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Algebra 2

Algebra 2 Practice Test: Practice Test 68

Practice Test 68 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Sketch p(x)=−x(x−3)2(x+2)p(x)=-x(x-3)^2(x+2)p(x)=−x(x−3)2(x+2) showing zeros with multiplicities, crossing vs touching, and end behavior.

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Question 1

Sketch p(x)=−x(x−3)2(x+2)p(x)=-x(x-3)^2(x+2)p(x)=−x(x−3)2(x+2) showing zeros with multiplicities, crossing vs touching, and end behavior.

  1. Zeros: x=0x=0x=0 (mult. 1), x=3x=3x=3 (mult. 2), x=−2x=-2x=−2 (mult. 1). Crosses at 000 and −2-2−2, touches at 333. End behavior: as x→−∞x\to-\inftyx→−∞, p(x)→−∞p(x)\to-\inftyp(x)→−∞ and as x→∞x\to\inftyx→∞, p(x)→−∞p(x)\to-\inftyp(x)→−∞. (correct answer)
  2. Zeros: x=0x=0x=0 (mult. 1), x=3x=3x=3 (mult. 2), x=−2x=-2x=−2 (mult. 1). Crosses at all three zeros. End behavior: as x→−∞x\to-\inftyx→−∞, p(x)→−∞p(x)\to-\inftyp(x)→−∞ and as x→∞x\to\inftyx→∞, p(x)→∞p(x)\to\inftyp(x)→∞.
  3. Zeros: x=0x=0x=0 (mult. 2), x=3x=3x=3 (mult. 1), x=−2x=-2x=−2 (mult. 1). Touches at 000, crosses at 333 and −2-2−2. End behavior: as x→−∞x\to-\inftyx→−∞, p(x)→∞p(x)\to\inftyp(x)→∞ and as x→∞x\to\inftyx→∞, p(x)→−∞p(x)\to-\inftyp(x)→−∞.
  4. Zeros: x=0x=0x=0 (mult. 1), x=3x=3x=3 (mult. 2), x=−2x=-2x=−2 (mult. 1). Crosses at 000 and −2-2−2, touches at 333. End behavior: as x→−∞x\to-\inftyx→−∞, p(x)→∞p(x)\to\inftyp(x)→∞ and as x→∞x\to\inftyx→∞, p(x)→∞p(x)\to\inftyp(x)→∞.

Explanation: This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = -x(x-3)²(x+2), we identify zeros at x = 0 (multiplicity 1, from -x), x = 3 (multiplicity 2, from (x-3)²), and x = -2 (multiplicity 1, from (x+2)). The graph crosses at x = 0 and x = -2 (odd multiplicities) and touches at x = 3 (even multiplicity). The polynomial has degree 1+2+1=4 (even) with negative leading coefficient (the -1 from -x), so both ends go down: as x→-∞, p(x)→-∞ and as x→∞, p(x)→-∞. Choice A correctly identifies all zeros with their multiplicities, states that the graph crosses at 0 and -2 while touching at 3, and has the correct end behavior for an even-degree polynomial with negative leading coefficient. Choice B has the same zero information but incorrectly shows both ends going up, which would be true for a positive leading coefficient. Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero.

Question 2

Write the system as AX=BAX=BAX=B (variable order x,y,zx, y, zx,y,z). Be sure to include 000 for any missing variable:

x+3y=5\\ 2x-z=1\\ y+2z=4 \end{cases}$$
  1. [13020−1012][xyz]=[514]\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}​120​301​0−12​​​xyz​​=​514​​ (correct answer)
  2. [13020−1012][xyz]=[541]\begin{bmatrix}1&3&0\\2&0&-1\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\4\\1\end{bmatrix}​120​301​0−12​​​xyz​​=​541​​
  3. [132−112][xyz]=[514]\begin{bmatrix}1&3\\2&-1\\1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}​121​3−12​​​xyz​​=​514​​
  4. [1032−10012][xyz]=[514]\begin{bmatrix}1&0&3\\2&-1&0\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\1\\4\end{bmatrix}​120​0−11​302​​​xyz​​=​514​​

Explanation: This question tests your ability to represent a system with missing variables as a matrix equation, requiring you to insert 0 coefficients where variables are absent. Matrix form AX = B requires the coefficient matrix A to have consistent dimensions: if using three variables (x, y, z), every row must have three entries, using 0 for missing variables. To convert x + 3y = 5, 2x - z = 1, and y + 2z = 4 to matrix form: (1) Equation 1 (x + 3y = 5) has no z term, so write as x + 3y + 0z = 5, giving row 1: [1, 3, 0]. (2) Equation 2 (2x - z = 1) has no y term, so write as 2x + 0y - z = 1, giving row 2: [2, 0, -1]. (3) Equation 3 (y + 2z = 4) has no x term, so write as 0x + y + 2z = 4, giving row 3: [0, 1, 2]. Therefore A = [[1, 3, 0], [2, 0, -1], [0, 1, 2]], X = [x, y, z], B = [5, 1, 4]. Choice A correctly includes all necessary zeros to maintain proper matrix structure. Choice B attempts a 3×2 matrix which cannot multiply with a 3×1 vector - matrix dimensions must be compatible for multiplication! When variables are missing from equations, always insert 0 coefficients in the appropriate positions to maintain rectangular matrix form. To verify: [[1, 3, 0], [2, 0, -1], [0, 1, 2]] times [x, y, z] gives [x + 3y + 0z, 2x + 0y - z, 0x + y + 2z] = [x + 3y, 2x - z, y + 2z], which equals [5, 1, 4], confirming our original system!

Question 3

Four functions are defined: linear f(x)=xf(x)=xf(x)=x, quadratic g(x)=x2g(x)=x^2g(x)=x2, polynomial h(x)=x6h(x)=x^6h(x)=x6, and exponential p(x)=2xp(x)=2^xp(x)=2x. Which function eventually grows fastest as x→∞x\to\inftyx→∞?

