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

davidgoes4wce · Feb 16, 2016

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

drsabz101 said:
Yup got this answer , but thanks for the help !

I thought you wanted help with all of them?

drsabz101 · Feb 16, 2016

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

How does this give x=0?, as a coordinate I have differentiated this correctly

davidgoes4wce · Feb 16, 2016

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

Anyway here is the last one,

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

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

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

$\frac{dy}{dx}=0$

$0=\frac{-2x}{(1+x^2)}$

$0=-2x$

$x=0$

$Substitute x=0, y=\frac{1}{(1+(0)^2}=1$

$\therefore (0,1) i$s a coordinate where the tangent is horizontal.$$

davidgoes4wce · Feb 16, 2016

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

drsabz101 said:
How does this give x=0?, as a coordinate I have differentiated this correctly
View attachment 32914

$When you cross multiply 0 with$ (1+x^2), $it eliminates the denominator term$

drsabz101 · Feb 17, 2016

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

Can someone explain why the answer is no coordinates , since x=1 is out of the domain. how did they know this? In relation to question 4 f

InteGrand · Feb 17, 2016

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

drsabz101 said:
Can someone explain why the answer is no coordinates , since x=1 is out of the domain. how did they know this? In relation to question 4 f
View attachment 32915

$\noindent Are you sure you mean 4 (f)? That is a sextic polynomial that looks basically like a steep parabola. It has a stationary point (point with horizontal tangent) at $x=5$.$

drsabz101 · Feb 17, 2016

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

Sorry Q.4K

InteGrand · Feb 17, 2016

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

drsabz101 said:
Sorry Q.4K

$\noindent For this one, $y=\sqrt{x^2 - 2x}$. This function is defined at points where $x^2-2x \geq 0$, and nowhere else. Since $x^2 - 2x = x(x-2)$, the function is defined where and only where $x\leq 0$ or $x\geq 2$. So $x=1$ is not in the domain.$

drsabz101 · Feb 17, 2016

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

Why is the function defined at points where x^2-2x>0? only.

InteGrand · Feb 17, 2016

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

drsabz101 said:
Why is the function defined at points where x^2-2x>0? only.

$\noindent This is because we don't want negative numbers under the square roots. In general, if $y=\sqrt{h(x)}$, this function is defined (assuming we want a real-valued function with domain a subset of $\mathbb{R}$) at real $x$-values where and only where $h(x) \geq 0$.$

drsabz101 · Feb 17, 2016

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

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.

InteGrand · Feb 17, 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

$\noindent To find the point, we need to sub. $x=1$ into the function (the thing given to us). This tells us the $y$-value for a given $x$-value. So for 5(a), the point in question has $y=(5\times 1-4)^4 = 1$. So the point is $(1,1)$. Subbing $x=1$ into the derivative function will tell us the slope of the tangent to the curve at this point.$

drsabz101 · Feb 17, 2016

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

Oh true! I just re did it and got the answer. thanks heaps

drsabz101 · Feb 17, 2016

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

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

InteGrand · Feb 17, 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

$\noindent We first find the expression for the derivative (this tells us the slope at any point $x$). We then equate this to the slope we want and solve for $x$. For (a), the slope we want is clearly 108. For (b), the given line has slope $-\frac{1}{12}$. Since we want the tangent to be \emph{perpendicular} to this, the slope we equate the derivative to is $12$.$

drsabz101 · Feb 17, 2016

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

When I differentiated it I got:
dy/dx= 12(4x-7)^2

I equated this with 108, and got 192x^2-672x+480=0?

InteGrand · Feb 17, 2016

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

drsabz101 said:
When I differentiated it I got:
dy/dx= 12(4x-7)^2

I equated this with 108, and got 192x^2-672x+480=0?

$\noindent There is a simpler way. We have $12(4x-7)^2= 108$. So $(4x-7)^2 = 9$ (dividing both sides by 12). So $4x-7 = \pm 3$. I think you can do it from here.$

drsabz101 · Feb 17, 2016

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

got the solution

drsabz101 · Feb 17, 2016

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

How do I approach this question after I differentiate it?

InteGrand · Feb 17, 2016

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

drsabz101 said:
How do I approach this question after I differentiate it?
View attachment 32918

$\noindent We know the value of the derivative at a given $x$-value, so sub. in this $x$-value to the derivative expression and equate it to what we're told its value is here. This gives us an equation to solve for $a$.$

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