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CBSE Class 12 Mathematics Part 2 Exercise Solutions

CBSE Class 12 Mathematics Part 2 NCERT exercise solutions covering Integrals, Applications of Integrals, Differential Equations, Vector Algebra, 3D Geometry, Linear Programming, and Probability with worked examples and exam tips.

Updated 15 June 202621 min read

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How to Use This Page

This page provides comprehensive exercise solutions for CBSE Class 12 Mathematics Part 2, covering Chapters 7 through 13 of the NCERT textbook. Each section is designed to strengthen your conceptual understanding and problem-solving speed for the board examination.

Start by attempting every question on your own before consulting the solutions. When you check your work, focus on the method rather than just the final answer — board examiners award marks for correct steps, proper notation, and logical flow. If your approach differs from the solution shown, verify whether it is mathematically valid; many problems admit multiple correct methods.

Pay special attention to the Common Mistakes to Avoid and Exam Checklist sections at the end. These consolidate the most frequent errors students make and the essential items to verify before submitting your paper. Use the checklist as a final review tool in the days leading up to the exam.

For maximum benefit, solve the MCQs, fill-in-the-blanks, and short-answer questions first to build speed, then tackle the detailed long-answer problems. Time yourself: aim to complete each section within the marks-weighted time allocation (roughly 1.5 minutes per mark).


Key Concepts (Quick Recap)

Chapter 7: Integrals

  • Indefinite integral: ∫ f(x) dx = F(x) + C, where F′(x) = f(x)
Substitution∫ f(g(x)) · g′(x) dx = ∫ f(t) dt, where t = g(x)
  • Integration by parts (ILATE rule): ∫ u dv = uv − ∫ v du; choose u by priority — Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential
  • Partial fractions: Decompose P(x)/Q(x) into simpler fractions based on factorisation of Q(x)
  • Linear factor (x − a): A/(x − a)
  • Repeated factor (x − a)²: A/(x − a) + B/(x − a)²
  • Irreducible quadratic (x² + bx + c): (Ax + B)/(x² + bx + c)
  • Definite integral properties:
  • ∫[a to b] f(x) dx = ∫[a to b] f(a + b − x) dx
  • ∫[0 to a] f(x) dx = ∫[0 to a] f(a − x) dx
  • ∫[−a to a] f(x) dx = 2∫[0 to a] f(x) dx if f is even; = 0 if f is odd
  • Fundamental Theorem: ∫[a to b] f(x) dx = F(b) − F(a)

Chapter 8: Applications of Integrals

  • Area under a curve: Area = ∫[a to b] y dx = ∫[a to b] f(x) dx
  • Area between two curves: Area = ∫[a to b] |f(x) − g(x)| dx
  • Area with y-axis: Area = ∫[c to d] x dy = ∫[c to d] g(y) dy
  • Circle: Area enclosed by x² + y² = a² is πa²
  • Ellipse: Area enclosed by x²/a² + y²/b² = 1 is πab

Chapter 9: Differential Equations

  • Order: Highest order derivative present
  • Degree: Power of the highest order derivative (when polynomial in derivatives)
  • Formation: Differentiate the given relation n times and eliminate constants
  • Variable separable: dy/dx = f(x)g(y) → dy/g(y) = f(x) dx → integrate both sides
  • Homogeneous DE: dy/dx = f(y/x); substitute y = vx, dy/dx = v + x dv/dx
  • Linear DE: dy/dx + P(x) · y = Q(x); integrating factor IF = e^(∫P dx); Solution: y · IF = ∫(Q · IF) dx + C
  • General solution: Contains as many arbitrary constants as the order
  • Particular solution: Obtained by substituting given conditions into the general solution

Chapter 10: Vector Algebra

  • Magnitude: |a⃗| = √(a₁² + a₂² + a₃²)
  • Unit vector: â = a⃗/|a⃗|
  • Direction cosines: l = a₁/|a⃗|, m = a₂/|a⃗|, n = a₃/|a⃗|; l² + m² + n² = 1
  • Dot product: a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃ = |a⃗||b⃗| cos θ
  • Cross product: a⃗ × b⃗ = |î ĵ k̂; a₁ a₂ a₃; b₁ b₂ b₃|; |a⃗ × b⃗| = |a⃗||b⃗| sin θ
  • Projection: Projection of a⃗ on b⃗ = (a⃗ · b⃗)/|b⃗|
  • Perpendicularity: a⃗ ⊥ b⃗ ⟺ a⃗ · b⃗ = 0
  • Parallelism: a⃗ ∥ b⃗ ⟺ a⃗ × b⃗ = 0⃗

