Monday, February 07, 2022

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2011)



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 2π/3.




1) 1/6
2) 1/3
3) 1/2
4) 2/3

Answer: 2) 1/3


The entire circle is 2π. Sector 2 is 2π/3, so Sector 4 is 2π/3. That means that the total of Sectors 1 and 4 must be 2π - (2π/3 + 2π/3) = 2π/3. Each of the two sectors would be half of that, or 2π/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).

x3 + x2 − 2x = 0

x(x2 + x − 2 = 0

x(x + 2)(x - 1) = 0

x = 0 or x = -2 or x = 1





9. If f(x) = x2 − 5 and g(x) = 6x, then g(f(x)) is equal to

1) 6x3 − 30x
2) 6x2 − 30
3) 36x2 − 5
4) x2 + 6x − 5

Answer: 2) 6x2 − 30


Since g(x) = 6x, if the input the g is f(x), then multiply f(x) by 6.

g(f(x)) = g(x2 − 5) = 6(x2 − 5) = 6x2 − 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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