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

drsabz101 · Feb 17, 2016

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

so does it come to :

-6-a=1/x+a

InteGrand · Feb 17, 2016

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

drsabz101 said:
so does it come to :

-6-a=1/x+a

$\noindent Observe that $y^\prime =-\frac{1}{\left(x+a\right)^2}$. Since $y^\prime =-1$ when $x=6$, we have $-1 = -\frac{1}{\left(6+a\right)^2}$. This is the equation to solve.$

davidgoes4wce · Feb 18, 2016

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

drsabz101 said:
Okay Thankyou. Also , for this question 5.a, I first differentiated the equation using the chain rule, to find the equation of the gradient. The because it said it was at x=1., I substituted in x=1 into the gradient getting points (1,20) . then used the formula for the equation of a tangent. i don't know where I went wrong, as my answer is incorrect.

View attachment 32916

$This would have been my solution step-by-step:$

$By substituting x=1 into the equation y=(5x-4)^3$, $ one can find the coordinate of 'y', y=[5(1)-4]^3=1$

$y=(5x-4)^4, \ \frac{dy}{dx}=4(5x-4)^3 (5)=20(5x-4)^3$

$$At x=1, one can find the tangent at that particular point, $ \frac{dy}{dx}=20[5(1)-4]^3=20$

$$Point Gradient Formula is then used$ , y-y_{1}=m(x-x_{1}) \textcircled{1}$

$$Given that we know the coordinate (x,y)=(1,1) by substituting this into \textcircled{1} we can find the linear equation at the tangent. y-1=20(x-1) \implies y=20x-19$

$Using the perpendicular rule of $ m_{1} \times m_{2}=-1, $ given that we know the slope of the tangent is $ m_{1}=20, $ we can then find the gradient of the normal, m_{2}.$

$m_{1} \times m_{2}= -1 , 20 \times m_{2}=-1. Hence, m_{2}=\frac{-1}{20}$

$$ Again using the Point Gradient Formula one can find the linear equation at the normal, but we now use $ m_{2}=\frac{-1}{20} $ as our gradient. $ ( $ use equation $ \textcircled{1})$

$y-1=\frac{-1}{20}(x-1) \implies x+20y=21$

davidgoes4wce · Feb 18, 2016

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

drsabz101 said:
Can someone outline the steps I need to take to solve this :

View attachment 32917

This is my solution
$Q6a$
$\frac{dy}{dx}=3(4x-7)^2 (4)=12(4x-7)^2$

$$ Gradient parallel to $ y=108x+7, is 108$

$Equate \frac{dy}{dx}=108 $ to the derivative $ 12(4x-7)^2$

$108=12(4x-7)^2, $ solve for x$$

$9=(4x-7)^2$

$\pm \sqrt{9}=(4x-7)$

$\pm 3=4x-7$

$x=\frac{5}{2} $ and $ x=1$

$Q6b$

$Rearrange to get 'y' as the subject$

$12y=-x-6 \implies y=\frac{-x-6}{12}$

$$ Using the perpendicular rule to solve for $ m_{2}$

$m_{1} \times m_{2} =-1$

$\frac{-1}{12} \times m_{2}=-1$

$\therefore m_{2}=12$

$Equate \frac{dy}{dx}=12 , 12=12(4x-7)^2$

$\therefore x=2, 1.5$

drsabz101 · Feb 18, 2016

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

Can you show how to do this question, I am assuming you use parametric differentiation?

drsabz101 · Feb 18, 2016

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

No need got the answer

drsabz101 · Feb 18, 2016

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

Need help with this:

davidgoes4wce · Feb 18, 2016

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

drsabz101 said:
Need help with this: View attachment 32921

$Question 9a$

$y=\frac{1}{x-4}=(x-4)^{-1}$

$\frac{dy}{dx}= -1(x-4)^{-2} = \frac{-1}{(x-4)^2}$

$At x=b, y=\frac{1}{b-4}. $ Point L is the coordinates $(b, \frac{1}{b-4})$

$At x=b, \frac{dy}{dx}=\frac{-1}{(b-4)^2}$

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

$$ By substituting the points of L into y $ - \frac{1}{(b-4)}=\frac{-1}{(b-4)^2} (x-b)$

$y- \frac{1}{b-4} = \frac{-x}{(b-4)^2}+\frac{b}{(b-4)^2}$

$\frac{y(b-4)^2}{(b-4)^2}-\frac{(b-4)}{(b-4)^2}=\frac{-x}{(b-4)^2}+\frac{b}{(b-4)^2}$

