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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4110 on: June 13, 2019, 01:15:44 pm »
+1
I just need a little assistance, I need an example equation of a Parabola with Two Real and Distinct Roots,
So for example, $y=x(x-1)$?

Any equation of the form $y=(x-\alpha)(x-\beta)$ where $\alpha$ and $\beta$ are two different real numbers will satisfy your property

#### Youssefh_

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##### Re: 3U Maths Question Thread
« Reply #4111 on: June 16, 2019, 01:08:23 pm »
0
Hey I am having some trouble finding the inverse function of
f(x) = (2^x)-1
and also I need help with proving the functions are mutually inverse can you please show me how to do it.

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4112 on: June 16, 2019, 03:19:05 pm »
+1
Hey I am having some trouble finding the inverse function of
f(x) = (2^x)-1
and also I need help with proving the functions are mutually inverse can you please show me how to do it.
$\text{If }y = \left(2^x\right)^{-1}\\ \text{then for the inverse, after swapping }x\text{ and }y,$
\begin{align*} x &= \left(2^y\right)^{-1}\\ \frac{1}{x} &= 2^y\\ \log_2 \frac1x &= y \end{align*}
$\therefore f^{-1}(x) = \log_2 \frac{1}{y}$
Ensure that you know that the exponential and logarithm are inverse functions. Because the logarithm is defined to be the inverse function to the exponential, the statements $a^x = b$ and $x = \log_a b$ are equivalent. i.e. we can go between one another.

Also, because they are inverses, the mutual inverses property is assumed to hold here. That is, we can always assume that $a^{\log_a x} = x$ for $x > 0$, and $\log_a (a^x) = x$ for all real $x$.

#### skintceaser

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##### Re: 3U Maths Question Thread
« Reply #4113 on: June 30, 2019, 04:19:38 pm »
0
Hey i need help doing this question..... (image attached)

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4114 on: June 30, 2019, 04:29:59 pm »
+1
Hey i need help doing this question..... (image attached)

Hey there!

Just note in the future we'd like you to tell us what you know or show us what you've done already to help us fill in the gaps in your understanding, and not just punch out answers. It'll definitely help you more

a) We can see there's a double root at x=2 and a single root at x=-1. Therefore, for some constant a, f(x)=a(x-2)2(x+1). Now, since f(0)=2, we know that a=1/2; ie. f(x)=0.5(x-2)2(x+1)
b) Essentially reflect the part of the graph that's to the left of the y-axis in the line y=x (this should be a relatively well known method of finding the inverse relation)
c) There's a few methods you could use to do this; you could find the gradient of the inverse function by finding the function then differentiating, you could use implicit differentiation, but alternately you could just use the reciprocal of the derivative of f(x) at the desired x value ie. 1/f'(-1), since the x-intercepts of f(x) become the y intercepts of the new function

Hope this helps
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#### skintceaser

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##### Re: 3U Maths Question Thread
« Reply #4115 on: June 30, 2019, 05:31:42 pm »
0
Hey there!

Just note in the future we'd like you to tell us what you know or show us what you've done already to help us fill in the gaps in your understanding, and not just punch out answers. It'll definitely help you more

a) We can see there's a double root at x=2 and a single root at x=-1. Therefore, for some constant a, f(x)=a(x-2)2(x+1). Now, since f(0)=2, we know that a=1/2; ie. f(x)=0.5(x-2)2(x+1)
b) Essentially reflect the part of the graph that's to the left of the y-axis in the line y=x (this should be a relatively well known method of finding the inverse relation)
c) There's a few methods you could use to do this; you could find the gradient of the inverse function by finding the function then differentiating, you could use implicit differentiation, but alternately you could just use the reciprocal of the derivative of f(x) at the desired x value ie. 1/f'(-1), since the x-intercepts of f(x) become the y intercepts of the new function

Hope this helps

sure my bad for this, its my first time posting. but thanks for the help.

#### skintceaser

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##### Re: 3U Maths Question Thread
« Reply #4116 on: July 01, 2019, 02:24:04 pm »
0
This question is posting a bit of a problem, as I am unsure as to how to find exact values of cos^-1 (2/3) and tan^-1 (3/4). I tried making cos= sin((pi/2)-A), but I am unsure how to do the same for tan^-1.......And that's where i got stuck.......

The image is attached containing the question.. but for safe measure......... here it is

find the exact value of sin(cos^-1(2/3)+tan^-1(-3/4))

#### julz_roha

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##### Re: 3U Maths Question Thread
« Reply #4117 on: July 01, 2019, 02:42:22 pm »
0
Hello. This is for simple harmonic motion.
How do you find the period of the equation in the image?

