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

InteGrand · Jan 11, 2016

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

appleibeats said:
View attachment 32782

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??

$\noindent Yes, that's correct. Observe that $\int \frac{x}{1+x^2} \text{ d}x = \frac{1}{2}\ln \left(1+x^2\right) + c$, for some arbitrary constant $c$ (remember to evaluate the antiderivative at the endpoints to get the value of the definite integral).$

appleibeats · Jan 11, 2016

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

Screen Shot 2016-01-11 at 5.43.21 pm.png

I understand how to do question a) and b). They are both odd so the area's cancel each other out and so the answer for both are 0.

However I am unsure about question c). It has no symmetry. The answer says 3pi/4

Ambility · Jan 11, 2016

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

appleibeats said:
View attachment 32783

I understand how to do question a) and b). They are both odd so the area's cancel each other out and so the answer for both are 0.

However I am unsure about question c). It has no symmetry. The answer says 3pi/4

Seeing as the inverse function has point symmetry around (0, pi/2), you can evaluate it by moving the area under the curve from -3/4 to 0 above the area under the curve from 0 to 3/4. This results in a rectangle, which has height pi and width 3/4. The total area is therefore 3pi/4.

Let me know if you still don't get it.

leehuan · Jan 11, 2016

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

Ambility said:
Seeing as the inverse function has point symmetry around (0, pi/2), you can evaluate it by moving the area under the curve from -3/4 to 0 above the area under the curve from 0 to 3/4. This results in a rectangle, which has height pi and width 3/4. The total area is therefore 3pi/4.

Let me know if you still don't get it.

After deleting my post and viewing the graph this was what I assumed. However what is the proof that the two areas superimpose to produce an actual rectangle?

I can't visualise it.

InteGrand · Jan 11, 2016

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

leehuan said:
After deleting my post and viewing the graph this was what I assumed. However what is the proof that the two areas superimpose to produce an actual rectangle?

I can't visualise it.

$\noindent Here's how to visualise it. Take that region (call it $\mathcal{R}$) bounded by the curve $y=\cos ^{-1} x$, the $x$-axis, and $x=0$ and $x=-\frac{3}{4}$. Then ``reflect'' this region about the point $\left(0,\frac{\pi}{2} \right)$. Due to point symmetry, the region $\mathcal{R}$ is a figure that is congruent to the region $\mathcal{R}^\prime$ that is the region bounded by the curve, and the lines $y=\pi$, $x=\frac{3}{4}$, and the $y$-axis. So when we ``flip'' $R$, it ends up exactly on $\mathcal{R}^\prime$, and the region $\mathcal{R}^\prime$ together with the area under the curve from $x=0$ to $\frac{3}{4}$ forms a rectangle $\Big{(}$bounded by the lines $y=0$, $y=\pi$, $x=0$ and $x=\frac{3}{4}$$\Big{)}$.$

Ambility · Jan 11, 2016

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

I made a diagram, hopefully this helps for visualising it.

http://imgur.com/E9yDMiq

Edit: Oops, theres a typo in the image. Width, not with.

InteGrand · Jan 11, 2016

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

$(By ``point symmetry'' about the point $\left(0,\frac{\pi}{2}\right)$, we are referring to the fact that the function $f(x)=\cos^{-1}x$ satisfies $f(-x) = \pi - f(x)$ for all $x$ in the domain of $f$.)$

InteGrand · Jan 11, 2016

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

$\noindent In general, a function $y=g(x)$ can have point symmetry about an arbitrary point $P(x_0,y_0)$ $\Big{(}$e.g. all cubic function graphs have point symmetry about their point of inflexion, the graph of $y=\cos^{-1}x$ has point symmetry about $\left(0,\frac{\pi}{2}\right)$, etc.$\Big{)}$. What this means is essentially, if we move transform the function so that the origin became where $P$ is, this transformed function would be an odd function (a.k.a. ``point symmetry about the origin'', as you may have heard in textbooks etc.).$