  1. h(x)=x6h(x)=x^6h(x)=x6
  2. g(x)=x2g(x)=x^2g(x)=x2
  3. p(x)=2xp(x)=2^xp(x)=2x (correct answer)
  4. f(x)=xf(x)=xf(x)=x

Explanation: This question tests your understanding of a fundamental mathematical principle: exponential functions eventually grow faster than any polynomial function (even very high-degree polynomials) when we look at sufficiently large x-values. The growth hierarchy is: exponential > any polynomial > linear (for large x). Even though x^6 grows very rapidly initially and might dominate 2^x for small values, the exponential 2^x will eventually exceed x^6 if you go far enough. This happens because exponential growth is multiplicative (multiply by same factor repeatedly, which compounds), while polynomial growth is essentially additive-based (even with acceleration). Multiplicative compounding always beats any additive pattern eventually—it's why compound interest (exponential) is so powerful long-term compared to simple interest (linear)! Let's verify when 2^x overtakes x^6: At x=10: h(10)=1,000,000, p(10)=1,024 (polynomial much larger). At x=20: h(20)=64,000,000, p(20)=1,048,576 (polynomial still larger). At x=30: h(30)=729,000,000, p(30)=1,073,741,824 (exponential just overtook!). At x=50: h(50)=15,625,000,000, p(50)=1,125,899,906,842,624 (exponential dominates massively!). Choice C correctly identifies p(x)=2^x as the eventual winner. Choice A incorrectly suggests the high-degree polynomial x^6 will dominate—this contradicts the fundamental growth hierarchy. Strategy insight: High-degree polynomials can fool you because they grow so fast initially. But remember, no matter how high the degree, exponentials always win eventually. It's like a race where one runner (polynomial) sprints early but tires, while the other (exponential) starts slow but accelerates forever!

Question 4

Use the conjugate property to simplify. What is (2+5i)(2−5i)?(2+5i)(2-5i)?(2+5i)(2−5i)?

  1. 4+25i4+25i4+25i
  2. 292929 (correct answer)
  3. 4−25i4-25i4−25i
  4. −21-21−21

Explanation: This question tests your understanding of complex numbers—numbers in the form a+bia + bia+bi where i is the imaginary unit with i2=−1i^2 = -1i2=−1—and using the conjugate property for multiplication. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}i=−1​, so i2=−1i^2 = -1i2=−1. The product (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2(a+bi)(a−bi)=a2+b2, always real: here (2+5i)(2−5i)=4+25=29(2 + 5i)(2 - 5i) = 4 + 25 = 29(2+5i)(2−5i)=4+25=29—outstanding! Choice B correctly applies this difference of squares pattern. Choice A might come from mistakenly subtracting instead of adding the squares, but remember it's plus b² from −(bi)2=−b2i2=+b2- (bi)^2 = -b^2 i^2 = +b^2−(bi)2=−b2i2=+b2. This is key for moduli and division too— you've nailed it! Build on this for more complex problems; you're unstoppable.

Question 5

Use the Rational Zeros Theorem to help factor completely: P(x)=x3−7x−6.P(x)=x^3-7x-6.P(x)=x3−7x−6. First list possible rational zeros, then test to find an actual zero, and factor the polynomial.

  1. (x−3)(x+1)(x+2)(x-3)(x+1)(x+2)(x−3)(x+1)(x+2) (correct answer)
  2. (x−3)(x2+3x+2)(x-3)(x^2 + 3x + 2)(x−3)(x2+3x+2)
  3. (x+1)(x2−x−6)(x+1)(x^2 - x - 6)(x+1)(x2−x−6)
  4. (x−2)(x2+2x+3)(x-2)(x^2 + 2x + 3)(x−2)(x2+2x+3)

Explanation: This question tests your understanding of the Rational Zeros Theorem—a powerful tool for finding possible rational zeros of polynomials, dramatically narrowing what values to test. The Rational Zeros Theorem states that if polynomial P(x)P(x)P(x) with integer coefficients has a rational zero p/qp/qp/q (in lowest terms), then ppp must be a factor of the constant term and qqq must be a factor of the leading coefficient. For P(x)=x3−7x−6P(x)=x^3 -7x -6P(x)=x3−7x−6, candidates ±1,±2,±3,±6±1, ±2, ±3, ±6±1,±2,±3,±6; testing finds P(−2)=0P(-2)=0P(−2)=0, P(−1)=0P(-1)=0P(−1)=0, P(3)=0P(3)=0P(3)=0, allowing complete factorization as (x+2)(x+1)(x−3)(x+2)(x+1)(x-3)(x+2)(x+1)(x−3). Choice D correctly provides the full linear factorization using these zeros, verified by multiplication back to original. Choice A stops at linear times quadratic, failing to factor further despite quadratic splitting into rationals—a common oversight. Rational Zeros Theorem application process: list and test candidates, factor out each zero found via synthetic division, repeat on lower-degree quotient until fully factored. This method efficiently reveals all rational factors—excellent work applying it step by step!

Question 6

Find the inverse function f−1(x)f^{-1}(x)f−1(x) for f(x)=4x+1f(x)=\dfrac{4}{x+1}f(x)=x+14​ with domain restriction x≠−1x\ne -1x=−1.​

  1. f−1(x)=4x+1f^{-1}(x)=\dfrac{4}{x+1}f−1(x)=x+14​ with x≠−1x\ne -1x=−1
  2. f−1(x)=4x−1f^{-1}(x)=\dfrac{4}{x}-1f−1(x)=x4​−1 with x≠0x\ne 0x=0 (correct answer)
  3. f−1(x)=x+14f^{-1}(x)=\dfrac{x+1}{4}f−1(x)=4x+1​
  4. f−1(x)=14x+1f^{-1}(x)=\dfrac{1}{4x+1}f−1(x)=4x+11​

Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = 4/(x+1), we write y = 4/(x+1), swap to get x = 4/(y+1), then solve for y: multiply both sides by (y+1) to get x(y+1) = 4, which gives xy + x = 4, then xy = 4 - x, so y = (4-x)/x = 4/x - 1. Therefore f⁻¹(x) = 4/x - 1. Choice B correctly finds f⁻¹(x) = 4/x - 1 with x ≠ 0 by swapping and solving properly. Choice A incorrectly gets 4/(x+1) which would make it its own inverse (not true here), while Choice D confuses inverse with reciprocal. The domain restriction x ≠ 0 comes from division by x in the inverse formula. Verify: f(f⁻¹(x)) = f(4/x - 1) = 4/((4/x - 1) + 1) = 4/(4/x) = x ✓.

