Part III Review

1: A. 4(5) − 3 = 20 − 3 = 17. Choice D reports 4(5) directly, forgetting to subtract 3.

2: G. g(−2)=(−2)²−1=3, then f(3)=3+3=6. Choice F reports the inner value g(−2) instead of completing the composition by evaluating f at that value.

3: C. The radical requires 2x−6≥0, so x≥3; the denominator requires x≠8. Choice A drops the denominator restriction, forgetting to exclude x=8.

4: J. f(g(x))=(3x+4)²=49, so 3x+4=±7. 3x+4=7 gives x=1; 3x+4=−7 gives x=−11/3. Choice H reports the values of 3x+4 (±7) directly instead of solving for x.

5: D. (20−4) ⁄ (5−1) = 16 ⁄ 4 = 4. Choice A reports the numerator of the slope calculation (20−4=16) without completing the division.

6: F. Factor: (x+7)(x−3)=0, giving x=−7 or x=3. Choice G reverses the sign of both roots.

7: C. Degree 4 is even, so both ends point the same direction; the leading coefficient is positive, so both ends rise. Choice A treats the leading coefficient as negative instead of positive, reversing the conclusion.

8: F. a₁₀ = −3 + (10 − 1)(6) = −3 + 54 = 51. Choice G uses one fewer step than required, an off-by-one error in the sequence formula.

9: B. The denominator is zero at x=−5, giving the vertical asymptote. Numerator and denominator have equal degree, so the horizontal asymptote is the ratio of leading coefficients, 4/1=4. Choice A drops the negative sign on the vertical asymptote.

10: F. 800(1.06)³ ≈ 952.81. Choice G comes from a rounding or computational slip in evaluating 1.06³.

11: D. 5³=125, so log₅(125)=3. Choice A reports the base of the logarithm instead of the exponent that solves it.

12: J. a₅ = 4 · 3⁴ = 4 · 81 = 324. Choice G uses one extra multiplication by the common ratio than required, computing a₆ instead of a₅.

13: A. g(−2)=f(−2+4)−2=f(2)−2=4−2=2. Choice B reverses the sign of the entire expression after evaluating f(2)−2.

14: G. p(−1)=2(−1)−3=−5 (first piece, x<0); p(2)=2²+1=5 (second piece, x≥0); −5+5=0. Choice F reports p(−1) alone instead of completing the requested sum p(−1) + p(2).

15: A. g(x)=−2f(x−3)+1 reflects f over the x-axis and stretches it vertically by a factor of 2 (from the −2), shifts it right 3 (from x−3), and shifts it up 1 (from the +1). Choice D misreads the vertical shift, treating +1 as a downward shift instead of upward.

16: H. g(3)=f(2·3)=f(6)=36+5=41. Choice F evaluates f at the original input (3) instead of first doubling it, computing f(3) instead of f(6).

17: B. Swap and solve: x=5y−8, so y=(x+8)/5. Choice A drops the sign while solving for the inverse, subtracting 8 instead of adding it.

18: H. f(4)=3(4)+9=21; g(21)=(21−9)/3=4. Choice F reports f(4) alone instead of completing the composition by evaluating g at that value.

19: B. x²−1=8 gives x²=9; since x≥0, x=3. Choice A reports the negative root, ignoring that the restricted domain x≥0 requires the positive value.

20: G. 4x³−2=30 gives 4x³=32, so x³=8, giving x=2. Choice F reports the value of x³ instead of completing the cube root to solve for x.

1: C. The terms are 3, 5, 7, 9, 11 (substituting k=1 through 5 into 2k+1); their sum is 35. Choice A starts the index one step early, using k=0 through 4 instead of k=1 through 5, giving terms 1,3,5,7,9, which sum to 25.

2: H. The 10th term is a₁₀=8+9(3)=35. The sum is S=10(8+35)/2=215. Choice G comes from an arithmetic slip while completing the sum formula.

3: A. Combine the left side using the product rule: log(4x)=log(36), so 4x=36, giving x=9. Choice B subtracts 4 from 36 instead of dividing, treating the equation as if it were log(x)+4=log(36).

4: J. Rewrite in exponential form: x−1=2⁴=16, so x=17. Choice H reports x−1 itself instead of completing the addition to solve for x.

5: A. Since sin θ=5/13, the adjacent side is √(13²−5²)=12, so cos θ=12/13. tan θ=sin θ/cos θ=(5/13)/(12/13)=5/12. Choice B inverts the ratio, reporting cot θ instead of tan θ.

6: G. The Pythagorean identity sin²θ+cos²θ=1 rearranges directly to cos²θ=1−sin²θ. Choice F confuses this with the related identity 1+tan²θ=sec²θ.