[20 Questions] Trigonometry, Locus&Parabola, Polynomials + More! :( (1 Viewer)

roadrage75 · Feb 11, 2009

P!xel said:
\Question 18)

use the remainder theorum.

P(x) leaves a remainder of 8 when divided by x+1.

So P(-1) = 8 ---> 1+ 3 +a -b -6 = 8.

ie a-b = 10 --- equation 1

and x-3 is a factor of P(x).

So P(3) = 0 ---> 81 - 81 + 9a +3b - 6 = 0

giving us the equation 9a + 3b = 6 which can be further simplified to form a second equation...

3a + b = 2 --- equation 2

solving equation 1 and 2 by inspection gives us a = 3, and b = -7

kurt.physics · Feb 11, 2009

$A circle has equation$ $x^2 + y^2 -4x + 2y = 0$ ,

$a) Find the centre and radius of the circle$
$b) the line $ x + 2y = 0 $ meets this circle at two points A and B.$
$i)Find the co-ordinates of A and B.$
$ii) Calculate the distance AB.$

$Okay, so i will complete the square to find the general equation of a circle.$

$x^2 + y^2 -4x + 2y = 0$

$x^2 -4x + y^2+ 2y = 0$

$x^2 - 4x + (\frac{-4x}{2x})^2 - (\frac{-4x}{2x})^2 + y^2+ 2y + (\frac{2y}{2y})^2 - (\frac{2y}{2y})^2 = 0$

$(x - 2)^2 -2^2 + (y + 1)^2 - 1 = 0$

$(x - 2)^2 + (y + 1)^2 = 5$

$\therefore (x - 2)^2 + (y + 1)^2 = 5$ $is the general equation of the circle.$

$\therefore c = (2,-1)$ and $r = \sqrt{5}$

$2. We will solve this by substitution;$

$x + 2y = 0$

$\therefore x = -2y$ $, we will sub this into the circles equation.$

$(-2y)^2 + y^2 -4(-2y) + 2y = 0$

$4y^2 + y^2 + 8y + 2y = 0$

$5y^2 + 10y = 0$

$\therefore y^2 + 2y = 0$

$y(y + 2) = 0$

$\therefore y = 0, -2$

$then$ $x = 0, 4$ $respectively, we did this by substituting back into$ $x + 2y = 0$ $, therefore the co-ordinates are;$

$A = (0,0)$

$B = (4,-2)$

$The distance, by the formula$

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$= \sqrt{(4 - 0)^2 + (-2 - 0)^2}$

$= \sqrt{16 + 4}$

$= \sqrt{20}$

$\therefore d_{AB} = 2\sqrt{5}$

edit: sorry gurmies, didnt see you did it

jet · Feb 11, 2009

1'll do question 10:
$2sin^2 \theta = sin 2\theta \\ 2sin^2 \theta - 2sin\theta cos\theta = 0 \\ 2sin\theta (sin\theta - cos\theta) = 0 \\ sin\theta = 0 \, \text{therefore, } \theta = 0, \pi, 2\pi \\ sin\theta - cos\theta = 0 \\ sin(\theta - 45^{\circ}) = 0 \therefore \theta = \frac{\pi}{4}, \frac{5\pi}{4}$

greentsquare · Feb 11, 2009

roadrage75 said:
use the remainder theorum.

P(x) leaves a remainder of 8 when divided by x+1.

So P(-1) = 8 ---> 1+ 3 +a -b -6 = 8.

ie a-b = 10 --- equation 1

and x-3 is a factor of P(x).

So P(3) = 0 ---> 81 - 81 + 9a +3b - 6 = 0

giving us the equation 9a + 3b = 6 which can be further simplified to form a second equation...

3a + b = 2 --- equation 2

solving equation 1 and 2 by inspection gives us a = 3, and b = -7

damn. you beat me to it.
and i already typed out the solution and stuff.
great to know we have the same answers, though.

kurt.physics · Feb 12, 2009

Question 9)

$Extend AQ arbitrarily in a west direction. Now if you could imagine "3d" compass (North, South, East, West thing), it would not be like a + , it would be at an angle.$

$Now, at point B, extend a line to the extended AQ such that it is perpendicular to it, let this point be M. We now have a right-triangle BMQ. Now $\angle MBQ = 51^{\circ}$, this is because this angle is the bearing. Therefore,$

$\angle BQM = 90^{\circ} - 51^{\circ} = 39^{\circ} $(angle sum$ \triangle AMQ)$

$\therefore \angle BQA = 180^{\circ} - 39^{\circ} = 141^{\circ} $(angles on a straight line)$$

$ii. In triangle APQ,$

$\angle APQ = 78^{\circ} $(angle sum of a triangle)$$

$\therefore tan 78^{\circ} = \frac{AQ}{h}$

$\therefore AQ = h \times tan 78^{\circ}$

$iii. Similarly in triangle BPQ,$

$\angle BPQ = 79^{\circ} $(angle sum triangle)$$

$\therefore tan 79^{\circ} = \frac{BQ}{h}$

$\therefore BQ = h \times tan 79^{\circ}$

$iv.$

$BQ = h \times tan 79^{\circ}$
$AQ = h \times tan 78^{\circ}$
$\angle BQA = 141^{\circ}$