Question 7

Divide and express the result in quotient-plus-remainder form: x3−8x−2=q(x)+r(x)x−2,deg⁡(r)<1.\frac{x^3-8}{x-2}=q(x)+\frac{r(x)}{x-2},\quad \deg(r)<1.x−2x3−8​=q(x)+x−2r(x)​,deg(r)<1.

  1. x2+2x+4+0x−2x^2+2x+4+\dfrac{0}{x-2}x2+2x+4+x−20​ (correct answer)
  2. x2+2x+4+8x−2x^2+2x+4+\dfrac{8}{x-2}x2+2x+4+x−28​
  3. x2−2x+4+0x−2x^2-2x+4+\dfrac{0}{x-2}x2−2x+4+x−20​
  4. x2+4x+4+0x−2x^2+4x+4+\dfrac{0}{x-2}x2+4x+4+x−20​

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Recognizing x³ - 8 as a difference of cubes, factor as (x - 2)(x² + 2x + 4) with no remainder, or use synthetic division with root 2: bring down 1, multiply by 2 for 2, add to 0 for 2, multiply by 2 for 4, add to 0 for 4, multiply by 2 for 8, add to -8 for 0, confirming quotient x² + 2x + 4 and remainder 0. Choice A correctly divides to get quotient x² + 2x + 4 and remainder 0 with deg(r) < 1. A distractor like choice B might add an unnecessary remainder from misfactoring, but the clean division shows exact fit. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (x - 2) (degree 1), remainder must be degree 0 (constant).

Question 8

Solve the system using the substitution method:

x - y = 1\\ y = 3x - 5 \end{cases}$$ What is the solution $(x, y)$?
  1. (2,1)(2, 1)(2,1) (correct answer)
  2. (3,4)(3, 4)(3,4)
  3. (1,−2)(1, -2)(1,−2)
  4. (2,−1)(2, -1)(2,−1)

Explanation: This question tests your ability to solve systems of linear equations using substitution when one equation is already solved for a variable. A system of linear equations can be solved by substitution: replace one variable with its expression from one equation into the other equation. We have x - y = 1 and y = 3x - 5. Since the second equation already gives y in terms of x, substitute this into the first equation: x - (3x - 5) = 1. Carefully distribute the negative sign: x - 3x + 5 = 1, which gives -2x + 5 = 1, so -2x = -4, thus x = 2. Now back-substitute into y = 3x - 5: y = 3(2) - 5 = 6 - 5 = 1. Choice A correctly gives the solution (2, 1), which we verify: first equation: 2 - 1 = 1 ✓; second equation: 1 = 3(2) - 5 = 6 - 5 = 1 ✓. Choice D might tempt students who get x = 2 correctly but make a sign error in back-substitution, getting y = -1 instead of y = 1. The substitution method is powerful when one equation is already solved for a variable: (1) substitute the expression into the other equation, (2) solve for the remaining variable, (3) back-substitute to find the other variable, (4) write as an ordered pair (x, y), (5) always verify in both original equations. Remember to use parentheses when substituting to avoid sign errors—this is where many mistakes happen!

Question 9

Add and simplify to lowest terms. (Rational expressions behave like fractions: find an LCD, combine, and reduce.)

3x−2+2x+2\frac{3}{x-2}+\frac{2}{x+2}x−23​+x+22​

(Assume x≠2,−2x\neq 2,-2x=2,−2.)

  1. 5x2−4\frac{5}{x^2-4}x2−45​
  2. 5x−2x2−4\frac{5x-2}{x^2-4}x2−45x−2​
  3. 5x+2x2−4\frac{5x+2}{x^2-4}x2−45x+2​ (correct answer)
  4. 3x2−4+2x2−4\frac{3}{x^2-4}+\frac{2}{x^2-4}x2−43​+x2−42​

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To add, find LCD (x−2)(x+2)(x-2)(x+2)(x−2)(x+2); rewrite as 3(x+2)+2(x−2)(x−2)(x+2)=3x+6+2x−4x2−4=5x+2x2−4\frac{3(x+2) + 2(x-2)}{(x-2)(x+2)} = \frac{3x+6 + 2x-4}{x^2-4} = \frac{5x+2}{x^2-4}(x−2)(x+2)3(x+2)+2(x−2)​=x2−43x+6+2x−4​=x2−45x+2​, which can't be factored further to cancel, so it's simplified. Choice C correctly finds the LCD, combines the numerators, and simplifies, giving 5x+2x2−4\frac{5x+2}{x^2-4}x2−45x+2​ in lowest terms. Choice A fails by incorrectly adding the numerators without distributing, perhaps treating it like same denominators when they're not—always use the LCD for different denominators! The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply (or flip and multiply for division). Addition and subtraction are harder—factor denominators, find LCD as the LCM of denominator factors, rewrite each fraction with LCD, then add/subtract numerators. Always simplify at the end by factoring the numerator and canceling any new common factors with the denominator. This systematic approach prevents errors!

Question 10

A car’s value is recorded at the end of each year.

Year: 0, 1, 2, 3 Value ($): 18000, 15300, 13005, 11054.25

Classify the pattern as exponential growth, exponential decay, or neither, and identify the percent rate per year.

  1. Exponential decay at 15% per year (decay factor b=0.85b=0.85b=0.85) (correct answer)
  2. Linear decay: decreases by $2700 each year
  3. Neither; the ratios between consecutive years are not constant
  4. Exponential growth at 15% per year (growth factor b=1.15b=1.15b=1.15)

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! Let's check the ratios: 15300/18000 = 0.85, 13005/15300 = 0.85, 11054.25/13005 = 0.85. All ratios equal 0.85, confirming exponential decay with base b = 0.85. Since 0.85 = 1 - 0.15, this represents 15% decay per year. Choice A correctly identifies exponential decay at 15% per year (decay factor b=0.85) through the constant ratio analysis. Choice C incorrectly suggests linear decay, but the differences aren't constant (18000-15300=2700, but 15300-13005=2295, not 2700). The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms!

