Question 1
Given the function f(x)=x2−3x+4, what is the value of f(−2)?
- 2
- 6
- 10
- 14
Explanation: The correct answer is D (14). Substitute x = −2 into the function: f(−2) = (−2)² − 3(−2) + 4 = 4 + 6 + 4 = 14. The three key steps are: (1) square the input: (−2)² = +4, not −4; (2) multiply: −3(−2) = +6, not −6; (3) add: 4 + 6 + 4 = 14. A (2) comes from treating −3(−2) as −6: 4 − 6 + 4 = 2. B (6) comes from treating (−2)² as −4: −4 + 6 + 4 = 6. C (10) comes from a partial sign error. Pro tip: when substituting a negative value, write every step explicitly — sign errors on squared terms and products are the most common mistakes in function evaluation.
Question 2
Let f(x) \= \begin{cases} 2x + 1 & \text{if } x \< 0 \ x^2 - 3 & \text{if } x \ge 0 \end{cases}. For how many values of x does f(x) \= 5?
- 0
- 1
- 2
- 3
Explanation: This is a piecewise functions question testing domain-aware equation solving. Choice B (1) is correct — solve f(x) = 5 in each piece separately, then check validity. Case 1 (x < 0): 2x + 1 = 5 → x = 2. Invalid: 2 ≥ 0, so it falls outside this piece's domain. Case 2 (x ≥ 0): x² − 3 = 5 → x² = 8 → x = ±2√2 ≈ ±2.83. Check: 2√2 ≥ 0 ✓ (valid); −2√2 < 0 ✗ (invalid for this piece). Exactly 1 valid solution. Choice A (0) results from making errors in both cases and concluding no solution exists. Choice C (2) counts both ±2√2 from Case 2 without checking that −2√2 fails the x ≥ 0 domain requirement. Choice D (3) counts the invalid solution x = 2 from Case 1 plus both ±2√2 from Case 2, accepting all algebraic results without domain verification. Pro tip: For piecewise functions, always solve each piece independently, then check whether each solution satisfies that piece's domain condition. An algebraically valid answer that violates the domain restriction must be rejected. This is the piecewise equivalent of checking for extraneous solutions in radical equations.
Question 3
A sequence is defined by a1=3 and an=2an−1−1 for n≥2. What is a5?
- 17
- 23
- 33
- 65
Explanation: This is a recursive sequences question testing careful step-by-step application of a rule. Choice C (33) is correct — apply the rule aₙ = 2aₙ₋₁ − 1 repeatedly: a₁ = 3, a₂ = 2(3) − 1 = 5, a₃ = 2(5) − 1 = 9, a₄ = 2(9) − 1 = 17, a₅ = 2(17) − 1 = 33. Choice A (17) stops at a₄ — off by one term, likely from miscounting or starting the index at n = 0. Choice B (23) results from an arithmetic error at one step — perhaps computing a₃ = 2(5) + 1 = 11 (adding instead of subtracting): then a₄ = 2(11) − 1 = 21, a₅ = 2(21) − 1 = 41... or some other mid-sequence error. Choice D (65) comes from using aₙ = 2aₙ₋₁ + 1 throughout (adding instead of subtracting): a₂ = 7, a₃ = 15, a₄ = 31, a₅ = 63 ≈ 65. Pro tip: For recursive sequences, build a table of values term by term — there's no shortcut to jumping ahead. Write out each term explicitly: write the term number, then compute carefully. The −1 in this rule is easy to misread as +1, which produces very different values.
Question 4
A model rocket's height in feet after t seconds is modeled by h(t)=−16t2+96t. What is the height, in feet, 4 seconds after launch?
- 128
- 256
- 384
- 512
Explanation: This is a function evaluation question testing substitution into a quadratic height model. Choice A (128) is correct — substitute t = 4: h(4) = −16(4²) + 96(4) = −16(16) + 384 = −256 + 384 = 128 feet. Choice B (256) comes from computing only the negative term: 16 × 16 = 256, or from adding the terms incorrectly: −256 + 384 treated as 256. Choice C (384) comes from evaluating only the positive term 96 × 4 = 384 and ignoring the −16t² term entirely. Choice D (512) likely comes from computing 96 × 4 + 16 × 4² but adding instead of subtracting: 384 + 256 = 640... or from −16(4) + 96(4) = −64 + 384 = 320, then some addition error. Pro tip: Evaluate each term of a polynomial function separately, then combine. The negative sign on the −16t² term is critical — squaring 4 first gives 16, then multiplying by −16 gives −256 (negative, not positive). The rocket is decelerating, so height at t = 4 should be less than at its peak, not more.
Question 5
A function is defined by f(x)=2−x. What is the value of f(1−x)?
- 1−x
- x−1
- 1+x
- 3−x
Explanation: We need to find f(1−x) where f(x)=2−x. This means replacing every x with (1-x): f(1−x)=2−(1−x)=2−1+x=1+x. When distributing the negative sign, −(1−x) becomes −1+x. Question 6
Which of the following represents f(x+4) if f(x)=2x2−x?
- 2x2+8x−4
- 2x2+32x
- 2x2+16x−x
- 2(x+4)2−(x+4)
Explanation: We need to find f(x + 4) when f(x) = 2x² - x. To find f(x + 4), we replace every x in the original function with (x + 4): f(x + 4) = 2(x + 4)² - (x + 4). This expression shows the direct substitution before any expansion. Choice D correctly represents this substitution form. Note that f(x + 4) means substituting (x + 4) for every x, not adding 4 to the result.
Question 7
What is f(−3) if f(x)=5x2−4x+1?
- 32
- 42
- 50
- 58
Explanation: We need to find f(-3) when f(x)=5x2−4x+1. To evaluate f(-3), we substitute x = -3 into the function: f(−3)=5(−3)2−4(−3)+1. Following order of operations, we calculate 5(−3)2=5(9)=45, then 4(−3)=−12, so f(−3)=45−(−12)+1=45+12+1=58. Therefore, f(−3)=58. Question 8
If f(x)=7−x, what is f(3)?
- 4
- 10
- 5
- 1
Explanation: We need to find f(3) when f(x) = 7 - x. To evaluate f(3), we substitute x = 3 into the function: f(3) = 7 - 3. Following order of operations, we calculate 7 - 3 = 4. Therefore, f(3) = 4.
Question 9
What is f(7) if f(x)=23x−1?
- 10
- 9
- 11
- 8
Explanation: We need to find f(7) when f(x)=23x−1. To evaluate f(7), we substitute x = 7 into the function: f(7)=23(7)−1. Following order of operations, we calculate the numerator: 3(7)−1=21−1=20, then 20/2=10. Therefore, f(7)=10. Question 10
A function is defined by f(x)=x2+2x−5. What is f(−1)?
- −8
- −6
- −2
- 6
Explanation: We need to find f(-1) where f(x) = x² + 2x - 5. Substituting x = -1 with parentheses: f(-1) = (-1)² + 2(-1) - 5 = 1 - 2 - 5 = -6. Remember that (-1)² = 1, not -1, since we're squaring the entire negative number.