Chapter 11: Three-dimensional Geometry

  • Direction cosines (l, m, n): cosines of angles with positive x, y, z axes; l² + m² + n² = 1
  • Direction ratios: [a, b, c] where l = a/√(a²+b²+c²), etc.
  • Equation of line (through point, parallel to direction): r⃗ = a⃗ + λb⃗ or (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
  • Equation of line through two points: (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−x₁) = (z−z₁)/(z₂−z₁)
  • Shortest distance between skew lines: d = |[(a⃗₂−a⃗₁) · (b⃗₁×b⃗₂)] / |b⃗₁×b⃗₂||
  • Angle between two lines: cos θ = |l₁l₂ + m₁m₂ + n₁n₂| = |b⃗₁ · b⃗₂|/(|b⃗₁||b⃗₂|)
  • Coplanarity condition: [(a⃗₂−a⃗₁) b⃗₁ b⃗₂] = 0

Chapter 12: Linear Programming

  • Objective function: Z = ax + by (to maximise or minimise)
  • Constraints: Linear inequalities defining the feasible region
  • Feasible region: Intersection of all constraint half-planes
  • Corner Point Method: Evaluate Z at each vertex of the feasible region; optimum occurs at a corner
  • Iso-profit/iso-cost lines: Parallel lines representing constant Z; slide to find optimum
  • Unbounded region: May have no optimal value; verify by checking neighbouring points

Chapter 13: Probability

  • Conditional probability: P(A|B) = P(A ∩ B)/P(B), P(B) ≠ 0
  • Multiplication theorem: P(A ∩ B) = P(A) · P(B|A) = P(B) · P(A|B)
  • Bayes' theorem: P(Eᵢ|A) = [P(Eᵢ) · P(A|Eᵢ)] / [Σ P(Eⱼ) · P(A|Eⱼ)]
  • Independence: A and B independent ⟺ P(A ∩ B) = P(A) · P(B)
  • Random variable: A real-valued function on the sample space
  • Probability distribution: P(X = xᵢ) = pᵢ, where Σpᵢ = 1
  • Mean: E(X) = μ = Σ xᵢpᵢ
  • Variance: Var(X) = Σ xᵢ²pᵢ − (Σ xᵢpᵢ)² = E(X²) − [E(X)]²
  • Binomial distribution: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, where q = 1 − p; Mean = np; Variance = npq
  • Total probability: P(A) = Σ P(Eᵢ) · P(A|Eᵢ)

I. Choose the Best Answer (MCQs)

1. ∫ x · eˣ dx equals:

(A) eˣ(x + 1) + C   (B) eˣ(x − 1) + C   (C) eˣ(x/2) + C   (D) xeˣ + C

Answer(B) eˣ(x − 1) + C Using integration by parts (ILATE: x is Algebraic, eˣ is Exponential, so u = x): ∫ x · eˣ dx = x · eˣ − ∫ 1 · eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C

2. The value of ∫[0 to π/2] sin³x dx is:

(A) 2/3   (B) 1/3   (C) π/3   (D) 0

Answer(A) 2/3 ∫[0 to π/2] sin³x dx = ∫[0 to π/2] sin²x · sinx dx Let cos x = t, dt = −sinx dx = ∫[0 to 1] (1 − t²) dt = [t − t³/3]₀¹ = 1 − 1/3 = 2/3

3. The order and degree of [d²y/dx²]³ + [dy/dx]² + y = 0 are respectively:

(A) 2, 3   (B) 3, 2   (C) 2, 2   (D) 1, 3

Answer(A) Order = 2, Degree = 3 The highest order derivative is d²y/dx² (order 2). It is raised to the power 3 (degree 3). The equation is polynomial in derivatives, so degree is well-defined.

4. If a⃗ = 2î − 3ĵ + k̂ and b⃗ = î + ĵ − 2k̂, then a⃗ · b⃗ equals:

(A) 7   (B) −7   (C) −3   (D) 3

Answer(C) −3 a⃗ · b⃗ = (2)(1) + (−3)(1) + (1)(−2) = 2 − 3 − 2 = −3

5. The direction cosines of the z-axis are:

(A) (1, 0, 0)   (B) (0, 1, 0)   (C) (0, 0, 1)   (D) (1, 1, 1)

Answer(C) (0, 0, 1) The z-axis makes angles 90°, 90°, 0° with x, y, z axes respectively. Hence l = cos90° = 0, m = cos90° = 0, n = cos0° = 1.

