1. Re: VCE Maths questions help

Originally Posted by InteGrand

Thanks. Also, do you know how to find the general solutions to the equations I posted above?
Thanks

2. Re: VCE Maths questions help

Originally Posted by boredsatan
Thanks. Also, do you know how to find the general solutions to the equations I posted above?
Thanks
Yes, I do.

3. Re: VCE Maths questions help

Originally Posted by InteGrand
Yes, I do.
Could you please explain how to do it as read my textbook but got confused with it's explanation

4. Re: VCE Maths questions help

Originally Posted by boredsatan
Also, can someone please explain how to find the general solution to the following equations.
Cos 2(x+pi/3) = 1/2

2 tan 2(x+pi/4) = 2√(3)

2 sin(x+pi/3) = -1

Thanks

Silly question, but is
sin 2(x+pi/3) the same as sin(2x+2pi/3)
Originally Posted by boredsatan
Could you please explain how to do it as read my textbook but got confused with it's explanation
$\noindent Are you able first of all to find the general solution to$

$\cos \theta = \frac{1}{2}?$

$\noindent If you can do this, you should be able to easily do the questions you wrote. But if you \emph{can't} do this, you should probably work on this first.$

5. Re: VCE Maths questions help

If (x-1)^3 = 8
then would x-1 = +- cube root of 8, or just + cube root of 8 and why?

Also, does anyone know how you would work out sin(11pi/4), which is sin(2.75pi), and is in quadrant 2, and the rule for quadrant 2 is pi - base, but since it's 2.75, would you do (2pi) + (pi-base)?
And how would you work out the value of the base?

How would you find the basic angle for a negative value?
For instance cos(x) = -1/2

How would you work out (x+1)^3 = -2 without a calculator?
Thanks

6. Re: VCE Maths questions help

Originally Posted by boredsatan
If (x-1)^3 = 8
then would x-1 = +- cube root of 8, or just + cube root of 8 and why?

Also, does anyone know how you would work out sin(11pi/4), which is sin(2.75pi), and is in quadrant 2, and the rule for quadrant 2 is pi - base, but since it's 2.75, would you do (2pi) + (pi-base)?
And how would you work out the value of the base?

How would you find the basic angle for a negative value?
For instance cos(x) = -1/2

How would you work out (x+1)^3 = -2 without a calculator?
Thanks
Anyone?

7. Re: VCE Maths questions help

Bump!
Any one know how to do the above questions?

Also, does anyone know how to solve the equation sin x = 1/2 for [-pi, to 2pi]
I'm confused with the negative domain
Thanks

Bump

9. Re: VCE Maths questions help

Originally Posted by boredsatan
Also, does anyone know how to solve the equation sin x = 1/2 for [-pi, to 2pi]
I'm confused with the negative domain
Thanks
The sine wave has symmetry across the origin, so it's an odd function, which means that $f(-x) = f(-x)$.

Basically that implies that the sine of a negative angle is the same as reversing the sign of the sine of the positive angle, i.e.

$\sin -\theta = -\sin \theta$.

When we go in the positive direction, the first angle $x$ for which $\sin x = -0.5$ is $7\pi / 6$.

So that means that when we go in the negative direction, the first angle $-x$ for which $\sin -x = 0.5$ is $-7\pi / 6$.

But $-7\pi / 6$ is outside the range of $[-\pi, 2\pi]$, so it's not a solution to the equation.