Question 11

Use structure to factor the expression completely. Notice that it can be seen as a difference of squares more than once:

x4−y4x^4 - y^4x4−y4

  1. (x+y)2(x−y)2(x+y)^2(x-y)^2(x+y)2(x−y)2
  2. (x2+y2)(x+y)(x−y)(x^2+y^2)(x+y)(x-y)(x2+y2)(x+y)(x−y) (correct answer)
  3. (x2−y2)(x2+y2)(x^2-y^2)(x^2+y^2)(x2−y2)(x2+y2)
  4. (x−y)(x3+y3)(x-y)(x^3+y^3)(x−y)(x3+y3)

Explanation: This question tests your ability to recognize mathematical patterns and structure in expressions—like difference of squares, sum and difference of cubes, or perfect square trinomials—and use those patterns to rewrite or factor the expression. Using structure means seeing an expression not just as a collection of terms, but as fitting a known pattern that enables transformation: for example, x⁴ - y⁴ might look intimidating, but seeing it as (x²)² - (y²)² reveals it's a difference of squares! Let's factor x⁴ - y⁴ step by step: (1) recognize it as (x²)² - (y²)² (difference of squares with a = x² and b = y²), (2) apply the pattern a² - b² = (a + b)(a - b) to get (x² + y²)(x² - y²), (3) notice that x² - y² is also a difference of squares!, (4) factor x² - y² further as (x + y)(x - y), giving us the complete factorization: (x² + y²)(x + y)(x - y). Choice C correctly recognizes both levels of the difference of squares pattern and factors completely—the order of factors doesn't matter in multiplication, so (x² + y²)(x + y)(x - y) is perfect. Choice B stops after the first application of difference of squares at (x² - y²)(x² + y²) without recognizing that x² - y² can be factored further—always check if your factors contain more patterns! Pattern recognition is like peeling an onion: sometimes there are multiple layers of structure, and recognizing each layer leads to complete factorization. When you see fourth powers minus fourth powers, think 'double difference of squares' and factor twice!

Question 12

Determine whether f(x)=x4−2x2f(x)=x^4-2x^2f(x)=x4−2x2 is even, odd, or neither.​

  1. Even (correct answer)
  2. Odd
  3. Neither even nor odd
  4. Both even and odd

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. To test f(x) = x⁴ - 2x², we find f(-x) = (-x)⁴ - 2(-x)² = x⁴ - 2x² = f(x), confirming it's even since f(-x) equals f(x). Choice A correctly identifies this function as even. The function contains only even powers of x (x⁴ and x²), which is a quick way to recognize even functions—all terms with even powers preserve sign when x is replaced by -x. For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither.

Question 13

Find the modulus rrr and an argument θ\thetaθ (in radians) for the complex number −3−i-\sqrt{3}-i−3​−i, then choose its polar form r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta)r(cosθ+isinθ). Use r=a2+b2r=\sqrt{a^2+b^2}r=a2+b2​ and θ=arctan⁡(ba)\theta=\arctan\left(\frac{b}{a}\right)θ=arctan(ab​) with quadrant adjustment. (Argument is measured counterclockwise from the positive real axis.)

  1. 2(cos⁡5π6+isin⁡5π6)2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)2(cos65π​+isin65π​)
  2. 3(cos⁡7π6+isin⁡7π6)\sqrt{3}\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right)3​(cos67π​+isin67π​)
  3. 2(cos⁡7π6+isin⁡7π6)2\left(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\right)2(cos67π​+isin67π​) (correct answer)
  4. 2(cos⁡(−π6)+isin⁡(−π6))2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)2(cos(−6π​)+isin(−6π​))

Explanation: This question tests your ability to convert complex numbers between rectangular form a + bi and polar form r(cos θ + i sin θ), which represent the same number using Cartesian coordinates versus magnitude and direction. Rectangular form a + bi uses horizontal (real) and vertical (imaginary) components, while polar form r(cos θ + i sin θ) uses distance from origin (modulus r) and angle from positive real axis (argument θ, measured counterclockwise). To convert -√3 - i to polar form: (1) Find modulus: r = √((-√3)² + (-1)²) = √(3 + 1) = √4 = 2. (2) Find argument: reference angle = arctan(|-1|/|-√3|) = arctan(1/√3) = π/6. Since a = -√3 < 0 and b = -1 < 0, we're in Quadrant 3, so θ = π + π/6 = 7π/6. (3) Write polar form: 2(cos(7π/6) + i sin(7π/6)). Choice B correctly calculates modulus using Pythagorean theorem and determines argument with proper quadrant adjustment for Q3. Choice A incorrectly places the complex number in Q2 with θ = 5π/6, but -√3 - i has both negative real and imaginary parts, placing it in Q3 where θ = 7π/6! Rectangular to polar recipe: (1) Calculate r = √(a² + b²). (2) Find reference angle. (3) Determine quadrant from signs. (4) For Q3 (--), use θ = π + reference. For -√3 - i, this gives 2(cos(7π/6) + i sin(7π/6)).

Question 14

Write an expression for f−1(x)f^{-1}(x)f−1(x) given f(x)=3x−7f(x)=3x-7f(x)=3x−7.​​​

  1. f−1(x)=3x+7f^{-1}(x)=3x+7f−1(x)=3x+7
  2. f−1(x)=x+73f^{-1}(x)=\dfrac{x+7}{3}f−1(x)=3x+7​ (correct answer)
  3. f−1(x)=x−73f^{-1}(x)=\dfrac{x-7}{3}f−1(x)=3x−7​
  4. f−1(x)=3x−7f^{-1}(x)=\dfrac{3}{x-7}f−1(x)=x−73​

Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For the linear function f(x) = 3x - 7, let's apply swap-and-solve: Start with y = 3x - 7, swap variables to get x = 3y - 7, then solve for y by adding 7 to both sides to get x + 7 = 3y, and finally divide by 3 to get y = (x + 7)/3, so f⁻¹(x) = (x + 7)/3. Choice B correctly finds f⁻¹(x) = (x + 7)/3 by properly swapping and solving—it undoes 'multiply by 3 then subtract 7' with 'add 7 then divide by 3.' Choice C shows a sign error (using x - 7 instead of x + 7), while Choice D incorrectly treats this as a reciprocal function rather than finding the true inverse. The swap-and-solve recipe: (1) Replace f(x) with y to get y = 3x - 7, (2) Swap every x with y and every y with x: x = 3y - 7, (3) Solve this equation for y using algebra (add 7, then divide by 3), (4) The expression for y is your f⁻¹(x) = (x + 7)/3. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation—if f(x) = 3x - 7 does 'multiply by 3, then subtract 7,' the inverse should do 'add 7, then divide by 3': (x + 7)/3!

