Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

davidgoes4wce · 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

$y=-\sqrt{a^2-b^2x^2}$

$Convert to indices notation$

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

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

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

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

drsabz101 · Feb 16, 2016

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

So the original equation?

InteGrand · Feb 16, 2016

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

drsabz101 said:
So the original equation?

Yeah

drsabz101 · Feb 16, 2016

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

Okay. You know for that second question, b) one of the coordinates is (2 , 256). I got whe I differentiated the equation :

dy/dx= 4(x^2-4x)^3

I made dy/dx = 0 then equation the equations, but I don't know how to get x=2?

davidgoes4wce · 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

Going on from what Integrand said

$y=(x^2-1)^3$
$\frac{dy}{dx}=3(x^2-1)^2(2x)$
$= 6x(x^2-1)^2$
$\frac{dy}{dx}=0$
$0=6x(x^2-1)^2$

$Solving for x$

$x=0,x=1,x=-1$

$Substitute $ x=0 into $ y=(x^2-1)^3$

$y=[(0)^2-1]^3=-1 . $One possible coordinate where the tangent is horizontal is (0,-1)$$
$y=[(1)^2-1]^3=0. $A second coordinate where the tangent is horizontal is (1,0) $$
$y=[(-1)^2-1]^3=0. $The final coordinate where the tangent is horizontal is (-1,0)$$

InteGrand · Feb 16, 2016

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

drsabz101 said:
Okay. You know for that second question, b) one of the coordinates is (2 , 256). I got whe I differentiated the equation :

dy/dx= 4(x^2-4x)^3

I made dy/dx = 0 then equation the equations, but I don't know how to get x=2?

$\noindent We have $4(x^2-4x)^3=0 \Longleftrightarrow x^2 - 4x = 0 \Longleftrightarrow x(x-4)=0$, so $x=0$ or $4$. So you must have an incorrect expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ (we need to multiply the expression you obtained by $(2x-4)$, which is the derivative of $\left(x^2 - 4x\right)$ due to the chain rule. From here, we see $x=2$ appears as a solution).$

drsabz101 · Feb 16, 2016

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

Yes, I know where I went wrong

drsabz101 · Feb 16, 2016

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

I forgot to complete the chain rule

davidgoes4wce · Feb 16, 2016

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

Q4b

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

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

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

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

$0=4(2x-4)(x^2-4x)^3$
$0=4(2x-4)[x(x-4)]^3$
$x=0,2,4 .$ Substitute these values back into the equation$ y=(x^2-4x)^4$

$y=[(0)^2-4(0)]^4=0.$ First coordinate which is a tangent is$ (0,0)$

$y=[2^2-4(2)]^4=256. $Second coordinate which is a tangent is (2,256)$$

$y=[4^2-4(4)]^4=0. $Final coordinate which is a tangent is (4,0)$$

drsabz101 · Feb 16, 2016

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

Thankyou

Also for d) , how come there an no intercepts?

InteGrand · Feb 16, 2016

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

drsabz101 said:
Thankyou

Also for d) , how come there an no intercepts?

Can't find (d) in your picture: http://community.boredofstudies.org...-cambridge-question-answer-thread-capture.png

drsabz101 · Feb 16, 2016

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

drsabz101 · Feb 16, 2016

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

It just continues from the previous questions

jathu123 · Feb 16, 2016

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

D is y=1/(5x+2)

davidgoes4wce · Feb 16, 2016

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

$y=\frac{1}{5x+2}$

$y=(5x+2)^{-1}$

$\frac{dy}{dx}=-1(5x+2)^{-2} (5)$

$\frac{dy}{dx}=\frac{-5}{(5x+2)^2}$

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

$If your cross multiply the $ (5x+2)^2 $ with 0 , there are no 'x' coefficients remaining. Hence, there are no points at which the tangent of the curve is horizontal$

$0=\frac{-5}{(5x+2)^2}$

InteGrand · Feb 16, 2016

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

drsabz101 said:
View attachment 32913

$\noindent There are no $x$-intercepts because $y$ can't equal 0, as it is equal to the reciprocal of something without singularities.$

drsabz101 · Feb 16, 2016

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

Okay thankyou guys!

davidgoes4wce · 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

Q4e.

$y=24-7(x-5)^2$

$\frac{dy}{dx}=-14(x-5)$

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

$0=-14(x-5)$
$x=5$

$$Substitute x=5 back into$ y=24-7(x-5)^2$

$y=24-7(5-5)^2 =24$

$\therefore (5,24)$ is a coordinate where the tangent is horizontal $$

drsabz101 · Feb 16, 2016

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

Yup got this answer , but thanks for the help !

davidgoes4wce · 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

Q4f. y=4+(x-5)^6

$\frac{dy}{dx} =6(x-5)^5$

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

$0=6(x-5)^5$

$x=5$

$Substitute x=5 back into the equation y=4+(5-5)^6 = 4$

$\therefore (5,4) $is a coordinate where the tangent is horizontal to the curve. $$

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