[High School - AP Calculus] Miscellaneous Concepts

[cry4eternity]

Soon we are having a statewide test in my AP Calculus class, and I'd just like to have a little leg up on it. My teacher gave the class some example questions usd on the test in 2004 to use as a guide. Basically, I'd just appreaciate it if somebody could show me how to even start on some of these. I realize a lot of the stuff hasn't even been covered in my class; and even though some of it has been, it hasn't been covered to the extent of mastery the test appears to demand. Here are six of the example questions I was not able to answer correctly. If possible, include a descriptive way to solve each type of question, and any problems I may run into if it were worded a little bit differently. What I mean is something like when taking the inverse tangent of a ratio, it gives you the angle in only the first or fourth quadrant, and you'd have to compensate accordingly. Thanks in advance.


The coefficient of x^9 when [x+(2/sqrt(x))]^30 is expanded and simplified is:
A: 30C9 * 2^9
B: 30C9 * 2^21
C: 30C16 * 2^14
D: 30C19
E: None of the above



If you begin with a circle of radius 1, inscribe an equilateral triangle in the circle, inscribe a circle in the triangle, inscribe an equialateral triangle in the smaller circle, and repeat this process indefinitely, then the sum of the areas of all of the circles is:
A: 2(pi)
B: 5(pi)/3
C: 3(pi)/2
D: 4(pi)/3
E: None of the above



How many integers between 200 and 700 consist of three distinct digits?
A: 350
B: 360
C: 365
D: 370
E: None of the above



The sum of the solutions of the equation 9^x-6*3^x+8=0 is
A: log(base 3)(2)
B: log(base 3)(6)
C: log(base 3)(8)
D: log(base 3)(4)
E: None of the above



How many diagonals does a regular polygon with 20 sides have?
A: 170
B: 180
C: 190
D: 200
E: None of the above



When five ordinary six sided dice are tossed simultaneously, the probability that a 1 shows on the top face of exactly two of the dice is
A: 625/3888
B: 1/3888
C: 2/5
D: 5/1944
E: None of the above

[dag12]

1.
Binomial theorem states:


In terms of the theorem, x = x in this case, and y = 2/sqrt(x).
So, given n = 30, we are trying to find the term k at which the power of x becomes 9.

k = 14 (I can elaborate on this if you want):
x ^ 16 + (2/sqrt(x)) ^ 14 = ___ x ^ 9
(___ is is our unknown coefficient)

So, moving on, using the binomial theorem, we know that the 16th term of the series can be expressed as:
(30 C 14) * (x ^ 16 * (2/sqrt(x)) ^ 14)

Answer: (E) - None of the Above

2.
(I'm assuming overlapping areas in more than one circle are still added over and over)
A quick diagram will show you that the radius of each subsequent circle is 1/2 that of the previous one.

so, the [k]th circle's area = [1 * 0.5 ^ (k-1) ] ^ 2 * [pi]
= 0.5 ^ (2k-2) * [pi]

Ignoring the [pi] for now, use infinite summation:
S∞ = a1 / (1-r )

a1 = 1
r = 1/4

1 / (1-1/4) = 4/3
Answer: D - 4(pi)/3

3.
Theoretically, the number of numbers from 200-299 with 3 distinct digits should equal those from 300-399, etc.
so, we have 5 "parts": 200-299, 300-399, 400-499, 500-599, 600-699 (We can disregard 700 because it doesn't have distinct digits)

Each "part" has 100 numbers, so just find the number of appropriate numbers in one "part" and multiply by 5.

Let's start with 200-299 as an example.
It's faster to count backwards, counting the number of numbers that do not have distinct digits.
First, we have numbers whose one's and ten's digits are the same:
200, 211, 222... 299 - 10 numbers

Next, we have numbers whose ten's and hundred's digits are the same:
220, 221, 222... 229 - 10 numbers

Finally, we have numbers whose one's and hundred's digits are the same:
202, 212, 222... 292 - 10 numbers

However, 222 is counted 3 times. so, we must subtract 2.
Total of inappropriate numbers: 28 numbers.
Total of appropriate numbers: 100 - 28 = 72

There are 5 "parts", so 72 * 5 = 360

Answer: B - 360

4.
Rewriting the equation:
9^x - 6*3^x + 8 = 0
(3^x)^2 - 6*3^x + 8 = 0

Let 3^x = A
A^2 - 6A + 8 = 0
(A - 2)(A - 4) = 0
A = 2, 4

Substituting back in for A:
3^x = 2
3^x = 4

x = log(base 3)(2), log(base 3)(4)

adding:
log(base 3)(2) + log(base 3)(4) = log(base 3)(8)

Answer: C - log(base 3)(8)

5.
Calculate the number of lines between any two points in the polygon:
20 C 2 = 190.
However, this includes adjacent points (sides of the polygon), which do not count as diagonals. So, subtract 20 from 190.

190-20 = 170

Answer: A - 170

6.
Use the binomial probability formula:


p = 1/6
x = 2
n = 5

5 C 2 (1/6) ^ 2 * (5/6) ^ 3
= 625/3888

Answer: A - 625/3888

Note: I did these in a hurry, so they might be wrong. Do you have the answers? if so, could you verify? I'm a little unsure of the answers, and I don't have time to check my work, so it's best to be safe.

[dooty_7]

just looking quickly, number 3 isnt much of a problem, simply find a pattern

between the numbers 200 and 209, 8 digits appear that are unique, so this holds true for every group of 10, (except the 20's in the 200s, 30s in the 300s...)

Therefore for every hundred, there are 9 groups of ten with 8 uniques in them (for the case of 200, 0s,1s,3s,4s,5s,6s,7s,8s,9s) giving 72 uniques per hundred (9 * 8). Since there are 5 hundreds to look at we get 360 (72 * 5) numbers with 3 uniques

Illl work on the others now as well and you can get the best answers from both me and dag.


EDIT:
For Question # 5 again look at patterns

draw a triangle, square, pentagon and hexagon and draw the diagonals.

in a square there are 2, in a pentagon there are 5 and in a hexagon there are 9
one can see a pattern here that for each repetition you add (n-1) from the previous number of diagonals. (So from a square where n=4 to a pentagon which is n = 5, add 3 (4-1)).
From here you could just put in 2+3+4+5+6+7+........+18 to get the answer of 170 or you could investigate more to get a better solution

Also from the drawings, notice the number of diagonals each point can make (count duplicates for now) in a square, only 1, for a pentagon, only 2, hexagon only 3.

each point can only make n-3 diagonals. Now multiply (n-3) by n (number of points) to get the diagonals created by an n sided polygon. Just divide this by 2 now to remove the duplicates.

therefore d = [n(n-3)] / 2

I will check out the others now ..

[MrRubix]

lol you can do these without paper

[cry4eternity]

The answers were
C: 30C16 * 2^14
D: 4(pi)/3
B: 360
C: log(base 3)(8)
A: 170
A: 625/3888

Thank you, dag and dooty. I'll let you guys know how I do on the test .

[dag12]

Originally Posted by cry4eternity

The answers were
C: 30C16 * 2^14
D: 4(pi)/3
B: 360
C: log(base 3)(8)
A: 170
A: 625/3888

Thank you, dag and dooty. I'll let you guys know how I do on the test .

oh hahaha number 1. Yea, number one is C... I had (E), 30C14 * 2^14 ... then again, it's the same thing as 30C16 * 2^14. Whoops :P