Miscellaneous Questions (1 Viewer)

pomsky · Jul 4, 2015

Yooo

So in prep for trials, i managed to stub my metaphorical brain into COUGHalmosteverythingCOUGH. Will be much appreciated if you guys could give me hints and tips to approaching these questions (and a walkthrough if you're feeling particularly generous). Or if you'd just like to have a mental workout..?

$$Q1. Find values of x for which $x+\frac{1}{|x|} > 0$ and $ x^2-x-2<0$ are simultaneously satisfied.\\ $$Q2. Solve $t^2+2t-1=0$ and using your results from this and the expansion for $tan2\theta$, find the exact value of $tan 22.5^o$. Simplify your answer.$\\$

InteGrand · Jul 4, 2015

pomsky said:
Yooo So in prep for trials, i managed to stub my metaphorical brain into COUGHalmosteverythingCOUGH. Will be much appreciated if you guys could give me hints and tips to approaching these questions (and a walkthrough if you're feeling particularly generous). Or if you'd just like to have a mental workout..?

$$Q1. Find values of x for which $x+\frac{1}{|x|} > 0$ and $ x^2-x-2<0$ are simultaneously satisfied.\\ $$Q2. Solve $t^2+2t-1=0$ and using your results from this and the expansion for $tan2\theta$, find the exact value of $tan 22.5^o$. Simplify your answer.$\\$

$Q.1 The quadratic inequality $x^2-x-2<0$ is solved by finding the roots of the L.H.S. quadratic. Using the quadratic formula, the roots are $\frac{1\pm \sqrt{1+4\times 2}}{2}=\frac{1\pm 3}{2}=2$ or $-1$. So this inequality becomes $(x-2)(x-(-1))<0$. If we sketch the L.H.S. parabola, we see that the solutions are $-1<x<2$.$

$Now we consider the inequality $x+\frac{1}{|x|}>0$. We only need consider $-1<x<2$, since we want both inequalities to be satisfied. Clearly for $x>0$, the left-hand side is always positive. So consider $-1<x<0$ (note $x\neq 0$). For $x<0$, $|x|=-x$, so the inequality becomes for $x<0$,$

$x+\frac{1}{-x}>0, -1<x<0$

$$\Rightarrow x^2 -1 < 0$ (multiplying through by $x<0$, so flipping the inequality sign). Now, this inequality is true for all $x\in (-1,0)$.$

$Hence both inequalities are satisfied precisely when $x\in (-1,2):x\neq 0$.$

$Q.2. $t^2+2t-1=0\Longleftrightarrow (t+1)^2-2=0$$

$\Rightarrow (t+1)^2 =2$

$\Rightarrow t+1=\pm \sqrt{2}$

$$\Rightarrow t=-1\pm \sqrt{2}$.$

$Now, we want to find $t=\tan \theta$, where $\theta = 22.5^\circ$ (so $t>0$ as $\theta$ is in the first quadrant).$

$Since $1=\tan 45^\circ = \tan 2\theta =\frac{2t}{1-t^2}$, we have $2t=1-t^2\Rightarrow t^2+2t - 1=0$. As we saw before, the positive solution to this is $t=-1+\sqrt{2}$. Hence the answer is $t=\tan 22.5^\circ = -1+\sqrt{2}$.$

pomsky · Jul 5, 2015

Thanks Integrand!

$$Q3. Find $\lim_{x\to 0}\left (\frac{x^2}{2-2cos2x}\right )$

Crisium · Jul 5, 2015

pomsky said:
Thanks Integrand!
$$Q3. Find $\lim_{x\to 0}\left (\frac{x^2}{2-2cos2x}\right )$

Factor out the two in the denominator

(x^2) / 2(1 - cos2x)

Since 1 - cos2x = 2sin^2(x)

(x^2) / 2(2sin^2(x))

(x^2) / 4sin^2(x)

Factor out the one-quarter

[1 / 4] x [ (x^2) / sin^2(x)]

Take the squared outside of the second bracket

[ 1 / 4] x [ x / sinx ]^2

You know that the limit of x / sinx approaches zero = 1 from the trigonometric function topic

NOTE: Keep that limit sign of x approaches zero should be placed in front of the working out until this point (I can't LaTex LOL)

Therefore

[ 1 / 4] x [ 1 ]^2

= [ 1 / 4] x 1

= 1 / 4

Is this right?