6. The integrating factor of the differential equation dy/dx + y/x = x² is:

(A) x   (B) eˣ   (C) 1/x   (D) e^(1/x)

Answer(A) x P(x) = 1/x, so IF = e^(∫1/x dx) = e^(ln x) = x

7. In a binomial distribution with n = 6 and p = 1/2, the mean is:

(A) 3   (B) 6   (C) 2   (D) 1

Answer(A) 3 Mean of binomial distribution = np = 6 × 1/2 = 3

8. The area bounded by the curve y = x² and the line y = 4 is:

(A) 16/3   (B) 32/3   (C) 8/3   (D) 4

Answer(B) 32/3 Points of intersection: x² = 4 → x = ±2. Area = ∫[−2 to 2] (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16/3 + 16/3 = 32/3

9. If P(A) = 1/2, P(B) = 1/3 and P(A ∩ B) = 1/6, then A and B are:

(A) Independent   (B) Mutually exclusive   (C) Dependent   (D) Complementary

Answer(A) Independent P(A) · P(B) = (1/2)(1/3) = 1/6 = P(A ∩ B). Hence A and B are independent.

10. The maximum value of Z = 3x + 2y subject to x + y ≤ 4, x ≥ 0, y ≥ 0 is:

(A) 8   (B) 12   (C) 6   (D) 10

Answer(B) 12 Corner points of feasible region: (0,0), (4,0), (0,4). At (0,0): Z = 0; at (4,0): Z = 12; at (0,4): Z = 8. Maximum Z = 12 at (4,0).

II. Fill in the Blanks

1. ∫ dx/√(1 − x²) = sin⁻¹x + C

2. The degree of the differential equation [d²y/dx² + (dy/dx)³]^(2/3) = 0 is 2 (after raising both sides to power 3 to make it polynomial in derivatives, the equation becomes (d²y/dx²)² + higher terms = 0)

3. If a⃗ × b⃗ = 0⃗ and a⃗ ≠ 0⃗, b⃗ ≠ 0⃗, then the angle between a⃗ and b⃗ is 0° or 180° (vectors are parallel)

4. The value of ∫[−1 to 1] x³ dx is 0 (since x³ is an odd function, the integral over a symmetric interval is zero)

5. The projection of a⃗ on b⃗ is given by **(a⃗ · b⃗)/b⃗**
6. The shortest distance between parallel lines r⃗ = a⃗₁ + λb⃗ and r⃗ = a⃗₂ + μb⃗ is **(a⃗₂ − a⃗₁) × b⃗/b⃗**
7. In Bayes' theorem, P(EᵢA) = **P(Eᵢ) · P(AEᵢ) / Σ P(Eⱼ) · P(AEⱼ)**

8. The variance of a binomial distribution with parameters n and p is npq where q = 1 − p

9. For a linear programming problem, the optimal value of the objective function occurs at a corner point (vertex) of the feasible region

10. The cross product a⃗ × b⃗ is maximum when the angle between a⃗ and b⃗ is 90° (π/2)


III. True or False

1. True. ∫[a to b] f(x) dx = ∫[a to b] f(t) dt — the variable of integration is a dummy variable.

2. False. The degree of a differential equation is always defined. → Correction: The degree is defined only when the differential equation is polynomial in its derivatives. For example, dy/dx + sin(dy/dx) = 0 has no defined degree.

3. True. Two vectors are perpendicular if and only if their dot product is zero.

4. False. The cross product of two vectors is commutative. → Correction: The cross product is anti-commutative: a⃗ × b⃗ = −(b⃗ × a⃗).

5. True. The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is (1/2)x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂).

6. True. If P(A ∪ B) = P(A) + P(B), then A and B are mutually exclusive.

7. False. A linear programming problem always has a unique optimal solution. → Correction: An LPP can have a unique optimal solution, infinitely many optimal solutions, or no feasible solution at all.