$\noindent To move the point $P(x_0,y_0)$ to the origin, the transformation to the function $g(x)$ is using $g(x+x_0)-y_0$. This shifts the graph of $y=g(x)$ down by $y_0$ (equivalently, shifts the $x$-axis \textsl{up} by $y_0$) and shifts the graph of $y=g(x)$ left by $x_0$ (equivalently, shifts the $y$-axis \textsl{right} by $y_0$), so that the origin has moved to where $(x_0,y_0)$ was. So for point-symmetry about $(x_0,y_0)$, we mean that $y=h(x):= g(x+x_0)-y_0$ is an odd function, i.e. $g(-x+x_0) - y_0 = - \left(g(x+x_0)-y_0 \right)$ for all $x$ in the domain.$

InteGrand · Jan 11, 2016

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

Here's an intuitive definition of point symmetry with diagrams: http://www.mathsisfun.com/geometry/symmetry-point.html

leehuan · Jan 11, 2016

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

Cool. I only really needed that first post.

InteGrand · Jan 11, 2016

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

leehuan said:
Cool. I only really needed that first post.

Haha yeah, I just added the extra stuff for any curious students and for generalisation purposes.

appleibeats · Jan 12, 2016

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

Screen Shot 2016-01-12 at 10.51.47 am.png

I am stuck on part g)

I have A = 2 Integral from 0 to infinity of 4/(x^2 + 4) dx

I continued on but the problem is I get inverse tan of infinity on 2

Not sure where to go from here.

InteGrand · Jan 12, 2016

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

appleibeats said:
View attachment 32784

I am stuck on part g)

I have A = 2 Integral from 0 to infinity of 4/(x^2 + 4) dx

I continued on but the problem is I get inverse tan of infinity on 2

Not sure where to go from here.

$\noindent Did you find the exact area from $-a$ to $a$? Use this expression and let $a\to \infty$. Also use the facts that $\lim_{\psi \to \pm \infty} \tan^{-1} \psi = \pm \frac{\pi}{2}$.$

appleibeats · Jan 12, 2016

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

Screen Shot 2016-01-12 at 11.14.33 am.png

I do the normal integration and get up to

tan^-1 3/5 - tan^-1 (-1/4)

Not sure where to go from here.

InteGrand · Jan 12, 2016

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

appleibeats said:
View attachment 32785

I do the normal integration and get up to

tan^-1 3/5 - tan^-1 (-1/4)

Not sure where to go from here.

$\noindent Use this inverse trigonometric identity: $\tan^{-1} A - \tan^{-1}B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$, which is true provided that $\tan^{-1} A - \tan^{-1}B$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. In our case, this clearly is the case, since$

$\begin{align*}\tan^{-1} \frac{3}{5} - \tan^{-1} \left(-\frac{1}{4}\right) &=\tan^{-1} \frac{3}{5} + \tan^{-1} \left(\frac{1}{4}\right) \quad (\text{as }\tan^{-1} \text{ is an odd function})\end{align*}$

$So $\tan^{-1} \frac{3}{5} - \tan^{-1} \left(-\frac{1}{4}\right)>0>-\frac{\pi}{2}$, and also$

$\begin{align*}\tan^{-1} \frac{3}{5} - \tan^{-1} \left(-\frac{1}{4}\right) &=\tan^{-1} \frac{3}{5} + \tan^{-1} \left(\frac{1}{4}\right) \\ &< \tan^{-1} 1 + \tan^{-1} 1 \quad (\text{as }\tan^{-1} \text{ is monotone increasing}) \\ &= \frac{\pi}{2},\end{align*}$

$so indeed we have $-\frac{\pi}{2}<\tan^{-1} \frac{3}{5} -\tan^{-1} \left(-\frac{1}{4}\right) < \frac{\pi}{2}$.$

InteGrand · Jan 12, 2016

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

appleibeats said:
View attachment 32785

I do the normal integration and get up to

tan^-1 3/5 - tan^-1 (-1/4)

Not sure where to go from here.