wiliaamnguuyen · Jul 5, 2015

Can you guys please help me with a question I'm currently stuck on? file:///Users/kim/Desktop/Outlook.com%202/IMG_5506.JPG
Specifically questions C) and G) and how do I tackle inequations and equations questions that has double or triple absolute values?

wiliaamnguuyen · Jul 5, 2015

Sorry, this is the attachment!

rand_althor · Jul 5, 2015

Part c:

$\begin{align*}\frac{1}{\left |x-1 \right |}&>\frac{1}{\left |x+1 \right |} \\\left |x-1 \right |&<\left |x+1 \right | \\\textup{Squaring both sides } \left |x-1 \right |^2&<\left |x+1 \right |^2 \\x^2-2x+1&<x^2+2x+1 \\ -4x&<0 \\x&>0 \end{align*}$

So x>0, where x does not equal 1.

wiliaamnguuyen · Jul 5, 2015

Could you explain why x doesn't equal to one? Is it because the denominator can't be zero or else it would be undefined?

rand_althor · Jul 5, 2015

wiliaamnguuyen said:
Could you explain why x doesn't equal to one? Is it because the denominator can't be zero or else it would be undefined?

Yep that is correct.

wiliaamnguuyen · Jul 5, 2015

What confuses me is that, there is two absolute values. Why isn't there two restrictions, but you only mentioned one "x cannot equal to 1", could you explain that to me?

rand_althor · Jul 5, 2015

I'm assuming you mean the restriction where x cannot be -1. I didn't consider it as when you end up with your final answer, you realise that x is always positive and hence it won't ever be -1 anyway.

wiliaamnguuyen · Jul 5, 2015

OOooOoO ty ty!!!!!

iamaloser17 · Jul 5, 2015

Part g (Sorry for no LaTeX):

|x-1| + |x+3| > 6

Check the critical points
x=1: |1-1| + |1+3| /> 6 --> x=1 is not a solution
x=-3: |-3-1| + |-3+3| /> 6 --> x=-3 is not a solution

For all x<-3, x-1<0 and x+3<0
-(x-1) - (x+3) > 6
-x+1 - x-3 >6
-2x > 8
x < -4
Therefore x<-4 is a solution.

For -3 < x < 1, x-1 < 0 and x+3 > 0
-(x-1) + x+3 >6
4 > 6
Therefore -3 < x < 1 is not a solution as 4 /> 6

For x>1, x-1>0 and x+3>0
x-1 + x+3 >6
2̶x̶ ̶>̶ ̶8̶ ̶
x̶ ̶>̶ ̶4̶ ̶
T̶h̶e̶r̶e̶f̶o̶r̶e̶ ̶x̶>̶4̶ ̶i̶s̶ ̶a̶ ̶s̶o̶l̶u̶t̶i̶o̶n̶.̶
Edit:
2x > 4
x > 2
Therefore x>2 is a solution.

Therefore the solution to |x-1| + |x+3| > 6 is x<-4, x̶>̶4̶ x>2. I hope I'm right.

