Prelim 2016 Maths Help Thread (2 Viewers)

Green Yoda · Apr 8, 2016

Oh yeah got it..I made silly mistakes while drawing in radian measure..I am really not used to that haha

Green Yoda · Apr 16, 2016

help with this:
shade the region -|x|+2>=-√4-x^2

Should I be looking it it terms of the y values or the x values?

InteGrand · Apr 16, 2016

Rathin said:
help with this:
shade the region -|x|+2>=-√4-x^2

Should I be looking it it terms of the y values or the x values?

If we're talking about sharing that as a region in the x-y plane, since y doesn't appear in that inequation, y can be anything at all. So you just need to find the set of all x-values satisfying that equation, and then essentially draw vertical lines at each of these places (which will end up shading some part of the plane).

Also there should be brackets under the square root, I initially thought it was just sqrt(4), before looking at the actual number and realising it would be just said 2 if that were the case.

Green Yoda · Apr 16, 2016

InteGrand said:
If we're talking about sharing that as a region in the x-y plane, since y doesn't appear in that inequation, y can be anything at all. So you just need to find the set of all x-values satisfying that equation, and then essentially draw vertical lines at each of these places (which will end up shading some part of the plane).

Also there should be brackets under the square root, I initially thought it was just sqrt(4), before looking at the actual number and realising it would be just said 2 if that were the case.

hmm I am not understanding this completely..What I do understand now is that we look at the x values..but how would I find the link to that inequality??

InteGrand · Apr 16, 2016

Rathin said:
hmm I am not understanding this completely..What I do understand now is that we look at the x values..but how would I find the link to that inequality??

Basically solving the inequation for x is the first step. Have you done that yet? Once you've done that, it's fairly easy.

Green Yoda · Apr 16, 2016

InteGrand said:
Basically solving the inequation for x is the first step. Have you done that yet? Once you've done that, it's fairly easy.

nope haven't done that..?
In the question they told us to draw the two graphs and then shade the region..so they are asking us to do it graphically.
But I still dont seem to understand what the questions means.. Are they asking us to find for which values of x in -|x|+2>=-√(4-x^2)

InteGrand · Apr 16, 2016

Rathin said:
nope haven't done that..?
In the question they told us to draw the two graphs and then shade the region..so they are asking us to do it graphically.
But I still dont seem to understand what the questions means.. Are they asking us to find for which values of x in -|x|+2>=-√(4-x^2)

Well essentially you'd first find those x-values that satisfy the inequation. Whether you do this algebraically or graphically is up to you. Once you've found these values, the region would be found by drawing vertical lines at each of these x-values.

Green Yoda · Apr 16, 2016

InteGrand said:
Well essentially you'd first find those x-values that satisfy the inequation. Whether you do this algebraically or graphically is up to you. Once you've found these values, the region would be found by drawing vertical lines at each of these x-values.

The value are 2 and -2 as thats where they intersect..Now what do I do now? and What do you mean by vertical lines?

em_paget · Apr 16, 2016

would someone be able to explain the 2t value things to me?
its part of theoretical triginometry
thanks

InteGrand · Apr 16, 2016

Rathin said:
The value are 2 and -2 as thats where they intersect..Now what do I do now? and What do you mean by vertical lines?

$\noindent Well note that $\sqrt{4-x^2}$ is only real for $x$ between $-2$ and $2$ (otherwise the thing under the square root (called the radicand) is negative). So we should only consider $x$ in the range $-2\leq x \leq 2$. Now it is actually clear that for any $x$ in this interval, $-|x|+2 \geq -\sqrt{4-x^2}$, because the LHS is always greater than or equal to 0, whilst the RHS is always \emph{less than} or equal to 0. (The equality occurs precisely at the endpoints of the interval, i.e. at $x=\pm 2$.) Now, here is a sketch of the two functions' graphs:$

http://www.wolframalpha.com/input/?i=sketch+y+=+-|x|+2+and+y+=+-sqrt(4-x^2) .

$\noindent I think what the question wants you to do is sketch in that region between the curves. In this case, the question should have asked to sketch the region given by $-\sqrt{4-x^2}\leq y \leq -|x|+2$. This seems to make more sense as a question.$

$\noindent If they actually meant to sketch the region in the $x-y$ plane where $-|x|+2 \geq -\sqrt{4-x^2}$, then the region would be the vertical band bounded by the lines $x=-2$ and $x=2$, with the inside shaded in (this is like drawing a vertical line at each of the infinitely many $x$-values between $-2$ and $2$).$

Green Yoda · Apr 16, 2016

InteGrand said:
$\noindent Well note that $\sqrt{4-x^2}$ is only real for $x$ between $-2$ and $2$ (otherwise the thing under the square root (called the radicand) is negative). So we should only consider $x$ in the range $-2\leq x \leq 2$. Now it is actually clear that for any $x$ in this interval, $-|x|+2 \geq -\sqrt{4-x^2}$, because the LHS is always greater than or equal to 0, whilst the RHS is always \emph{less than} or equal to 0. (The equality occurs precisely at the endpoints of the interval, i.e. at $x=\pm 2$.) Now, here is a sketch of the two functions' graphs:$

http://www.wolframalpha.com/input/?i=sketch+y+=+-|x|+2+and+y+=+-sqrt(4-x^2) .