$So basically we want to show that $\tan^{-1} \frac{3}{5}-\tan^{-1} \left(-\frac{1}{4} \right)=\frac{\pi}{4}$. So what we do is, we show that $\tan LHS = \tan RHS$ (using the compound angle formula for tan on the L.H.S., which allows us to cancel out with the inverse tangents), and then we show that the L.H.S. and R.H.S. lie in the same branch of the tan graph $\Big{(}$i.e. between $-\frac{\pi}{2}$ and $\frac{\pi}{2}\Big{)}$, which allows us to conclude that $LHS=RHS$, since their tans are the same and they're in the same branch (and the tan function is one-to-one in any of its branches).$

InteGrand · Jan 12, 2016

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

appleibeats said:
View attachment 32785

I do the normal integration and get up to

tan^-1 3/5 - tan^-1 (-1/4)

Not sure where to go from here.

$So in other words, let $\alpha = \tan^{-1} \frac{3}{5}-\tan^{-1} \left( -\frac{1}{4}\right)\Rightarrow \alpha=\tan^{-1} \frac{3}{5} + \tan^{-1} \left( \frac{1}{4}\right) $. Then$

$\begin{align*}\tan \alpha &= \frac{\tan \left(\tan^{-1} \frac{3}{5} \right) + \tan \left(\tan^{-1} \left( \frac{1}{4}\right) \right)}{1-\tan \left(\tan^{-1} \frac{3}{5} \right)\tan \left(\tan^{-1} \left( \frac{1}{4}\right) \right)} \quad (\text{compound angle formula}) \\ &= \frac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\cdot \frac{1}{4}} \quad (\text{using the fact that }\tan \left(\tan^{-1}\varrho \right) = \varrho \text{ for all real }\varrho) \\ &= \frac{12 + 5}{20 - 3}\quad (\text{multiplying top and bottom by 20}) \\ &= \frac{17}{17} \\ &= 1\\ &= \tan \frac{\pi}{4}. \end{align*}$

$Hence $\alpha$ and $\frac{\pi}{4}$ have the same tangent. Also, using the argument in my earlier post, $\alpha$ and $\frac{\pi}{4}$ are in the same branch of the tan graph. Hence $\alpha = \frac{\pi}{4}$, as required.$

InteGrand · Jan 12, 2016

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

$The other method is to \textsl{first} argue that $\alpha$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and then use the inverse trigonometric identity. We can use the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB} \right)$, which is valid iff the left-hand side is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (which it is indeed, as we showed earlier). Then using $A=\frac{3}{5}$ and $B=\frac{1}{4}$ gives us$

$\begin{align*}\alpha &=\tan^{-1} \left(\frac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\cdot \frac{1}{4}} \right) \\ &= \tan^{-1} 1 \\ &= \frac{\pi}{4},\end{align*}$

$as desired.$

appleibeats · Jan 12, 2016

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

Screen Shot 2016-01-12 at 12.11.28 pm.png

Stuck on question b)

I let u = e^-x

and got u' = - 1/e^x

Is that correct? Not sure where to go now.

InteGrand · Jan 12, 2016

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

appleibeats said:
View attachment 32786

Stuck on question b)

I let u = e^-x

and got u' = - 1/e^x

Is that correct? Not sure where to go now.

$\noindent For (b), first multiply top and bottom of the integrand by $\mathrm{e}^{x}$. So we obtain$

$\begin{align*}\int_0 ^1 \frac{1}{\mathrm{e}^{-x}+\mathrm{e}^{x}}\text{ d}x &=\int_0 ^1 \frac{\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^2} \text{ d}x \\ &= \Big{[} \tan^{-1} \left( \mathrm{e}^{x}\right)\Big{]}_0 ^1 \\ &= \tan^{-1} \left( \mathrm{e}^{1}\right)-\tan^{-1} \left( \mathrm{e}^{0}\right)\\ &= \tan^{-1} \mathrm{e} - \frac{\pi}{4}.\end{align*}$

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

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