10. Re: VCE Maths questions help

Originally Posted by boredsatan
Also, does anyone know how you would work out sin(11pi/4)
The period of $\sin(x)$ is $2\pi$. So,

$\sin\left(\frac{11\pi}{4}\right) = \sin\left(\frac{11\pi}{4} - 2\pi \right) = \sin\left(\frac{3\pi}{4}\right)$

Also, $\sin(x) = \sin(\pi-x)$. So,

$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt 2}{2}$

Originally Posted by boredsatan
How would you work out (x+1)^3 = -2 without a calculator?
Thanks
If you want the answer in exact form, then

$x+1 = \sqrt[3]{-2}$

$x+1 = -\sqrt[3]{2}$

$x = -1 -\sqrt[3]{2}$

11. Re: VCE Maths questions help

The equation of the image of the graph of y = sin x under a transformation of a dilation of factor 1/2 from the y-axis followed by a translation of pi/4 units in the positive direction of the x axis is:
I'm narrowed it down two 2 options, but don't know which one is correct. Can someone please explain which one is correct and why
y = sin(2x - pi/4) or y = sin2(x-pi/4)

12. Re: VCE Maths questions help

let f: [0, pi/2] → R, where f(x) = cos(3x) -2. The graph of f is transformed by a reflection in the x-axis followed by a dilation of factor 3 from the y-axis. How would you work out the resulting graph, super confusing so wold greatly appreciate if someone helped out

13. Re: VCE Maths questions help

Originally Posted by boredsatan
The equation of the image of the graph of y = sin x under a transformation of a dilation of factor 1/2 from the y-axis followed by a translation of pi/4 units in the positive direction of the x axis is:
I'm narrowed it down two 2 options, but don't know which one is correct. Can someone please explain which one is correct and why
y = sin(2x - pi/4) or y = sin2(x-pi/4)
I'm not familiar with that specific terminology, but if "dilation of factor 1/2 from the y-axis" means compressing the function horizontally, then you replace $x$ with $2x$ and to shift it right you would replace $x$ with $x - \pi/4$, in that order.

$y = \sin x$

Dilate:

$y = \sin 2x$

Shift graph:

$y = \sin 2\left(x-\frac{\pi}{4}\right)$

Originally Posted by boredsatan
let f: [0, pi/2] → R, where f(x) = cos(3x) -2. The graph of f is transformed by a reflection in the x-axis followed by a dilation of factor 3 from the y-axis. How would you work out the resulting graph, super confusing so wold greatly appreciate if someone helped out
$f(x) = \cos 3x -2$

Reflect across x axis:

$f(x) = -(\cos 3x -2)$

Dilate:

$f(x) = -\left(\cos 3\left(\frac{x}{3}\right) -2\right)$

$= 2 - \cos x$

14. Re: VCE Maths questions help

An open tank is to be made from a sheet of metal 84 cm by 40 cm by cutting congruent squares of side length x cm from each of the corners.
I've found the volume to be = (84-2x)(40-2x)(x)

State the maximal domain for V when it is considered as a function of x. Confused how to do this part
Any help would be appreciated

15. Re: VCE Maths questions help

Originally Posted by boredsatan
An open tank is to be made from a sheet of metal 84 cm by 40 cm by cutting congruent squares of side length x cm from each of the corners.
I've found the volume to be = (84-2x)(40-2x)(x)

State the maximal domain for V when it is considered as a function of x. Confused how to do this part
Any help would be appreciated
bump

16. Re: VCE Maths questions help

A rectangle is defined by vertices N and P(x,y) on the curve with equation y = 16-x^2 and vertices M and Q on the x axis
a.i. Find the area, A of the rectangle in terms of x
ii. state the implied domain for the function defined by the rule given in part i

17. Re: VCE Maths questions help

A metal worker is required to cut a circular cylinder from a solid sphere of radius 5 cm. Express r in terms of h, where r cm is the radius of the cylinder and h cm is the height of the cylinder. Hence show that the volume, V cm^3, of the cylinder is given by V = (1/4)(pi)(h)(100-h^2)

18. Re: VCE Maths questions help

n a tidal river, the time between high tides and low tides is 12 hours. The average depth of water at a point in the river is 5m. At high tide the depth is 8 m. Assume that the depth of the water, h(t) m, at this point is given by
h(t) = A sin(nt+e)+b, where t is the number of hours after noon. At noon there is a high tide.

a. find the values of A, n,b, and e

b. at what times is the depth of the water 6 m?

19. Re: VCE Maths questions help

Bump

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