Question 15

A sequence can be viewed as a discrete function fff whose domain is the positive integers n∈{1,2,3,… }n \in \{1,2,3,\dots\}n∈{1,2,3,…}. For the Fibonacci sequence defined by f(1)=f(2)=1f(1)=f(2)=1f(1)=f(2)=1 and f(n)=f(n−1)+f(n−2)f(n)=f(n-1)+f(n-2)f(n)=f(n−1)+f(n−2) for n≥3n\ge 3n≥3, what is f(7)f(7)f(7)?​

  1. 111111
  2. 212121
  3. 888
  4. 131313 (correct answer)

Explanation: This question tests your understanding that sequences are special functions where the domain is a subset of the integers—meaning they're only defined for whole number inputs like 1, 2, 3, not for values in between. A sequence is a function whose inputs are integers (often starting at 1 or 0) and whose outputs are the sequence terms: f(1) = first term, f(2) = second term, etc. Unlike continuous functions that have values for all real numbers, sequences have values only at integer positions. If you graphed a sequence, you'd see separate dots (like (1, a₁), (2, a₂), (3, a₃)), not a connected curve. This discrete nature makes sequences fundamentally different from functions like f(x) = x²! Recursive definitions specify how to get each term from previous term(s): the Fibonacci sequence is the classic example with f(1) = 1, f(2) = 1 and f(n) = f(n-1) + f(n-2) for n ≥ 3, meaning each term equals the sum of the two before it. To find f(7), you can't jump there directly—you must calculate f(3) = f(2) + f(1) = 1 + 1 = 2, then f(4) = 2 + 1 = 3, then f(5) = 3 + 2 = 5, f(6) = 5 + 3 = 8, finally f(7) = 8 + 5 = 13. Choice C correctly evaluates recursively as 13. A common distractor like D (21) might come from miscounting terms or starting from f(0), but remember to use the given initial values precisely. Evaluating recursive sequences systematically: (1) Write down the initial value(s) clearly: f(1) = 1, f(2) = 1, (2) Set up a table or list: n = 1, 2, 3, ... down one side, (3) Apply the recursive rule one step at a time: for Fibonacci, f(3) = f(2) + f(1), so look up f(2) and f(1) from what you've already calculated, add them, write the result, (4) Continue until you reach the desired term. Don't skip steps—recursion is sequential by nature! Keep practicing, and you'll master these quickly!

Question 16

What is the coefficient of x2y3x^2y^3x2y3 in the expansion of (x+y)5?(x+y)^5?(x+y)5? (You may use Pascal's Triangle.)​

  1. 5
  2. 10 (correct answer)
  3. 20
  4. 6

Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. To find the coefficient of x^2y^3 in (x + y)^5, we need the term where x has power 2 and y has power 3 (note: 2 + 3 = 5, which matches our exponent). In the expansion, this is the 4th term (counting from term 1), and row 5 of Pascal's Triangle is 1, 5, 10, 10, 5, 1—so the 4th coefficient is 10. Choice B correctly identifies this coefficient as 10, while Choice A (5) would be the coefficient of x^4y or xy^4, and Choice C (20) doesn't appear in row 5 at all. Quick Pascal's Triangle construction: write 1s down both edges, then for interior numbers, add the two directly above—row 5 builds from row 4 (1, 4, 6, 4, 1) to get 1, 1+4=5, 4+6=10, 6+4=10, 4+1=5, 1!

Question 17

A one-to-one function fff is shown by the table, so fff has an inverse. Use the table to evaluate f−1(21)f^{-1}(21)f−1(21). (Because inverse pairs swap, if f(x)=21f(x)=21f(x)=21 then f−1(21)=xf^{-1}(21)=xf−1(21)=x.)

xxxf(x)f(x)f(x)
29
413
617
821
1025
  1. f−1(21)=8f^{-1}(21)=8f−1(21)=8 (correct answer)
  2. f−1(21)=6f^{-1}(21)=6f−1(21)=6
  3. f−1(21)=10f^{-1}(21)=10f−1(21)=10
  4. f−1(21)=21f^{-1}(21)=21f−1(21)=21

Explanation: This question tests your ability to read values of an inverse function from a table or graph using the fundamental property that inverses swap inputs and outputs. If f has a table or graph showing its (x, y) pairs, then f⁻¹ has the same pairs but swapped: every (x, y) on f becomes (y, x) on f⁻¹. So to find f⁻¹(a), you don't need the inverse's table or graph—just find where y = a in f's representation and read the corresponding x-value. That x is your answer! For example, if f's table shows x = 3 gives y = 7, then f⁻¹(7) = 3. The inverse reverses the input-output relationship. To evaluate f⁻¹(21), I scan the f(x) column for 21: f(2) = 9, f(4) = 13, f(6) = 17, f(8) = 21, f(10) = 25. Found it! When x = 8, f(x) = 21. Therefore, f⁻¹(21) = 8. Choice C correctly finds f⁻¹(21) = 8 by locating where f(x) = 21 in the table. Choice A incorrectly assumes f⁻¹(21) = 21, which would require f(21) = 21, but 21 isn't even an input value in the table. Table reading strategy for f⁻¹(a): (1) Look in the f(x) column (the outputs) for the value a, (2) When you find a, look at the x-value in that same row, (3) That x-value is f⁻¹(a). The swap happens automatically when you read this way!

Question 18

Apply the sum of cubes identity to factor the polynomial x3+8x^3+8x3+8 completely over the integers.​

  1. (x+2)(x2+2x+4)(x+2)(x^2+2x+4)(x+2)(x2+2x+4)
  2. (x+8)(x2−8x+64)(x+8)(x^2-8x+64)(x+8)(x2−8x+64)
  3. (x+2)(x2−2x+4)(x+2)(x^2-2x+4)(x+2)(x2−2x+4) (correct answer)
  4. (x−2)(x2+2x+4)(x-2)(x^2+2x+4)(x−2)(x2+2x+4)

Explanation: This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. The sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²) is a powerful factoring tool. To apply it to x³ + 8, we need to recognize that 8 = 2³, so we have x³ + 2³. Using the identity with a = x and b = 2: x³ + 2³ = (x + 2)(x² - x(2) + 2²) = (x + 2)(x² - 2x + 4). Choice C correctly applies the sum of cubes pattern, giving (x + 2)(x² - 2x + 4). Choice A has the wrong sign in the first factor (should be + not just +), Choice B treats 8 as if it equals 8³ instead of 2³, and Choice D uses the wrong sign pattern (that would be for difference of cubes). To apply sum/difference of cubes identities: (1) Recognize perfect cubes (8 = 2³, 27 = 3³, 64 = 4³, etc.), (2) Identify whether it's sum (+) or difference (-), (3) Apply the correct pattern: sum uses (a + b)(a² - ab + b²), difference uses (a - b)(a² + ab + b²), (4) Simplify the resulting factors. These identities are invaluable for factoring cubic expressions that would otherwise be difficult to handle!

