ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: dinosaur93 on March 25, 2012, 09:40:08 pm
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1. If (http://latex.codecogs.com/gif.latex?\mathit{f}:(-1,4)\rightarrow \textit{R, f(x)} =|x^3 + 4x^2 - 10x - 4|), then state the range of f.
2. Find the equation of the tangent and the perpendicular to the curve (http://latex.codecogs.com/gif.latex?y=x^3 + 3x^2 - 6x + 9) at (2, 17). How do you find the perpendicular via CAS Calculator or by hand?
3. The graphs of (http://latex.codecogs.com/gif.latex?y=\frac{a}{x}) and (http://latex.codecogs.com/gif.latex?\displaystyle y=x+2) intersects at 2 distinct point for (http://latex.codecogs.com/gif.latex?a \geqslant 1) Is this right?
4. Given that (http://latex.codecogs.com/gif.latex?$\displaystyle \mathit{f(x)} = \left\{\begin{matrix}\frac{1}{(x-2)^3} & x>2 \\ -4 & x\leqslant 2 \end{matrix}\right.$)
Given that (http://latex.codecogs.com/gif.latex?$\displaystyle a\neq 1$), find f(a + 1) in terms of 'a' for all other values of 'a'. (HINT: Express your answer as a hybrid function)
5. A function with the equation (http://latex.codecogs.com/gif.latex?$\displaystyle y = A \cos (n(x - b)) + c $) having a period of 12, a range of [-9, 15] and contains the point (2, 15). Determine the values of 'A', 'n', 'b' and 'c' given they are all positive.
6. The function (http://latex.codecogs.com/gif.latex?$\displaystyle y = x^2$) is altered by the following transformations (http://latex.codecogs.com/gif.latex?$\displaystyle (x, y) \to (2x, -y + 2)$) Write the equation of the transformed function.
7. (http://latex.codecogs.com/gif.latex?$\displaystyle g: D \to R, g(x) = (x-1)^2 + 2$)
a.) Find D, a suitable domain of g(x) such that g-1 (x) exists.
b.) State the domain and range of g for the answer in a)
domg = ____
rang = ____
c.) State the equation of g-1 (x), using the proper function notation.
d.) Find the exact value of the intersection point.
8. If the function (http://latex.codecogs.com/gif.latex?$\displaystyle f(t) = at^3 + bt^2 +ct+d$) is divisible by (t - f)(f - t) and (ht + g), then identify the incorrect statement
A. a equals -h
B. d equals -f2g
C. The three t intercepts are located at t = -f, f, and -g/h
D. The cubic touches the t-axis at t = f
E. f(0) - d
Am I right?
9. The graph of (http://latex.codecogs.com/gif.latex?$\displaystyle y = |x-1| + |x+1|$) is sketched. The domain and range respectively are:
A. R and R
B. R and R+
C. R and [2, \infty)
D. R and (2, \infty)
E. (2, \infty) and R
10. The function f: [\frac{-1}{2}, \frac{1}{2}] \to R, f(x) = 2 tan (3\pi x) has an asymptotes for this restricted domain at ____?____ Show working out...
11. Define the hybrid function of the graph (http://latex.codecogs.com/gif.latex?$\displaystyle y = |x-1| + |x+2|$).
12. A probe use to measure the temperature of a substance. The values recorded follow the function (http://latex.codecogs.com/gif.latex?$\displaystyle T(t) = \cos (\frac{2 \pi t}{23}) + 35$), where t is time in minutes and T is temperature in 0C. State he initial temperature and the next 2 times the substance will have this temperature
Is it 700 C initial temperature and 23 mins, 46 mins?
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1. Sketch the graph as normal, remembering to put in the endpoints. Range = [0,84).
2. Define f(x) = (your function). Go to menu 9 4 for tangent (or 5 for normal). In the bracket after tangntLine(f(x),x=2). This will give 18x-19. Just put in y= in front. By hand, the gradient of the perp will be 1/18 and use the given point to find the equation.
3. Equating will give a/x = x + 2
a = x^2 + 2x
x^2 +2x - a = 0
x = (-2 +/- sqrt(4 + 4a))/2
x = (-2 +/- 2sqrt(1 + a))/2
x = -1 +/- sqrt(1 + a) therefore 2 solutions for all values a > -1.
4. f(a + 1) ={1/(a + 1 - 2)^2, a + 1>2
-4, a + 1<=2
={1/(a - 1)^2, a>1
-4, a<1 (since a is not equal to 1)
5. Period = 12 = 2pi/n so n = pi/6
range means there is a vertical translation of (-9 + 15)/2 = 3 so c = 3 and amp = 12. Since A > 0, A = 12.
x = 2 gives a y = 15 which is the max so pi/6(2 - b) = 0 so b = 2.
6. x' = 2x so x = x'/2 and y' = -y + 2 so y = -y' + 2. Substituting gives
-y + 2 = (x/2)^2 so y = -x^2/4 + 2
7. a) Need either left half or right half. D = [1,inf)
b) Range = [2,inf)
c) g^-1:[1,inf) arrow [2,inf), g^-1 = sqrt(x - 2) + 1
d) Normally you would solve (x - 1)^2 + 2 = x as both graphs meet on the y = x line IF they meet. These graphs do not meet. You've made an error.
8. f(t) = -(t - f)^2(ht + g).
When this is expanded, the first term will be -ht^3 so A is correct.
The last term will be -f^2g so B is correct.
As there is a repeated factor, there will be only 2 intercepts at t = f and -g/h so C is INCORRECT
D is correct (see above)
Of course E is correct.
9. As hybrid:
y={-(x - 1) - (x - 1) = -2x, x<-1
{-(x - 1) + (x +1) = 2, -1<=x<1
{x - 1 + x + 1 = 2x,x>=1
So Dom = R and ran = [2,inf) C
10. You have this wrong and it's too late at night for me to suggest that pi/3 should be the coeff.
Solve cos(pix/3) = 0 for asymptotes.
11. Use method in Q9.
12. T(0) = 36. The next time will be a period later at t = 23 minutes and again at t = 46 minutes.