faisalabdul16 · Jul 5, 2015

For g, these are the conditions for which the inequality holds

<a href="http://www.codecogs.com/eqnedit.php?latex=|x-1|&space;+&space;|x+3|&space;>&space;6&space;\\&space;(|x-1|&space;+&space;|x+3|)^{2}&space;>&space;36&space;\\&space;\\&space;x^{2}&space;-2x&space;+&space;1&space;+&space;2|x-1||x+3|&space;+&space;x^{2}+6x+9>36&space;\\&space;$Now,&space;middle&space;term&space;is&space;positive&space;for&space;any&space;value&space;of&space;x&space;because&space;of&space;the&space;absolute&space;value&space;signs&space;\\&space;\\&space;\therefore&space;x^{2}&space;-2x&space;+&space;1+&space;2(x^{2}+2x-3)&space;+&space;x^{2}+6x+9>36&space;\\&space;\\4x^{2}+&space;8x-32>&space;0&space;\\&space;4(x^{2}+2x&space;-8)>&space;0&space;\\&space;4(x+4)(x-2)>0&space;\\&space;\therefore&space;x>2\\&space;x<-4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?|x-1|&space;+&space;|x+3|&space;>&space;6&space;\\&space;(|x-1|&space;+&space;|x+3|)^{2}&space;>&space;36&space;\\&space;\\&space;x^{2}&space;-2x&space;+&space;1&space;+&space;2|x-1||x+3|&space;+&space;x^{2}+6x+9>36&space;\\&space;$Now,&space;middle&space;term&space;is&space;positive&space;for&space;any&space;value&space;of&space;x&space;because&space;of&space;the&space;absolute&space;value&space;signs&space;\\&space;\\&space;\therefore&space;x^{2}&space;-2x&space;+&space;1+&space;2(x^{2}+2x-3)&space;+&space;x^{2}+6x+9>36&space;\\&space;\\4x^{2}+&space;8x-32>&space;0&space;\\&space;4(x^{2}+2x&space;-8)>&space;0&space;\\&space;4(x+4)(x-2)>0&space;\\&space;\therefore&space;x>2\\&space;x<-4" title="|x-1| + |x+3| > 6 \\ (|x-1| + |x+3|)^{2} > 36 \\ \\ x^{2} -2x + 1 + 2|x-1||x+3| + x^{2}+6x+9>36 \\ $Now, middle term is positive for any value of x because of the absolute value signs \\ \\ \therefore x^{2} -2x + 1+ 2(x^{2}+2x-3) + x^{2}+6x+9>36 \\ \\4x^{2}+ 8x-32> 0 \\ 4(x^{2}+2x -8)> 0 \\ 4(x+4)(x-2)>0 \\ \therefore x>2\\ x<-4" /></a>

DatAtarLyfe · Jul 5, 2015

iamaloser17 said:
Part g (Sorry for no LaTeX):

|x-1| + |x+3| > 6

Check the critical points
x=1: |1-1| + |1+3| /> 6 --> x=1 is not a solution
x=-3: |-3-1| + |-3+3| /> 6 --> x=-3 is not a solution

For all x<-3, x-1<0 and x+3<0
-(x-1) - (x+3) > 6
-x+1 - x-3 >6
-2x > 8
x < -4
Therefore x<-4 is a solution.

For -3 < x < 1, x-1 < 0 and x+3 > 0
-(x-1) + x+3 >6
4 > 6
Therefore -3 < x < 1 is not a solution as 4 /> 6

For x>1, x-1>0 and x+3>0
x-1 + x+3 >6
2x >8
x > 4
Therefore x>4 is a solution.

Therefore the solution to |x-1| + |x+3| > 6 is x<-4, x>4. I hope I'm right.

Minor error in your algebra. The answer is x<-4, x>2

wiliaamnguuyen · Jul 5, 2015

Do you guys have any advice on tackling double/triple absolute equations/inequations??/

iamaloser17 · Jul 5, 2015

DatAtarLyfe said:
Minor error in your algebra. The answer is x<-4, x>2

Whoops. Sorry about that. Thank you.

iamaloser17 · Jul 5, 2015

wiliaamnguuyen said:
Do you guys have any advice on tackling double/triple absolute equations/inequations??/

Make sure you understand the concepts and the methods in tackling these questions. Look for a variety of questions, doing them and just practicing will help. And don't make silly mistakes like me lel. Nothing is better than simply practicing honestly.

pomsky · Jul 11, 2015

Solve the equation $tan^{-1}3x - tan^{-1}2x= tan^{-1}\frac{1}{5}$

rand_althor · Jul 11, 2015

pomsky said:
Solve the equation $tan^{-1}3x - tan^{-1}2x= tan^{-1}\frac{1}{5}$

$\begin{align*}\tan^{-1}3x-\tan^{-1}2x&=\tan^{-1}\frac{1}{5} \\\textup{Let }\alpha &= \tan^{-1}3x,\, \beta =\tan^{-1}2x \rightarrow \tan\alpha=3x, \, \tan\beta=2x\\\tan{(\alpha-\beta)} &= \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\\&=\frac{3x-2x}{1+(3x)(2x)} \\&=\frac{x}{1+6x^2} \\\therefore \alpha-\beta &= \tan^{-1}\left ({\frac{x}{1+6x^2}} \right ) \\\tan^{-1}3x-\tan^{-1}2x &= \tan^{-1}\left ({\frac{x}{1+6x^2}} \right ) \\ \therefore \frac{1}{5}&=\frac{x}{1+6x^2} \\5x&=1+6x^2 \\6x^2-5x+1&=0 \\ (3x-1)(2x-1)&=0 \\x&=\frac{1}{3}, \, \frac{1}{2}\end{align*}$

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