^2-28a , 4(1)c=[1-(-1)]^2-28(1), c=-6. Hence, y=x^2-x-6 $ is the quadratic equation.$$

drsabz101 · Feb 16, 2016

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

How do I differentiate this , using the chain rule , the constants are confusing me :

InteGrand · Feb 16, 2016

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

drsabz101 said:
How do I differentiate this , using the chain rule , the constants are confusing me :
View attachment 32911

$\noindent If it's just the constants that are confusing you, you can write $A = a^2$ and $B = b^2$, and then differentiate. Then at the end, replace $A$ and $B$ with $a^2$ and $b^2$.$

drsabz101 · Feb 16, 2016

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

okay, yeah it's just the constants, I know how to use the formula

drsabz101 · Feb 16, 2016

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

Ahhh! I hate myself for doing this. I copied the question down wrong , no wonder

drsabz101 · Feb 16, 2016

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

I have already differentiated this question a , but what do i do next ?

InteGrand · Feb 16, 2016

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

drsabz101 said:
I have already differentiated this question a , but what do i do next ?
View attachment 32912

$\noindent The tangents being horizontal means that the derivative is 0, so set the derivative equal to 0 and solve. The solutions are the $x$-values where the curve has a horizontal tangent. Plug these $x$-values into the function to obtain the corresponding $y$-values of these points. Then you will have found their coordinates.$

drsabz101 · Feb 16, 2016

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

Do I sub it back into the differentiated equation, to get y?

InteGrand · Feb 16, 2016

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

drsabz101 said:
Do I sub it back into the differentiated equation, to get y?

No, into the equation for y, which is given to us.

Cambridge Prelim MX1 Textbook Marathon/Q&A (9 Viewers)

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