Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

InteGrand · Jan 10, 2016

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

appleibeats said:
Find the derivative of in the simplest form :

sin^-1 1/4 . ( 2x - 3)

Answer says 2/ (7 + 12x - 4x^2)^1/2

Not sure what to do with the 1/4
Do you multiply it with the numbers in the brackets??

$We have$

$$ $\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left( \sin^{-1} \left(\frac{2x-3}{4}\right) \right)&= \frac{1}{\sqrt{16-\left(2x-3\right)^2}}\times 2, \end{align*}$

$\noindent using the chain rule and the rule for differentiating $\sin^{-1} \left(\frac{x}{a}\right)$. Now just simplify that expression to get to the answer.$

appleibeats · Jan 10, 2016

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

I have tried the chain rule and I am not getting your result above.

I let u = (2x - 3) / 4 and did the usual du/dx and dy/du

InteGrand · Jan 10, 2016

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

appleibeats said:
I have tried the chain rule and I am not getting your result above.

I let u = (2x - 3) / 4 and did the usual du/dx and dy/du

$\noindent To get my result, it's better to let $u=2x-3$ instead. :-)$

InteGrand · Jan 10, 2016

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

appleibeats said:
I have tried the chain rule and I am not getting your result above.

I let u = (2x - 3) / 4 and did the usual du/dx and dy/du

This'll work too, it's just a bit more algebraically tedious when simplifying it.

appleibeats · Jan 10, 2016

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

Still very confused. I still am not getting the result. Also where does the 1/4 go now?

InteGrand · Jan 10, 2016

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

appleibeats said:
Still very confused. I still am not getting the result. Also where does the 1/4 go now?

$\noindent Let $y=\sin^{-1} \left(\frac{2x-3}{4} \right)$. We want to find $\frac{\mathrm{d} y}{\mathrm{d} x}$. Let $u=2x-3$, so $\frac{\mathrm{d} u}{\mathrm{d} x}=2$. Also, $y=\sin^{-1} \left(\frac{u}{4}\right)\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} u}=\frac{1}{\sqrt{16-u^2}}$. So we have$

$\begin{align*}\frac{\mathrm{d} y}{\mathrm{d} x}&= \frac{\mathrm{d} y}{\mathrm{d} u}\cdot \frac{\mathrm{d} u}{\mathrm{d} x} \\ &= \frac{1}{\sqrt{16-u^2}} \times 2 \\ &= \frac{2}{\sqrt{16-\left(2x-3 \right)^2}} \\ &= \frac{2}{\sqrt{16 - \left(4x^2 - 12 x + 9 \right)}} \\ &= \frac{2}{\sqrt{7 + 12x - 4x^2}}.\end{align*}$

InteGrand · Jan 10, 2016

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

$\noindent \textbf{Note.} We used the result that $\frac{\mathrm{d}}{\mathrm{d}u}\left(\sin ^{-1} \left(\frac{u}{a} \right) \right)=\frac{1}{\sqrt{a^2 - u^2}}$ (where $a>0$ is a constant). This result is a standard HSC 3U differentiation formula and should be in your textbook.$

appleibeats · Jan 10, 2016

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

I don't know that formula. I tried searching for it in the cambridge 3u textbook and can't find it.
If I don't use the formula, is there another method to solving the question?

InteGrand · Jan 10, 2016

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

appleibeats said:
I don't know that formula. I tried searching for it in the cambridge 3u textbook and can't find it.
If I don't use the formula, is there another method to solving the question?

$\noindent That formula is found on the standard integrals sheet of HSC papers $\Big{(}$written in the form $\int \frac{1}{\sqrt{a^2-x^2}}\text{ d}x =\sin ^{-1}\left(\frac{x}{a}\right)\Big{)}$. It should be in the Year 12 3U Pender (Cambridge) textbook.$

leehuan · Jan 10, 2016

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

The sine inverse integral is also on the new formula sheet anyway. http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths-ref-sheet.pdf

And it is expected that you know that the derivative and the integral are inverse operations.

InteGrand · Jan 10, 2016

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

appleibeats said:
If I don't use the formula, is there another method to solving the question?