...

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4118 on: July 01, 2019, 05:09:20 pm »
+1
This question is posting a bit of a problem, as I am unsure as to how to find exact values of cos^-1 (2/3) and tan^-1 (3/4). I tried making cos= sin((pi/2)-A), but I am unsure how to do the same for tan^-1.......And that's where i got stuck.......

The image is attached containing the question.. but for safe measure......... here it is

find the exact value of sin(cos^-1(2/3)+tan^-1(-3/4))

Hey again

$\text{Let} \cos^{-1} \frac{2}{3} = \alpha \\ \text{Then,} \ \cos \alpha = \frac{2}{3} \\ \text{Similarly, let} \ \tan^{-1} \frac{-3}{4} = \beta \\ \tan \beta = \frac{-3}{4} \\ \text{Draw triangles with this new information, and include the length of the hypotenuse} \\ \text{Now,} \ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \\ = \frac{\sqrt{5}}{3} \times \frac{4}{5} + - \frac{3}{5} \times \frac{2}{3} \\ =\frac{4\sqrt{5}-6}{15}$

Hello. This is for simple harmonic motion.
How do you find the period of the equation in the image?

$\text{Given this is simple harmonic, and seeing a} \ v^2 \ \text{this is a clear marker for using the identity} \\ v^2=n^2(a^2-x^2) \\ \text{For} \ v^2 = -5+6x-x^2 \\ v^2 = 4-9+6x-x^2 = 1(2^2-(x-3)^2) \\ \text{Since} \ T = \frac{2\pi}{n}, \ T=2\pi$

Hope this helps
« Last Edit: July 01, 2019, 05:16:14 pm by fun_jirachi »
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#### DrDusk

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##### Re: 3U Maths Question Thread
« Reply #4119 on: July 01, 2019, 05:10:23 pm »
+2
Hello. This is for simple harmonic motion.
How do you find the period of the equation in the image?
We know that:
$T = \frac{2\pi}{n} \\ v^2 = -5+6x-x^2 \\ \frac{1}{2}v^2 = \frac{1}{2}(-5+6x-x^2) \\ \text{but}\hspace{1mm} a = \frac{d}{dx}(\frac{1}{2}v^2) \\ a = \frac{1}{2}(6-2x) = -(x-3) \\ \text{This is in the form}\hspace{1mm}a = -n^2(x-a) \\ \text{Giving:} \\ T = 2\pi$
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#### julz_roha

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##### Re: 3U Maths Question Thread
« Reply #4120 on: July 01, 2019, 05:46:01 pm »
0
Thank you.
We know that:
$T = \frac{2\pi}{n} \\ v^2 = -5+6x-x^2 \\ \frac{1}{2}v^2 = \frac{1}{2}(-5+6x-x^2) \\ \text{but}\hspace{1mm} a = \frac{d}{dx}(\frac{1}{2}v^2) \\ a = \frac{1}{2}(6-2x) = -(x-3) \\ \text{This is in the form}\hspace{1mm}a = -n^2(x-a) \\ \text{Giving:} \\ T = 2\pi$
...

#### julz_roha

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##### Re: 3U Maths Question Thread
« Reply #4121 on: July 01, 2019, 09:27:04 pm »
0
Another simple harmonic equation.
Question in the image.
...

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4122 on: July 02, 2019, 07:06:09 am »
+1
There's two ways you can do this:
Method 1: (Same way as yours above (but your integral was wrongly computed))
$v=\int -12\sin 2t \ dt \\ = 6\cos 2t + C \\ \text{When} \ t=0, v=6 \\ v=6\cos 2t \\ v=6(1-2\sin^2t) = -12\sin^2t+6 \ \text{(By the double angle formula)}$
Method 2:
$v=\int -12\sin 2t \ dt \\ = \int -24\sin t\cos t \ dt \\ = -24 \times \frac{1}{2} \sin^2t + C \ \text{(By the reverse chain rule)} \\ \text{When} \ t=0, v=6 \\ v= -12\sin^2t +6$

You can use the definite integral in both cases as well to make it easier if it helps.
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#### skintceaser

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##### Re: 3U Maths Question Thread
« Reply #4123 on: July 02, 2019, 10:48:22 am »
0
well I did this question, but i feel like this is very wrong, and when i put it in maths calculators.. they dont use the same u sub too. pls help.

question and image containing ans i did attached. edit: for some reason the file with my working out is not being able to be uploaded............

qn is  u= e^3x, integral(1/3,0) e^3x/e^6x+1 dx.
« Last Edit: July 02, 2019, 10:54:51 am by skintceaser »