$c^2 = a^2 + b^2 - 2ab \times cos C$

$1000^2 = (h \times tan 79^{\circ})^2 + (h \times tan 78^{\circ})^2 - 2(h \times tan 79^{\circ})(h \times tan 78^{\circ})(cos 141^{\circ})$

$\therefore h^2 \times tan^2 79^{\circ} + h^2 \times tan^2 78^{\circ} -2 \times h^2 \times tan 78^{\circ} \times tan 79^{\circ} \times cos 141^{\circ} = 1000^2$

$h^2 (tan^2 79^{\circ} + tan^2 78^{\circ} -2 \times tan 78^{\circ} \times tan 79^{\circ} \times cos 141^{\circ}) = 1000^2$

$\therefore h^2 = \frac{1000^2}{tan^2 79^{\circ} + tan^2 78^{\circ} -2 \times tan 78^{\circ} \times tan 79^{\circ} \times cos 141^{\circ}}$

$\therefore h = \sqrt{\frac{1000^2}{tan^2 79^{\circ} + tan^2 78^{\circ} -2 \times tan 78^{\circ} \times tan 79^{\circ} \times cos 141^{\circ}} }$

$\boxed {h = 108 m}$

...damn im tired!

kurt.physics · Feb 12, 2009

Question 14

$1 + 3 + 6 + . . . + \frac{1}{2} n(n+1) = \frac{1}{6} n(n+1)(n+2)$

$n = 1$

$LHS = \frac{1}{2} 1(2)$
$= 1$

$RHS = \frac{1}{6} 1(2)(3)$
$= 1$

$\therefore LHS = RHS$

$assume$ $n = k$, ie$$

$1 + 3 + 6 + . . . + \frac{1}{2} k(k+1) = \frac{1}{6} k(k+1)(k+2)$

$prove$ $n = k + 1$, ie$$

$1 + 3 + 6 + . . . + \frac{1}{2} (k + 1)(k + 2) = \frac{1}{6} (k+1)(k+2)(k+3)$

$LHS = 1 + 3 + . . . + \frac{1}{2} k(k+1) + \frac{1}{2} (k + 1)(k + 2)$

$= \frac{1}{6} k(k+1)(k+2) + \frac{1}{2} (k + 1)(k + 2)$

$= \frac{k(k+1)(k+2)}{6} + \frac{3 \times (k + 1)(k + 2)}{3 \times 2}$

$= \frac{k(k+1)(k+2) + 3(k + 1)(k + 2)}{6}$

$= \frac{(k+1)(k+2)\times (k + 3)}{6}$

$= \frac{(k+1)(k+2)(k + 3)}{6}$

$= \frac{1}{6} (k+1)(k+2)(k+3)$

$= RHS$

Then just right a statement

roadrage75 · Feb 12, 2009

kurt.physics said:
Question 14

$1 + 3 + 6 + . . . + \frac{1}{2} n(n+1) = \frac{1}{6} n(n+1)(n+2)$

$n = 1$

hey mate, how do you type out your answers like that?? Is there a special program?

kurt.physics · Feb 12, 2009

roadrage75 said:
hey mate, how do you type out your answers like that?? Is there a special program?

its called LaTeX, you can learn it here About LaTeX

but when you use latex in this forum, you must put the code in $bla bla whatever [*tex] but where it says * you have to put /$

jet · Feb 12, 2009

Question 5:
$\sqrt{3}cosx - sinx =1\Rightarrow Rcos(x + \alpha) \\ \text{Where, } \\ R = \sqrt{3 + 1}= 2 \, \text{and} \, \alpha = tan^{-1} \frac{1}{\sqrt{3}}= \frac{\pi}{6} \\ \therefore 2cos(x + \frac{\pi}{6}) = 1 \\ cos(x + \frac{\pi}{6}) = \frac{1}{2} \\ x + \frac{\pi}{6} = \frac{\pi}{3}, \frac{5\pi}{3} \\ x = \frac{\pi}{6}, \frac{3\pi}{2} \\ \text{The general solution to this equation will be:} \\ x + \frac{\pi}{6} = 2k\pi \pm \frac{\pi}{3} \\ x = \frac{12k\pi - \pi \pm 2\pi}{6} \\ = \frac{(12k - 1)\pi \pm 2\pi}{6}, \, \text{k integral}$

jet · Feb 12, 2009

Question 12
$\text{You should be able to work out part a) by yourself, and atleast sketch it - intercepts 0 and 4} \\ \text{Solving the two equations together: } \\ mx - 4 = x^2 - 4x \\ x^2 -(m + 4)x + 4 = 0 \\ \text{Since this is a tangent, the discriminant must be equal to 0} \\ \Delta = 0 \\ (m + 4)^2 - 16 = 0 \\ (m + 4)^2 = 16 \\ m + 4 = \pm 4 \\ m = -8, 0$

jet · Feb 12, 2009

Here is question 15

Lol. This was in a 3 unit exam we did once. We didn't have that many parts. There was only part b, d and e, as it was the last question. I was the only one who could get it.

