Grade 12 Pre-Calculus Practice — Medium

Question 1 of 28Score 0/0Medium

Question 1 of 28: Find the 8th term of the arithmetic sequence: a₁ = 7, d = −2.

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Answer key for parents & teachers (28 questions)
  1. Find the 8th term of the arithmetic sequence: a₁ = 7, d = −2.−7. a₈ = 7 + (8−1)(−2) = 7 − 14 = −7.
  2. The sum of the first n natural numbers formula is:n(n+1)/2. Σᵢ₌₁ⁿ i = n(n+1)/2. This is Gauss's famous formula for the sum of consecutive integers.
  3. For the geometric series 3 + 6 + 12 + 24, what is S₆?189. a₁=3, r=2. S₆ = 3·(1−2⁶)/(1−2) = 3·(1−64)/(−1) = 3·63 = 189.
  4. What is the common ratio of the geometric sequence 81, 27, 9, 3, …?1/3. r = 27/81 = 1/3. Each term is divided by 3.
  5. Find S∞ for the series 12 + 4 + 4/3 + 4/9 + …18. a₁=12, r=1/3. S∞ = 12/(1−1/3) = 12/(2/3) = 18.
  6. A repeating decimal 0.333… = 0.3̄ can be expressed as a fraction using:Infinite geometric series with r=1/10. 0.333… = 3/10 + 3/100 + … This is a geometric series with a₁=3/10, r=1/10. Sum = (3/10)/(1−1/10) = (3/10)/(9/10) = 1/3.
  7. Evaluate Σᵢ₌₁⁴ (2i + 1).24. Terms: i=1: 3, i=2: 5, i=3: 7, i=4: 9. Sum = 3+5+7+9 = 24.
  8. Which sequence is neither arithmetic nor geometric?1, 1, 2, 3, 5, 8. The Fibonacci sequence 1, 1, 2, 3, 5, 8 has no constant difference or ratio.
  9. Find the number of terms in the arithmetic sequence 5, 9, 13, …, 101.25. aₙ = 5 + (n−1)·4 = 101 → (n−1)·4 = 96 → n−1 = 24 → n = 25.
  10. Insert two arithmetic means between 4 and 22 (find them).10 and 16. The sequence is 4, a₂, a₃, 22 — four terms with a₁=4, a₄=22. d = (22−4)/3 = 6. So terms are 4, 10, 16, 22.
  11. Convert the polar point (6, π) to rectangular coordinates.(−6, 0). x = 6·cos(π) = 6·(−1) = −6. y = 6·sin(π) = 0. Rectangular: (−6, 0).
  12. For the parametric curve x = t², y = t³, find dy/dx in terms of t.3t/2. dy/dt = 3t², dx/dt = 2t. So dy/dx = 3t²/(2t) = 3t/2.
  13. Convert the rectangular equation x² + y² = 16 to polar form.r = 4. Since r² = x² + y², we get r² = 16, so r = 4 (taking r > 0). This is a circle of radius 4.
  14. At what values of t does the parametric curve x = sin t, y = cos(2t) cross the x-axis (y = 0)?t = π/4, 3π/4. y = cos(2t) = 0 when 2t = π/2 + nπ, i.e. t = π/4, 3π/4, 5π/4, 7π/4.
  15. What is the rectangular form of r = 2cos θ?(x−1)² + y² = 1. Multiply both sides by r: r² = 2r cos θ → x²+y² = 2x → (x−1)²+y² = 1. A circle of radius 1 centered at (1, 0).
  16. Eliminate the parameter from x = 2t − 1 and y = t + 3.y = (x+7)/2. From x = 2t−1: t = (x+1)/2. Substitute: y = (x+1)/2 + 3 = (x+1+6)/2 = (x+7)/2.
  17. The polar curve r = a(1 − cos θ) is called a:Cardioid. r = a(1 − cos θ) is a cardioid — a heart-shaped curve symmetric about the x-axis passing through the origin.
  18. For the curve r = 3sin(2θ) (a 4-petal rose), how many petals does it have?4. r = a·sin(nθ) gives 2n petals when n is even. For n=2: 4 petals.
  19. Find the slope of the tangent to x = t², y = 2t at t = 3.1/3. dy/dt = 2, dx/dt = 2t = 6 at t=3. Slope = dy/dx = 2/6 = 1/3.
  20. Find the unit vector in the direction of ⟨3, 4⟩.⟨3/5, 4/5⟩. |v⃗| = 5. Unit vector = ⟨3/5, 4/5⟩.
  21. Find the angle between ⟨1, 0⟩ and ⟨0, 1⟩.90°. Dot product = 0. cos θ = 0/(1·1) = 0. θ = 90°. These are the standard x and y unit vectors, which are perpendicular.
  22. Are ⟨3, −2⟩ and ⟨4, 6⟩ perpendicular?Yes, because their dot product is 0. Dot product = 3·4 + (−2)·6 = 12 − 12 = 0. Since the dot product is 0, the vectors are perpendicular.
  23. A vector v⃗ has magnitude 10 and makes a 30° angle with the x-axis. Find its components.⟨5√3, 5⟩. vₓ = 10cos30° = 10·(√3/2) = 5√3. v_y = 10sin30° = 10·(1/2) = 5. So v⃗ = ⟨5√3, 5⟩.
  24. Find the dot product of ⟨5, 2, −1⟩ and ⟨3, −4, 2⟩ (3D vectors).5. Dot product = 5·3 + 2·(−4) + (−1)·2 = 15 − 8 − 2 = 5.
  25. A force of 50 N acts at 60° to a displacement of 8 m. Find the work done (W = F⃗ · d⃗).200 J. W = |F||d|cos θ = 50·8·cos60° = 400·(1/2) = 200 J.
  26. If u⃗ = ⟨a, b⟩ and v⃗ = ⟨c, d⟩, which property confirms u⃗ · v⃗ = v⃗ · u⃗?Commutativity. The dot product is commutative: u⃗ · v⃗ = v⃗ · u⃗, since ac+bd = ca+db.
  27. What is the magnitude of the zero vector ⟨0, 0⟩?0. |⟨0, 0⟩| = √(0²+0²) = 0. The zero vector is the only vector with magnitude 0.
  28. Find the angle between ⟨2, 2⟩ and ⟨1, 0⟩.45°. Dot product = 2. |⟨2,2⟩| = 2√2, |⟨1,0⟩| = 1. cos θ = 2/(2√2) = 1/√2. θ = 45°.