Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.
More Regents problems.
Algebra 2/Trigonometry Regents, January 2011
Part I: Each correct answer will receive 2 credits.
6. What is the solution set of the equation 4a + 6 − 4a = −10?
1) ∅
2) {0}
3) {1/2}
4) {0, 1/2}
Answer: 1) ∅
You can plug in the last 3 choices, or you can work backward.
If should be obvious that 0 cannot be a choice because 6 =/= 10.
So you can check 1/2 or solve.
4 (1/2) + 6  4(1/2) = 2 + 6  2 = 8  2 = 6, not 10.
While that isn't enough to prove that the answer is the Null Set, ∅, it does eliminate all of the choices we were given.
Working backward:
4a + 6 − 4a = −10
4a + 6 = 10 + 4a
4a + 6 = 10 + 4a OR 4a + 6 = (10 + 4a)
6 = 10 (impossible) or 4a + 6 = 10 4a
Impossible or 8a = 4
Impossible or a = 1/2
But we have already check a = 1/2, and it is an extraneous solution that was created when we removed the absolute value symbols. If you substitute a = 1/2 into the second line above, you would have an absolute value equaling a negative number, which isn't possible.
So the answer is Null.
7. If sin A = 2/3 where 0° < A < 90°, what is the value of sin 2A?
1) 2√(5) / 3
2) 2√(5) / 9
3) 4√(5) / 9
4) 4√(5) / 9
Answer: 3) 4√(5) / 9
If 0° < A < 90° then 0° < 2A < 180°, so sin 2A is positive. Eliminate Choice (4).
If sin A = 2/3, then imagine a right triangle with one leg = 2 and the hypotenuse = 3. The other leg would be √(9  4) = √(5). So the cos A = √(5) / 3.
Sin 2A = 2 sin A cos A = 2 (2/3) (√(5)/3) = 4√(5) / 9, which is Choice (3).
You could have also found the approximate value for sin^{2} 2/3, which is about 41.81 degrees. Multiply that by 2 to get 83.62. The sine of that value would be close to 1. It's 0.993.
If you evaluate Choices (1) and (3), you'll see that (3) is the correct answer.
8. A dartboard is shown in the diagram below. The two lines intersect at the center of the circle, and the central angle in sector 2 measures 2Ï€/3.
1) 1/6
2) 1/3
3) 1/2
4) 2/3
Answer: 2) 1/3
The entire circle is 2Ï€. Sector 2 is 2Ï€/3, so Sector 4 is 2Ï€/3. That means that the total of Sectors 1 and 4 must be 2Ï€  (2Ï€/3 + 2Ï€/3) = 2Ï€/3. Each of the two sectors would be half of that, or 2Ï€/6 each.
So the probability that a dart hitting the circle randomly will fall in 1 or 3 is 1/3 because 1 and 3 together have the same area as 2 and 4 have individually. This is Choice (2).
x^{3} + x^{2} − 2x = 0
x(x^{2} + x − 2 = 0
x(x + 2)(x  1) = 0
x = 0 or x = 2 or x = 1
9. If f(x) = x^{2} − 5 and g(x) = 6x, then g(f(x)) is equal to
1) 6x^{3} − 30x
2) 6x^{2} − 30
3) 36x^{2} − 5
4) x^{2} + 6x − 5
Answer: 2) 6x^{2} − 30
Since g(x) = 6x, if the input the g is f(x), then multiply f(x) by 6.
g(f(x)) = g(x^{2} − 5) = 6(x^{2} − 5) = 6x^{2} − 30.
That is Choice (2).
10. Which arithmetic sequence has a common difference of 4?
1) {0, 4n, 8n, 12n, . . .}
2) {n, 4n, 16n, 64n, . . .}
3) {n + 1, n + 5, n + 9, n + 13, . . .}
4) {n + 4, n + 16, n + 64, n + 256, . . .}
Answer: 3) {n + 1, n + 5, n + 9, n + 13, . . .}
If there is a common difference of 4, then each term should be 4 more than the term before it (except for the first term, of course).
Choice (1) looks like it has a common ratio of 4, not a common difference. However, it's not a geometric series because of the 0 first term. You can't multiply 0 by any number and get anything other than 0. Eliminate it.
Choice (2) has a common ratio of 4, not a common difference. Eliminate it.
Choice (3) has a common difference of 4. The expression n + 5 is 4 more than n + 1, or n + 5  4 = n + 1, and n + 9  4 = n + 5, etc. Choice (3) is the answer.
Choice (4) has neitehr a common difference nor a common ratio. The constant has a ratio of 4, but the variable has a ratio of 1. This isn't allowed. Eliminate it.
More to come. Comments and questions welcome.
More Regents problems.
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