Yeah, definitely. You can do it with your original substitution too, it's just more tedious when simplifying it. Maybe you made an algebraic mistake when doing it, so you might want to post your working.

appleibeats · Jan 10, 2016

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

Finally worked it out.

Now stuck on the Show part of part c) in this question. Not sure how to do the branching of it.

Screen Shot 2016-01-10 at 7.24.26 pm.png

InteGrand · Jan 10, 2016

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

appleibeats said:
Finally worked it out.

Now stuck on the Show part of part c) in this question. Not sure how to do the branching of it.

View attachment 32779

$\noindent Since the derivative is 0 for all $x>0$, the function is some constant for $x >0$. Sub. in a convenient value of $x$ ($x=1$) to find this constant. Do $x=-1$ to find the constant value for negative $x$.$

appleibeats · Jan 10, 2016

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

Screen Shot 2016-01-10 at 10.30.22 pm.png

Could you explain how do find the domain. part a)

InteGrand · Jan 10, 2016

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

appleibeats said:
View attachment 32780

Could you explain how do find the domain. part a)

$\noindent The domain is (the set of all $x$ such that) $\boxed{x\leq -1\text{ or }x\geq 1}$. We have $y=\cos ^{-1}\frac{1}{x}$, so for the domain, we must have $-1\leq \frac{1}{x}\leq 1$, since the input to an inverse cosine is always between $-1$ and $1$. In other words, $\left | \frac{1}{x}\right | \leq 1 \Longleftrightarrow \frac{1}{|x|}\leq 1 \Longleftrightarrow |x| \geq 1$. So we must have $x\geq 1$ or $x\leq -1$ for the domain. In set notation, the domain is $\{x\in \mathbb{R}:x\geq 1 \text{ or }x\leq -1 \}$.$

appleibeats · Jan 11, 2016

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

For part b), I get the answer they have in the question except for no absolute sign around the x in the denominator. Why is that? I used chain rule to differentiate the fn.

leehuan · Jan 11, 2016

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

$\\f\left( x \right) =\cos ^{ -1 }{ \frac { 1 }{ x } } \\ f^{ \prime }\left( x \right) =-\frac { 1 }{ { x }^{ 2 } } \cdot \frac { -1 }{ \sqrt { 1-{ \left( \frac { 1 }{ x } \right) }^{ 2 } } } \\ f^{ \prime }\left( x \right) =\frac { 1 }{ { x }^{ 2 }\sqrt { \frac { 1 }{ { x }^{ 2 } } \left( { x }^{ 2 }-1 \right) } } \\ f^{ \prime }\left( x \right) =\frac { 1 }{ { x }^{ 2 }\left| \frac { 1 }{ x } \right| \sqrt { { x }^{ 2 }-1 } } \\ f^{ \prime }\left( x \right) =\frac { 1 }{ \left| x \right| \sqrt { { x }^{ 2 }-1 } }$

The fact that x^2/|x| = |x| can be determined by simply realising that |x|^2=x^2

Extra fact: That function is actually f(x)=sec^-1(x)

appleibeats · Jan 11, 2016

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

Screen Shot 2016-01-11 at 2.29.49 pm.png

Don't know how to do part b)

Answer is 1 - 1/2 (3)^1/2

I have A = integral from 0 to pi/6 sin^2 y du

Is that correct?

then I got integral from 0 to pi/6 1/2 (1 - cos2y) dy

I further continued on but got an incorrect answer.

InteGrand · Jan 11, 2016

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

appleibeats said:
View attachment 32781

Don't know how to do part b)

Answer is 1 - 1/2 (3)^1/2

I have A = integral from 0 to pi/6 sin^2 y du

Is that correct?

then I got integral from 0 to pi/6 1/2 (1 - cos2y) dy

I further continued on but got an incorrect answer.

$\noindent The area is given by $A = \int _0 ^{\frac{\pi}{6}} \sin y\text{ d}y$.$

appleibeats · Jan 11, 2016

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

Screen Shot 2016-01-11 at 4.23.45 pm.png

Having trouble with part b

Using part a), I took the integral of both side and I got this

Integral of tan^-1 x dx = xtan^-1 x - Integral of x/(1 + x^2) dx

So I need to find the integral of x/(1 + x^2)

Is this the correct method??

Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

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