jet · Feb 12, 2009

Question 19:
$\Sigma \alpha = \alpha + \beta + \gamma = -2 \\ \Sigma \alpha \beta = \alpha \beta + \alpha \gamma + \beta \gamma = -3 \\ \Pi \alpha = \alpha \beta\gamma = -5 \\ \Therefore (\alpha - 1)(\beta - 1)(\gamma - 1) = (\alpha\beta - \alpha - \beta + 1)(\gamma - 1) \\ = \alpha \beta \gamma - \alpha \gamma - \beta \gamma + \gamma - \alpha \beta + \alpha + \beta - 1 \\ = \alpha \beta \gamma - (\alpha \beta + \alpha \gamma + \beta \gamma) + (\alpha + \beta + \gamma) - 1 \\ =-5 -(-3) + (-2) - 1 = -5$

jet · Feb 12, 2009

Question 17:
$\text{The chord PQ has gradient} \frac{1}{2}(p + q) \text{ and equation} \\ y - Ap^2 = \frac{1}{2}(p + q)(x - 2Ap) \\ \text{it passes through the focus, } (0, A): \\ A - Ap^2 = \frac{1}{2}(p + q)(-2Ap) \\ A - Ap^2 = -Ap^2 - Apq \\ Apq = -A \\ pq = -1 \\ \\ \text{The two tangents are given by the equations: } \\ y = px - Ap^2 \\ y = qx - Aq^2 \\ \text{Solving the two, } px - qx = A(p^2 - q^2) \\ x(p - q) = A(p - q)(p + q) \\ x = A(p + q) \\ y = Ap(p + q) - Ap^2 \\ = Apq = -A \\ \\ PQ^2 = (2Ap - 2Aq)^2 + (Ap^2 - Aq^2)^2 \\ = 4A^2p^2 + 4A^2q^2 - 8A^2pq + A^2p^4 + A^2q^4 - 2A^2(pq)^2 \\ = 4A^2p^2 + 4A^2q^2 + 8A^2 + A^2p^4 + A^2q^4 - 2A^2 \\ =A^2[p^4 + 4p^2 + 6 + 4q^2 + q^2] \\ \text{but, } q = \frac{-1}{p} \\ \therefore PQ^2 = A^2[p^4 + 4p^2 + 6 + \frac{4}{p^2} + \frac{1}{p^4}] \\ = A^2(p + \frac{1}{p})^4 \\ \therefore PQ = A(p + \frac{1}{p})^2$

The expansion is just something you should know, or will know. If you don't see it, expand it and see the result. A knowledge of the expansions of (a + b)^2, (a + b)^3 and (a + b)^4 are helpful, granted you will learn them at the end of 3 unit.

kurt.physics · Feb 12, 2009

ZOMG!

Question 3 is the crappiest question ever! I have spend half an hour on it!

What i came up with the diagram that i have attached. The algebra with those stupid decimals makes me sick!

So i got 3 equations, two being the similar triangles ratios (of the two different triangles) and one of the cosine rule.

Then i had the unimaginable job of solving them, i came up with (though the quadratic formula)

h = 6.9983 or 0.0632!

to the OP, do you have the answer???

I think this is a stupid question for a test!

kurt.physics · Feb 12, 2009

here it is

kurt.physics · Feb 12, 2009

In fact, i changed some of the values, the man i made 2 m, his first shadow i made 6, his second, 10, this gave me

h = 6.3 as the only answer,

so i think it was a bit unnecessary for them to have all those crap decimals >:Z lol

tommykins · Feb 12, 2009

kurt.physics said:
In fact, i changed some of the values, the man i made 2 m, his first shadow i made 6, his second, 10, this gave me

h = 6.3 as the only answer,

so i think it was a bit unnecessary for them to have all those crap decimals >:Z lol

6units for hsc.

what a dream.

jet · Feb 12, 2009

Did question 3. It took so long and I dont think you would find a question like that in an HSC exam EVER. Those decimals ruined my life. Here it is:

Someone else do the rest of the questions lol.
If im not right, tell me and ill change it.

EDIT: Thanks to lolokay for finding my mistake

tacogym27101990 · Feb 13, 2009

and you honestly couldn't do any of these?
sure you weren't just being lazy???

Prashant Sallan · Feb 13, 2009

hey- i can only do 16 part (i) and (ii)
but im not sure if its the correct answer. :mad1:

[20 Questions] Trigonometry, Locus&Parabola, Polynomials + More! :( (1 Viewer)

Member

Member

Banned

Member

Member

Member

Member

Member

Banned

Banned

Banned

Banned

Banned

Member

Member

Member

i am number -e^i*pi

Banned

Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)