$\noindent I think what the question wants you to do is sketch in that region between the curves. In this case, the question should have asked to sketch the region given by $-\sqrt{4-x^2}\leq y \leq -|x|+2$. This seems to make more sense as a question.$

$\noindent If they actually meant to sketch the region in the $x-y$ plane where $-|x|+2 \geq -\sqrt{4-x^2}$, then the region would be the vertical band bounded by the lines $x=-2$ and $x=2$, with the inside shaded in (this is like drawing a vertical line at each of the infinitely many $x$-values between $-2$ and $2$).$

So -|x|+2>=-√4-x^2
basically means y>=-|x|+2 and y<=-√(4-x^2)

InteGrand · Apr 16, 2016

Rathin said:
So -|x|+2>=-√4-x^2
basically means y>=-|x|+2 and y<=-√(4-x^2)

It doesn't mean that. I was just saying that would have made a better Q. If that was what was being meant, it would have been written as -√(4-x^2) ≤ y ≤ -|x|+2 .

Green Yoda · Apr 16, 2016

InteGrand said:
It doesn't mean that. I was just saying that would have made a better Q. If that was what was being meant, it would have been written as -√(4-x^2) ≤ y ≤ -|x|+2 .

what does it mean then??
would then shading -|x|+2>=-√(4-x^2) be the same as y>=-|x|+2 and y<=-√(4-x^2)??

InteGrand · Apr 16, 2016

Rathin said:
what does it mean then??
would then shading -|x|+2>=-√(4-x^2) be the same as y>=-|x|+2 and y<=-√(4-x^2)??

No, because the former has no mention of y. It is simply an inequation in x, which has a solution: -2 ≤ x ≤ 2. If we wanted to sketch this on the number line, we'd sketch -2 ≤ x ≤ 2 on the number line, i.e. we'd shade in the part of the number line between -2 and 2 and include the endpoints. But if we wanted to sketch this on the x-y plane, we'd need to shade in all possible places in the x-y plane where -2 ≤ x ≤ 2. This becomes the vertical band between the vertical lines x = -2 and x = 2 (including these boundary lines).

It's like how an equation like x = 3 in the x-y plane is represented by sketching all places where x = 3. This becomes a vertical line because we just set x = 3 and can take any y-value (which means we can go up and down by any amount, but not across, since x needs to be 3). This is why something like x = 3 represents a vertical line in the x-y plane.

Green Yoda · Apr 16, 2016

InteGrand said:
No, because the former has no mention of y. It is simply an inequation in x, which has a solution: -2 ≤ x ≤ 2. If we wanted to sketch this on the number line, we'd sketch -2 ≤ x ≤ 2 on the number line, i.e. we'd shade in the part of the number line between -2 and 2 and include the endpoints. But if we wanted to sketch this on the x-y plane, we'd need to shade in all possible places in the x-y plane where -2 ≤ x ≤ 2. This becomes the vertical band between the vertical lines x = -2 and x = 2 (including these boundary lines).

It's like how an equation like x = 3 in the x-y plane is represented by sketching all places where x = 3. This becomes a vertical line because we just set x = 3 and can take any y-value (which means we can go up and down by any amount, but not across, since x needs to be 3). This is why something like x = 3 represents a vertical line in the x-y plane.

Another silly question; But I now get that it's between +-2 but why do we only share between the two curves
http://www.wolframalpha.com/input/?i=-|x|+2>=-sqrt(4-x^2)

why not outside them as we are not bound by the y values??

InteGrand · Apr 16, 2016

Rathin said:
Another silly question; But I now get that it's between +-2 but why do we only share between the two curves
http://www.wolframalpha.com/input/?i=-|x|+2>=-sqrt(4-x^2)

why not outside them as we are not bound by the y values??

We do go outside them. I was just saying I think it'd have been a better question if they made it bound by the y-values too.

bobmyself · Apr 16, 2016

Given that a<0, find the solution to the inequality 1<abs(ax+1)<2, leaving your answer in terms of a.

Green Yoda · Apr 17, 2016

help:
(sinθ/1+cosθ) + (1+cosθ/sinθ)

InteGrand · Apr 17, 2016

Rathin said:
help:
(sinθ/1+cosθ) + (1+cosθ/sinθ)

$\noindent I am assuming it is $\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$. $\Big{(}$If writing without \LaTeX, it is best to put brackets around the numerators and denominators, i.e. $\left(\sin \theta \right)/\left(1+\cos \theta\right)$, etc.$\Big{)}$.$

$\noindent Assuming none of the denominators are 0, we have$

$\begin{align*}\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta} &= \frac{\sin^2 \theta + \left(1+\cos \theta\right)^2}{\sin \theta \left(1+\cos \theta\right)} \\ &= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta \left(1+\cos \theta\right)} \\ &= \frac{2 + 2\cos \theta }{\sin \theta \left(1+\cos \theta\right)} \quad (\text{here we have made use of the Pythagorean trigonometric identity}) \\ &= \frac{2 \left(1+ \cos \theta \right)}{\sin \theta \left(1+\cos \theta\right)}\\ &= \frac{2}{\sin \theta} = 2\csc \theta. \end{align*}$

Green Yoda · Apr 17, 2016

InteGrand said:
$\noindent I am assuming it is $\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$. $\Big{(}$If writing without \LaTeX, it is best to put brackets around the numerators and denominators, i.e. $\left(\sin \theta \right)/\left(1+\cos \theta\right)$, etc.$\Big{)}$.$

$\noindent Assuming none of the denominators are 0, we have$

$\begin{align*}\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta} &= \frac{\sin^2 \theta + \left(1+\cos \theta\right)^2}{\sin \theta \left(1+\cos \theta\right)} \\ &= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta \left(1+\cos \theta\right)} \\ &= \frac{2 + 2\cos \theta }{\sin \theta \left(1+\cos \theta\right)} \quad (\text{here we have made use of the Pythagorean trigonometric identity}) \\ &= \frac{2 \left(1+ \cos \theta \right)}{\sin \theta \left(1+\cos \theta\right)}\\ &= \frac{2}{\sin \theta} = 2\csc \theta. \end{align*}$

omg that was easy af..I just didnt pick up the Pythagoras identity..
Thanks

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