$y(b-4)^2-(b-4)=-x+b$

$\therefore x+y(b-4)^2=2b-4$

davidgoes4wce · Feb 18, 2016

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

rand_althor said:
$\begin{align*}x+y(b-4)^2&=2b-4 \\\textup{Tangent through the origin:} \\\textup{Sub in }x&=0\, ,y=0 \\0+0(b-4)^2&=2b-4 \\2b&=4 \\b&=2 \\\textup{When }b=2 &\textup{ the tangent equation is:} \\x+y(2-4)^2&=2(2)-4 \\x+4y&=0 \\\textup{Tangent through the W(6,0):} \\\textup{Sub in }x&=6\, ,y=0\\6+0(b-4)^2&=2b-4\\2b&=10\\b&=5 \\\textup{When }b=5 &\textup{ the tangent equation is:} \\x+y(5-4)^2&=2(5)-4 \\x+y&=6 \\\end{align*}$

^^ This guy deserves the credit for Part 9b (i) and (ii)
I did part (a) my method though.

drsabz101 · Feb 18, 2016

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

Help solving this:

davidgoes4wce · Feb 18, 2016

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

drsabz101 said:
Help solving this:
View attachment 32927

$Q11(a) \frac{dy}{dx}=2a(x+b)$

$At \ P(4,8), $The Gradient is 2 (from the equation y=2x)$

$At x=4, \frac{dy}{dx}=2$

$$ Substitute (4,8) into $ y=a(x+b)^2-8 \implies 16=a(4+b)^2 $ which can be rewritten as 16=[a(4+b)](4+b) $ \textcircled{1}$

$Substitute \ x=4 , \frac{dy}{dx}=2 $ into the derivative $ \frac{dy}{dx}=2a(x+b) \implies 2=2a(4+b) $ which can be rewritten as $ 1=a(4+b) \textcircled{2}$

$Substitute \ \textcircled{2} \ into \textcircled{1} \implies 16=1(4+b) \implies 16=4+b \therefore b=12 $ and by subbing it back into either $ \textcircled{1} or \textcircled{2} a=\frac{1}{16}$

davidgoes4wce · Feb 18, 2016

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

Q11b

$y=a(x+b)^2-8$

$\frac{dy}{dx}=2a(x+b)$

$$ The derivative of the common tangent $ y=2-x^2 \ at \ x=1, \frac{dy}{dx}=-2x=-2(1)=-2$

$$ Substituting (1,1) into the equation $ y=a(x+b)^2-8 $ gives $ 1=a(1+b)^2-8 \implies 9=a(1+b)^2 $ which could be rewritten as $ 9=a(1+b)(1+b) \ \textcircled{1}$

$$ Substituting (1,1) into the derivative function $ \frac{dy}{dx}=2a(x+b) \implies -2=2a(1+b) . $ Simplifying gives $ -1=a(1+b) \ \textcircled{2}$

$Substituting \ \textcircled{2} into \textcircled{1} 9=[a(1+b)](1+b) \implies 9=-1(1+b)$

$9=-1-b \therefore b=-10. $ Subbing 'b' back into either $ \textcircled{1} or \textcircled{2} gives \ a=\frac{1}{9}$

Blitz_N7 · Feb 20, 2016

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

I need help on how to evaluate:

drsabz101 · Feb 20, 2016

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

from q11, where did u get y=2x+1 from?

InteGrand · Feb 20, 2016

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

Blitz_N7 said:
I need help on how to evaluate:

$\noindent Observe that the summand is identically equal to $\frac{\frac{1}{2}}{r} + \frac{-1}{r+1} + \frac{\frac{1}{2}}{r+2}$, and use telescoping.$

InteGrand · Feb 20, 2016

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

drsabz101 said:
from q11, where did u get y=2x+1 from?

$\noindent He meant $y=2x$.$

davidgoes4wce · Feb 20, 2016

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

drsabz101 said:
from q11, where did u get y=2x+1 from?

It should have been y=2x, the gradient is still 2.

davidgoes4wce · Feb 20, 2016

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

Fixed

drsabz101 · Feb 20, 2016

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

how does this happen (sorry I am still unsure of what occurred in this step in q11)

davidgoes4wce · Feb 20, 2016

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

drsabz101 said:
how does this happen (sorry I am still unsure of what occurred in this step in q11) View attachment 32929

Basically , look at the common link between the 1st equation and 2nd equation. Here, I have highlighted it in pink to show how the substituting occurs. (in pink. Find out what they both have in common, that's why the 1 is replaced)

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