Question 19

The function f(x)=x−2f(x)=\sqrt{x-2}f(x)=x−2​ can be used to model the side length (in meters) of a square when xxx represents the measured area (in square meters) plus 2. What is the realistic domain for xxx?​

  1. x∈(−∞,∞)x\in(-\infty,\infty)x∈(−∞,∞)
  2. x∈(2,∞)x\in(2,\infty)x∈(2,∞)
  3. x∈[2,∞)x\in[2,\infty)x∈[2,∞) (correct answer)
  4. x∈(−∞,2]x\in(-\infty,2]x∈(−∞,2]

Explanation: This question tests your understanding of how a function's domain relates both to its graph (which x-values have points) and to the real-world context (which values make sense practically). The domain is the set of allowed inputs, and for real-world functions, we distinguish: (1) mathematical domain (what the formula allows—like 'all reals' for polynomials), and (2) realistic domain (what makes sense in context—like 'positive integers' for number of items or 't ≥ 0' for time). The realistic domain is usually more restrictive because real-world constraints (can't have negative time, can't produce -5 items, can't work 1.7 hours if discrete) limit what mathematical values are meaningful. Discrete vs continuous domains depend on what you're measuring: if counting distinct objects (people, tickets, items sold), the domain is discrete—only integers work, graph as separate dots. If measuring continuous quantities (time, distance, temperature, money as a continuous quantity), the domain is continuous—any value in an interval works, graph as connected line/curve. The physical nature of the quantity determines which! You can have 2.7 hours but not 2.7 people. For f(x)=√(x-2) modeling side length, both mathematical and realistic domains require x ≥ 2 to ensure non-negative under the square root and meaningful area (x = area + 2 ≥ 2). Choice C correctly identifies the domain as x ∈ [2,∞) based on the square root constraint and non-negative side length. Choice B fails by excluding x=2, where side length is 0, which could be a degenerate case but is mathematically allowed. Domain determination framework: (1) Start with mathematical domain—what does the formula allow? (polynomials: all reals; √(expression): expression ≥ 0; 1/expression: expression ≠ 0; log(expression): expression > 0), (2) Apply context constraints—can variable be negative? Are there upper limits? Must it be integer?, (3) Combine all restrictions—domain is where ALL conditions are met. Example: f(x) = √(x - 3) for measuring length has mathematical domain x ≥ 3, and realistic domain also x ≥ 3 (lengths are non-negative, and formula requires x ≥ 3, so both agree). Discrete vs continuous quick-check: ask 'can there be in-between values?' If you're modeling number of students in a classroom, going from 20 to 21 students happens instantly (no 20.5 students exist)—discrete! If you're modeling temperature change from 20°C to 21°C, every value in between occurs—continuous! Context tells you: countable/distinct objects → discrete (dots on graph), measurable/varying quantities → continuous (line/curve on graph). This determines how you graph and what domain notation to use!

Question 20

Use technology to solve AX=BAX=BAX=B by computing X=A−1BX=A^{-1}BX=A−1B.

A=[210131024],B=[51016]A=\begin{bmatrix}2&1&0\\1&3&1\\0&2&4\end{bmatrix},\quad B=\begin{bmatrix}5\\10\\16\end{bmatrix}A=​210​132​014​​,B=​51016​​

What is XXX?

  1. [231]\begin{bmatrix}2\\3\\1\end{bmatrix}​231​​
  2. [123]\begin{bmatrix}1\\2\\3\end{bmatrix}​123​​ (correct answer)
  3. [132]\begin{bmatrix}1\\3\\2\end{bmatrix}​132​​
  4. [213]\begin{bmatrix}2\\1\\3\end{bmatrix}​213​​

Explanation: This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. When a system is written as matrix equation AX=BAX = BAX=B (where AAA is coefficient matrix, XXX is variable vector, BBB is constant vector), we can solve for XXX by multiplying both sides by the inverse matrix A−1A^{-1}A−1 (if it exists): A−1×AX=A−1×BA^{-1} \times AX = A^{-1} \times BA−1×AX=A−1×B. The left side simplifies to (A−1×A)×X=I×X=X(A^{-1} \times A) \times X = I \times X = X(A−1×A)×X=I×X=X (the identity matrix III acts like the number 1), giving X=A−1×BX = A^{-1} \times BX=A−1×B. This single matrix multiplication produces the solution vector! The method works if and only if AAA has an inverse, which requires the determinant of AAA to be nonzero. For 3×3 or larger matrices, we use technology (graphing calculator or computer) to find the inverse because hand calculation is extremely tedious—the standard explicitly acknowledges technology use for these sizes. Given A=[210131024]A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 2 & 4 \end{bmatrix}A=​210​132​014​​ and B=[51016]B = \begin{bmatrix} 5 \\ 10 \\ 16 \end{bmatrix}B=​51016​​, we use technology to find A−1A^{-1}A−1, then compute X=A−1×BX = A^{-1} \times BX=A−1×B. Enter matrix AAA into your calculator, find its inverse using the INV or x−1x^{-1}x−1 function, then multiply by vector BBB. The calculation yields X=[123]X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}X=​123​​, meaning x=1x = 1x=1, y=2y = 2y=2, z=3z = 3z=3. Choice A correctly identifies this solution vector. Choice B would give x=2x = 2x=2, y=1y = 1y=1, z=3z = 3z=3, which doesn't satisfy the original system—always verify your answer! The matrix inverse solving process: (1) Write system as AX=BAX = BAX=B (represent as matrix equation), (2) Check if AAA is invertible: for 3×3, use technology to attempt finding inverse, (3) Find A−1A^{-1}A−1: use calculator inverse function, (4) Multiply: compute A−1×BA^{-1} \times BA−1×B to get solution vector XXX, (5) Extract solutions: read xxx, yyy, zzz values from the result vector. Technology makes matrix inverse method practical: for 3×3 systems, the calculator handles finding the inverse (which involves computing 9 cofactors and dividing by the determinant). This lets you focus on understanding the method rather than getting bogged down in arithmetic!

