Qs (1 Viewer)

=)(= · Oct 16, 2021

and

ii pls

ExtremelyBoredUser · Oct 16, 2021

=)(= said:
View attachment 32778
and
View attachment 32779
ii pls

For the first question, straight off the bat I would say (D). The rest of the choices seem wrong. Just square the equations and simultaneously solve them. A,B are dead giveaways that they're not a circle so its really between C and D and obviously C would produce extra remainder so its D.

=)(= · Oct 16, 2021

ExtremelyBoredUser said:
For the first question, straight off the bat I would say (D). The rest of the choices seem wrong. Just square the equations and simultaneously solve them. A,B are dead giveaways that they're not a circle so its really between C and D and obviously C would produce extra remainder so its D.

wouldn't c be right because you can make cos t and sin t the subject and then use the Pythagorean idenitity

ExtremelyBoredUser · Oct 16, 2021

=)(= said:
wouldn't c be right because you can make cos t and sin t the subject and then use the Pythagorean idenitity

No because when you square the x and y terms.

$x^2 = 4cos^2t$
$y^2 = sin^2t$
$x^2 + y^2 = 4cos^2t + sin^2t$
$x^2 + y^2 = 4(cos^2t + \frac{1}{4} sin^2t)$

Hence you can't apply Pythagorean identity to simplify it to get a number value. The dead giveaway for this is that there's a 2 in front of cost so there's no way it would be able to get a clean factorisation which would eliminate the cos and sins.

For (D) you would get $8cos^2t + 8sin^2t$ if you simplify and factorising it $8(sin^2t+cos^2t) = 8(1).$

Hence the eqn would be $x^2 + y^2 = 8$ so it would be the correct answer.

Life'sHard · Oct 16, 2021

ExtremelyBoredUser said:
No because when you square the x and y terms.

x^2 = 4cos^2t
y^2 = sin^2t
--------------
x^2 + y^2 = 4cos^2t + sin^2t
x^2 + y^2 = 4(cos^2t + 1/4sin^2t)

Hence you can't apply Pythagorean identity to simplify it to get a number value. The dead giveaway for this is that there's a 2 in front of cost so there's no way it would be able to get a clean factorisation which would eliminate the cos and sins.

For (D) you would get 8cos^2t + 8sin^2t if you simplify and factorising it 8(sin^2t+cos^2t) = 8(1).

Hence the eqn would be x^2 + y^2 = 8 so it would be the correct answer.

Might as well start learning latex my guy. You're already typing it in latex format hahaha.

CM_Tutor · Oct 16, 2021

For (C): Make cos t the subject and apply a Pythagorean identity:

$\begin{align*} \sin^2{t} + \cos^2{t} &= 1 \\ y^2 + \left(\cfrac{x}{2}\right)^2 &= 1 \\ x^2 + 4y^2 &= 4 \end{align*}$

The shape is an ellipse.

For (D):

$\begin{align*} \sin^2{t} + \cos^2{t} &= 1 \\ \left(\cfrac{x - y}{4}\right)^2 + \left(\cfrac{x + y}{4}\right)^2 &= 1 \\ x^2 - 2xy + y^2 + x^2 + 2xy + y^2 &= 16 \\ x^2 + y^2 &= 8 \end{align*}$

The shape is a circle.

ExtremelyBoredUser · Oct 16, 2021

Life'sHard said:
Might as well start learning latex my guy. You're already typing it in latex format hahaha.

The latex format doesn't show up on my screen? I know how to use it but it doesn't appear on my screen at all so I presume its the same for everyone?

5uckerberg · Oct 17, 2021

For part ii the general strategy is this given $y=(x-1)^{2}, x > 1,$ we see there we are required to find where the points of the inverse are the same as the given graph. Now we will apply the mirror $y=x$ to make x the main ingredient. In this case what we will have is $x=x^{2}-2x+1$ becoming $x^{2}-3x+1$ . There use the quadratic formula and we should obtain two solutions. The rest should be trivial because you just have to rely on what the question specified and one of the solutions will be within the domain and the other will not.

In general when they ask "find any value for x where $f(x)=f^{-1}(x)$ ," for me the general strategy will be to turn y into x because we are focusing on x and also for inverses the line $y=x$ is used as a mirror to sketch the inverted graph. At this moment using the quadratic equation find the values of x and then if they give you the domain from the question apply the domain and you will be done. However, if you are required to sketch the inverse, different story. Now you have to swap x & y from the original function and make y the subject and why I say swap x & y is because the inverse is reflected in the line $y=x$ . Is there someone who can clarify what I said or make any corrections?

5uckerberg · Oct 17, 2021

ExtremelyBoredUser said:
The latex format doesn't show up on my screen? I know how to use it but it doesn't appear on my screen at all so I presume its the same for everyone?

There is the f(x) button and press that this is what you will obtain
$y=x^{2}, x > 0$

CM_Tutor · Oct 17, 2021

I agree with @5uckerberg on solving $f(x)=f^{-1}(x)$ by noting that the intersections must occur on $y=x$ , and thus by solving $f(x) = x$ .

