More Algebra 2 problems.

__January 2020, Part I__

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.

*16. As θ increases from -π/2 to 0 radians, the value of cos θ will
(1) decrease from 1 to 0
(2) decrease from 0 to -1
(3) increase from -1 to 0
(4) increase from 0 to 1
*

**Answer: (4) increase from 0 to 1 **

Cos(-π/2) = 0, Cos(0) = 1

*17. Consider the following patterns:
*

I. 16, -12, 9, -6.75, …

II. 1, 4, 9, 16, …

III. 6, 18, 30, 42, …

IV. 1/2, 2/3, 3/4, 4/5, …

I. 16, -12, 9, -6.75, …

II. 1, 4, 9, 16, …

III. 6, 18, 30, 42, …

IV. 1/2, 2/3, 3/4, 4/5, …

Which pattern is geometric?

(1) I

(2) II

(3) III

(4) IV

Which pattern is geometric?

(1) I

(2) II

(3) III

(4) IV

**Answer: (1) I**

The terms in pattern I have a common ratio: -12/16 = -3/4; 9/-12 = -3/4

Pattern II is quadratic. The terms are squares.

Pattern III is arithmetic. Each term is 12 more than the prior term.

Pattern IV is none of these. It has neither a common ratio or common difference.

*18. Consider the system below.
*

x + y + z = 9

x - y - z = -1

x - y + z 21

x + y + z = 9

x - y - z = -1

x - y + z 21

Which value is not in the solution, (x,y,z), of the system?

(1) -8

(2) -6

(3) 11

(4) 4

Which value is not in the solution, (x,y,z), of the system?

(1) -8

(2) -6

(3) 11

(4) 4

**Answer: (1) -8 **

Combine the first two equations:

x + y + z = 9

__x - y - z = -1__

2x = 8

x = 4

Combine the second and third equations:

x - y - z = -1

__x - y + z 21__

-2z = -22

z = 11

Solve for y: (4) + y + (11) = 9

y + 15 = 9

y = -6

*19. Which statement regarding polynomials and their zeros is true?
(1) f(x) = (x*

^{2}- 1)(x + a) has zeros of 1 and -a, only (2) f(x) = x

^{3}- ax

^{2}+ 16x - 16a has zeroes of 4 and a, only. (3) f(x) = (x

^{2}+ 25)(x + a) has zeros of

__+__5 and -a. (4) f(x) = x

^{3}- ax

^{2}- 9x + 9a has zeros at

__+__3 and a.

**Answer: (4) f(x) = x ^{3} - ax^{2} - 9x + 9a has zeros at +3 and a.**

Choice (1) has factors (x + 1)(x - 1) and a zero at -1.

In choice (2), f(4) = 64 - 16a + 64 - 16a = 128 - 32a, not zero.

In choice (3), x

^{2}+ 25 cannot be factored in (x + 5)(x - 5), so they are not zeros.

Choice (4) can be factored as follows:

f(x) = x

^{3}- ax

^{2}- 9x + 9a

f(x) = x

^{2}(x - a) - 9(x - a)

f(x) = (x

^{2}- 9)(x - a)

f(x) = (x + 3)(x - 3)(x - a), x =

__+__3 or a.

*20. If a solution of 2(2x - 1) = 5x ^{2} is expressed in simplest a + bi form,
the value of b is
*

**Answer: (2) SQRT(6)/5 **

^{2}

4x - 2 = 5x

^{2}

5x

^{2}- 4x + 2 = 0

x = ( -(-4) __+__ SQRT( (-4)^{2} - 4(5)(2) ) / (2(5))

x = ( 4 __+__ SQRT( 16 - 40 ) ) / (10)

x = ( 4 __+__ SQRT( -24 ) ) / (10)

x = ( 4 __+__ 2i SQRT( 6 ) / (10)

x = 2/5 __+__ i SQRT( 6 ) / 5

The answer is choice (2) because they only wanted the value of *b*, not the entire term. Choice one includes *i*.

Comments and questions welcome.

More Algebra 2 problems.

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