Question 21

A physics model for distance is d(t)=v (t−2)2+k,d(t)=v\,(t-2)^2+k,d(t)=v(t−2)2+k, where vvv and kkk are constants and ttt is time. By chunking (t−2)2(t-2)^2(t−2)2 as a single entity, which statement correctly identifies the part independent of vvv?

  1. The term kkk is independent of vvv; changing vvv scales only the chunk (t−2)2(t-2)^2(t−2)2 and does not change kkk. (correct answer)
  2. No part is independent of vvv because vvv appears in the expression.
  3. The chunk (t−2)2(t-2)^2(t−2)2 depends on vvv because it is multiplied by vvv.
  4. Both (t−2)2(t-2)^2(t−2)2 and kkk are independent of vvv because vvv is outside the parentheses.

Explanation: This question tests your ability to interpret complicated expressions by viewing one or more parts as a single entity—a powerful technique for understanding structure and relationships in complex algebraic forms. When expressions get complex, chunking (viewing sub-expressions as single units) reveals structure: for example, in d(t) = v (t - 2)^2 + k, viewing (t - 2)^2 as a chunk shows it's v times chunk plus k, with k independent of v. The key insight: k doesn't depend on v! Changing v scales the chunk term but leaves k unchanged. This shows d as a scaled quadratic plus constant. Viewing parts separately highlights additive independence. In d(t) = v (t - 2)^2 + k, let's view (t - 2)^2 as a single entity—call it S for squared term. Then d = v S + k, a linear combination. Now we can see: (1) v scales S, (2) S depends on t but not v, (3) k is constant and independent of v—changing v affects v S but not k. This chunking reveals the parabolic structure shifted by k. Understanding this aids in physics interpretations! Choice B correctly identifies k as independent of v, and changing v scales only the chunk without changing k, capturing the additive structure. Choice A claims the chunk depends on v because multiplied by it, but dependence means v is inside the chunk—it's not; multiplication doesn't embed v in S. Independence is about contents! Chunking strategy for complicated expressions: (1) For sums, view each term as a unit, especially if one is constant, (2) Check independence—if a term lacks a variable, it's independent, (3) Use substitution: let s = chunk, rewrite as v s + k, (4) Ask: changing v affects s? No, only the product. Common chunking patterns: In a(expression)^power: view (expression)^power as single factor independent of a. In (polynomial) divided by (polynomial): view numerator and denominator as separate entities. In sum of similar terms like 5(x + 1)^2 - 3(x + 1): view (x + 1) as single unit (substitute u = x + 1 gives 5u^2 - 3u, revealing quadratic structure). In nested function f(g(x)): view g(x) as the input entity to outer function f. Chunking isn't arbitrary—chunk in ways that reveal structure, independence, or simplify analysis! Superb job—you're excelling!

Question 22

The expression x+1⋅(x+1)3/2+5x+1−2(x+1)5/2\sqrt{x+1} \cdot (x+1)^{3/2} + 5\sqrt{x+1} - 2(x+1)^{5/2}x+1​⋅(x+1)3/2+5x+1​−2(x+1)5/2 involves multiple powers of the same radical. Which substitution best reveals the polynomial structure?

  1. Let u=xu = xu=x, transforming to u+1⋅(u+1)3/2+5u+1−2(u+1)5/2\sqrt{u+1} \cdot (u+1)^{3/2} + 5\sqrt{u+1} - 2(u+1)^{5/2}u+1​⋅(u+1)3/2+5u+1​−2(u+1)5/2 with no simplification
  2. Let u=x+1u = x+1u=x+1, transforming to u⋅u3/2+5u−2u5/2=u2+5u1/2−2u5/2\sqrt{u} \cdot u^{3/2} + 5\sqrt{u} - 2u^{5/2} = u^2 + 5u^{1/2} - 2u^{5/2}u​⋅u3/2+5u​−2u5/2=u2+5u1/2−2u5/2
  3. Let u=(x+1)1/2u = (x+1)^{1/2}u=(x+1)1/2, transforming to u⋅u3+5u−2u5=u(u3+5−2u4)u \cdot u^3 + 5u - 2u^5 = u(u^3 + 5 - 2u^4)u⋅u3+5u−2u5=u(u3+5−2u4)
  4. Let u=x+1u = \sqrt{x+1}u=x+1​, transforming to u⋅u3+5u−2u5=u4+5u−2u5=u(u3+5−2u4)u \cdot u^3 + 5u - 2u^5 = u^4 + 5u - 2u^5 = u(u^3 + 5 - 2u^4)u⋅u3+5u−2u5=u4+5u−2u5=u(u3+5−2u4) (correct answer)

Explanation: When you encounter expressions with multiple powers of the same radical, substitution can reveal hidden polynomial structure and make the expression much easier to work with. The key insight is identifying what's being repeated throughout the expression. Here, every term contains some power of x+1\sqrt{x+1}x+1​. Let's set u=x+1u = \sqrt{x+1}u=x+1​, which means u2=x+1u^2 = x+1u2=x+1. Now we can rewrite each term: x+1=u\sqrt{x+1} = ux+1​=u, (x+1)3/2=(x+1)3=u3(x+1)^{3/2} = (\sqrt{x+1})^3 = u^3(x+1)3/2=(x+1​)3=u3, and (x+1)5/2=(x+1)5=u5(x+1)^{5/2} = (\sqrt{x+1})^5 = u^5(x+1)5/2=(x+1​)5=u5. The expression becomes: u⋅u3+5u−2u5=u4+5u−2u5u \cdot u^3 + 5u - 2u^5 = u^4 + 5u - 2u^5u⋅u3+5u−2u5=u4+5u−2u5. Factoring out uuu: u(u3+5−2u4)u(u^3 + 5 - 2u^4)u(u3+5−2u4). This reveals a clean polynomial structure in terms of uuu. Choice A accomplishes nothing since u=xu = xu=x doesn't eliminate the radicals. Choice B sets u=x+1u = x+1u=x+1, but this leaves fractional exponents like u1/2u^{1/2}u1/2 and u5/2u^{5/2}u5/2, failing to achieve polynomial form. Choice C uses the correct substitution u=(x+1)1/2u = (x+1)^{1/2}u=(x+1)1/2 but makes an algebraic error in the final simplification, incorrectly writing u⋅u3+5u−2u5=u(u3+5−2u4)u \cdot u^3 + 5u - 2u^5 = u(u^3 + 5 - 2u^4)u⋅u3+5u−2u5=u(u3+5−2u4) instead of properly combining u⋅u3=u4u \cdot u^3 = u^4u⋅u3=u4. Choice D correctly applies u=x+1u = \sqrt{x+1}u=x+1​ and shows the accurate algebraic steps, transforming the radical expression into a polynomial in uuu. Study tip: When you see mixed powers of the same radical, substitute the radical itself as your new variable to eliminate all fractional exponents.