A useful prior step can be to quickly sketch $y = f(x) = (x-1)^2$ and $y = x$ . Since $y = f(x)$ is a concave up parabola with vertex at $(1,\ 0)$ and $y$ -intercept at $y = 1$ , sketching on $y=x$ will immediately show that there should be two points of intersection to find, one somewhere in the domain $0<x<1$ , and the other in $2<x<3$ . This allows a quick quick on whether the results you find are reasonable.

5uckerberg · Oct 17, 2021

To be fair one can do Q9 in a different way. I can say with parametric equations for a circle usually it can be written like this $x^{2}+y^{2}=n^{2}$ . All you gotta do is square both terms and see if they give something like this $x^{2}+y^{2}=n^{2}$ , if not then it is not a circle.

CM_Tutor · Oct 17, 2021

PS... That (D) in the MCQ is a circle can also be found by examining

$\begin{align*} x^2 + y^2 &= (2\cos{t} + 2\sin{t})^2 + (2\cos{t} - 2\sin{t})^2 \\ &= 4\cos^2{t} + 8\cos{t}\sin{t} + 4\sin^2{t} + 4\cos^2{t} - 8\cos{t}\sin{t} + 4\sin^2{t} \\ &=8\left(\cos^2{t} + \sin^2{t}\right) \\ &= 8 \qquad \text{as $\sin^2{\theta} + \cos^2{\theta} = 1$ for all $\theta$} \\ \\ x^2 + y^2 &= 8 \qquad \text{is a circle, centre $(0,\ 0)$, radius $2\sqrt{2}$} \end{align*}$

but this wouldn't work as well as the method I posted above if the circle had a centre anywhere but the origin.

5uckerberg · Oct 17, 2021

CM_Tutor said:
PS... That (D) in the MCQ is a circle can also be found by examining

$\begin{align*} x^2 + y^2 &= (2\cos{t} + 2\sin{t})^2 + (2\cos{t} - 2\sin{t})^2 \\ &= 4\cos^2{t} + 8\cos{t}\sin{t} + 4\sin^2{t} + 4\cos^2{t} - 8\cos{t}\sin{t} + 4\sin^2{t} \\ &=8\left(\cos^2{t} + \sin^2{t}\right) \\ &= 8 \qquad \text{as $\sin^2{\theta} + \cos^2{\theta} = 1$ for all $\theta$} \\ \\ x^2 + y^2 &= 8 \qquad \text{is a circle, centre $(0,\ 0)$, radius $2\sqrt{2}$} \end{align*}$

but this wouldn't work as well as the method I posted above if the circle had a centre anywhere but the origin.

Thank you CM tutor for debunking my idea, even though it works for one case.
I would now bring up a schema for locus,

For circles $(x-h)^{2}+(y-k)^{2}=n^{2}, h, \ k \in \mathbb{R}$ well for the circle that is anywhere but the origin if one were given constants, bring the constants on the side with the x and the y and square the LHS.
For ellipses $(x-h)^{2}/a^{2}+(y-k)^{2}/b^{2}=1, h, \k, \a, \b \in \mathbb{R}$
For hyperbolas $(x-h)^{2}/a^{2}-(y-k)^{2}/b^{2}=1, h, \k, \a, \b \in \mathbb{R}$
For lines $y=mx+b, m, \b \in \mathbb{R}$
For the well-known asymptote $c^{2}=xy, c \in \mathbb{R}$

These are what you want to get at the end.

CM_Tutor · Oct 17, 2021

@5uckerberg, parametrised equations of lines aren't restricted to going through the origin. Option (B), above, has parametric equations $x = 2t$ and $y = t + 2$ , which are the equivalent of the Cartesian equation is $x - 2y + 4 = 0$ ... this doesn't fit the $y = mx$ model that you have stated, so you might want to edit your post to expand the forms to $y = mx + b$ where $m,\ b \in \mathbb{R}$ ... and FYI, the code I have used here is "m,\ b \in \mathbb{R}", rather than using \epsilon as you have.

Also, I think picking out the forms of ellipses and hyperbolae is likely beyond most MX1 expectations.

Finally, if you had something like $x = 3 + 2\cos{\theta}$ and $y = 1 - 2\sin{\theta}$ , the technique that is easiest is to recognise that:

$\cos{\theta} = \cfrac{x-3}{2} \qquad \qquad \text{and} \qquad \qquad \sin{\theta} = \cfrac{1-y}{2}$

and thus use the Pythagorean identity:

$\begin{align*} \sin^2{\theta} + \cos^2{\theta} &= 1 \\ \left(\cfrac{x-3}{2}\right)^2 + \left(\cfrac{1-y}{2}\right)^2 &= 1 \\ (x-3)^2 + (1 - y)^2 &= 4 \\ (x-3)^2 + (y - 1)^2 &= 4 \qquad \qquad \text{as $(1 - y)^2 = 1 - 2y + y^2 = (y - 1)^2$} \end{align*}$

to recognise the parametric equations describe a circle with centre at (3, 1) and radius = 2.

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