Question 23

For the system below (variable order x,y,zx, y, zx,y,z), identify the coefficient matrix AAA in AX=BAX=BAX=B: [ \begin{cases} -x+4y+0z=2\ 3x-y+5z=-1\ 2x+0y-z=7 \end{cases} ]

  1. A=[−1403−1520−1]A=\begin{bmatrix}-1&4&0\\3&-1&5\\2&0&-1\end{bmatrix}A=​−132​4−10​05−1​​ (correct answer)
  2. A=[−1324−1005−1]A=\begin{bmatrix}-1&3&2\\4&-1&0\\0&5&-1\end{bmatrix}A=​−140​3−15​20−1​​
  3. A=[140315201]A=\begin{bmatrix}1&4&0\\3&1&5\\2&0&1\end{bmatrix}A=​132​410​051​​
  4. A=[−143−120]A=\begin{bmatrix}-1&4\\3&-1\\2&0\end{bmatrix}A=​−132​4−10​​

Explanation: This question tests your ability to identify the coefficient matrix A from a system of linear equations written with explicit zero coefficients. Matrix form AX = B requires the coefficient matrix A to contain all coefficients from the left sides of equations, including zeros for missing variables - this ensures proper matrix dimensions and correct multiplication. For the system -x + 4y + 0z = 2, 3x - y + 5z = -1, and 2x + 0y - z = 7, the coefficient matrix A is built by: (1) Row 1 uses coefficients from equation 1: [-1, 4, 0] (note the explicit 0 for z), (2) Row 2 uses coefficients from equation 2: [3, -1, 5], (3) Row 3 uses coefficients from equation 3: [2, 0, -1] (note the 0 for missing y). Therefore A = [[-1, 4, 0], [3, -1, 5], [2, 0, -1]]. Choice A correctly identifies this coefficient matrix with all proper signs and zeros included. Choice B transposes the matrix (swaps rows and columns), putting coefficients in wrong positions - for example, it has -1, 3, 2 in the first column instead of as the first row. Remember: rows correspond to equations, columns to variables! Matrix construction tip for systems with missing variables: When a variable doesn't appear in an equation, its coefficient is 0, not blank or omitted - you must include this 0 in the matrix to maintain proper dimensions (3×3 for three variables). For example, if equation 2 were just 3x + 5z = -1 (no y term), you write it as 3x + 0y + 5z = -1, giving row [3, 0, 5] in matrix A. To verify coefficient matrix A, check that each row matches its equation: row 1 [-1, 4, 0] represents -1·x + 4·y + 0·z = -x + 4y, row 2 [3, -1, 5] represents 3x - y + 5z, and row 3 [2, 0, -1] represents 2x + 0y - z = 2x - z, perfectly matching our original system!

Question 24

Solve the system algebraically. The solution(s) are the intersection point(s) of the line and the parabola.

y = -x + 1 \\ y = x^2 + 2x - 3 \end{cases}$$ How many real solutions does the system have?​
  1. 0 real solutions
  2. 1 real solution
  3. 2 real solutions (correct answer)
  4. Infinitely many solutions

Explanation: This question tests your ability to determine the number of real solutions in a linear-quadratic system, corresponding to intersection points of a line and parabola. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in y = mx + b form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system y = -x + 1 and y = x² + 2x - 3, substitute: -x + 1 = x² + 2x - 3, rearrange to x² + 3x - 4 = 0, discriminant 9 + 16 = 25 > 0, so two real solutions (x = 1 and x = -4). Choice C correctly identifies there are 2 real solutions based on the positive discriminant. Excellent—using the discriminant is a quick way to preview the number without full solving!

Question 25

Use the inverse property log⁡b(bx)=x\log_b(b^x)=xlogb​(bx)=x to simplify log⁡3(38)\log_3\left(3^8\right)log3​(38).

  1. 383^838
  2. 888 (correct answer)
  3. 333
  4. 242424

Explanation: This question tests your understanding that logarithms and exponentials are inverse operations—logarithms undo exponentiation and vice versa, just like square roots undo squaring. As inverse operations, logarithms and exponents cancel each other: log_b(b^x) = x for any x (the log undoes the exponent), and b^(log_b(x)) = x for x > 0 (the exponent undoes the log). These inverse properties are incredibly useful for simplification: log₃(3⁵) immediately simplifies to 5, and 7^(log₇(20)) immediately simplifies to 20. No calculation needed—they just undo each other! For log₃(3^8), since the argument is the base raised to a power, the log undoes it directly to give the exponent 8. Choice B correctly applies the inverse property log_b(b^x)=x to get 8. A distractor like choice A might forget to apply the inverse and leave it as 3^8, but that's the argument, not the simplified log value. Converting between forms: identify the three parts—base, exponent, and result. In b^y = x: base is b, exponent is y, result is x. In log_b(x) = y: base is b (subscript), argument is x (inside the log), value is y (what log equals). The exponent in exponential form BECOMES the log value, and the result BECOMES the argument. Example: 5³ = 125 has base 5, exponent 3, result 125, so log₅(125) = 3 has base 5, argument 125, value 3. The positions shift but the numbers stay the same! Evaluating logs using the definition: to find log₂(64), ask 'what power of 2 gives 64?' Think through powers of 2: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64. Found it! 2⁶ = 64, so log₂(64) = 6. This works for any log with a perfect power. If it's not a perfect power (like log₂(10)), you'd need a calculator, but for problems designed for hand calculation, you can find the answer by listing powers of the base until you hit the argument.