8. True. For an odd function f(x), ∫[−a to a] f(x) dx = 0.


IV. Match the Following

Column AColumn B
(a) ∫ eˣ sinx dx(iii) eˣ(sin x − cos x)/2 + C
(b) ∫ dx/(x² + a²)(v) (1/a) tan⁻¹(x/a) + C
(c) a⃗ · (b⃗ × c⃗)(vi) Scalar triple product [a⃗ b⃗ c⃗]
(d) Direction cosines of a line(i) l² + m² + n² = 1
(e) P(AB)(iv) P(A ∩ B)/P(B)
(f) ∫[0 to a] f(a − x) dx(ii) ∫[0 to a] f(x) dx (by substitution x → a − x)

V. Answer Briefly

1. Evaluate ∫ sin²x dx.
Answer

∫ sin²x dx = ∫ (1 − cos 2x)/2 dx = (1/2) ∫ dx − (1/2) ∫ cos 2x dx = x/2 − (sin 2x)/4 + C

2. Find the area bounded by y = x², x-axis, and x = 2.
Answer

Area = ∫[0 to 2] x² dx = [x³/3]₀² = 8/3 square units

3. Find the order and degree of (d³y/dx³)² + (d²y/dx²)⁴ + y = 0.
Answer

Order = 3 (highest derivative is d³y/dx³). Degree = 2 (power of the highest order derivative). The equation is polynomial in derivatives, so the degree is well-defined.

4. If a⃗ = î + ĵ + k̂ and b⃗ = 2î − ĵ + 3k̂, find a⃗ × b⃗.
Answer

a⃗ × b⃗ = |î ĵ k̂; 1 1 1; 2 −1 3| = î(3 − (−1)) − ĵ(3 − 2) + k̂(−1 − 2) = 4î − ĵ − 3k̂

5. Find the direction cosines of the line passing through (2, −3, 1) and (4, 1, 5).
Answer

Direction ratios: a = 4 − 2 = 2, b = 1 − (−3) = 4, c = 5 − 1 = 4 √(a² + b² + c²) = √(4 + 16 + 16) = √36 = 6 Direction cosines: l = 2/6 = 1/3, m = 4/6 = 2/3, n = 4/6 = 2/3

6. State the condition for a linear programming problem to have no feasible solution.
Answer

An LPP has no feasible solution when the constraints are contradictory — that is, there is no common point satisfying all the inequalities simultaneously. The feasible region is empty.

**7. If P(A) = 0.6, P(B) = 0.4 and P(A ∪ B) = 0.8, find P(A ∩ B) and P(AB).**

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.4 − 0.8 = 0.2 P(A|B) = P(A ∩ B)/P(B) = 0.2/0.4 = 0.5

8. Find the integrating factor of dy/dx + 2y = eˣ.
Answer

P(x) = 2, so IF = e^(∫2 dx) = e^(2x)


VI. Answer in Detail

1. Evaluate ∫ x² · eˣ dx using integration by parts. Show all steps and identify the correct ILATE choice.
Answer

We apply integration by parts twice. By ILATE, x² (Algebraic) is chosen as u and eˣ (Exponential) as dv. First application: ∫ x² · eˣ dx Let u = x², dv = eˣ dx → du = 2x dx, v = eˣ = x²eˣ − ∫ 2x · eˣ dx = x²eˣ − 2∫ x · eˣ dx Second application: For ∫ x · eˣ dx Let u = x, dv = eˣ dx → du = dx, v = eˣ = xeˣ − ∫ eˣ dx = xeˣ − eˣ + C Combining: ∫ x² · eˣ dx = x²eˣ − 2[xeˣ − eˣ] + C = x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C Verification: Differentiate eˣ(x² − 2x + 2): using product rule, we get eˣ(x² − 2x + 2) + eˣ(2x − 2) = eˣ(x² − 2x + 2 + 2x − 2) = x²eˣ ✓

2. Find the area enclosed between the curves y = x² and y = 2x.
Answer

Step 1 — Find points of intersection: Set x² = 2x → x² − 2x = 0 → x(x − 2) = 0 → x = 0 or x = 2. The curves intersect at (0, 0) and (2, 4). Step 2 — Determine which curve is above: For 0 < x < 2: 2x > x² (e.g., at x = 1: 2 > 1). So the line y = 2x is above the parabola y = x² in this interval. Step 3 — Compute area: Area = ∫[0 to 2] (2x − x²) dx = [x² − x³/3]₀² = (4 − 8/3) − 0 = 4/3 square units Result: The area enclosed between y = x² and y = 2x is 4/3 square units.

3. Solve the linear differential equation: (x + 1) dy/dx − y = (x + 1)².
Answer

Step 1 — Rewrite in standard form: dy/dx − y/(x + 1) = (x + 1) Compare with dy/dx + P(x)y = Q(x): P(x) = −1/(x + 1), Q(x) = x + 1 Step 2 — Find integrating factor: IF = e^(∫ P dx) = e^(∫ −1/(x+1) dx) = e^(−ln|x+1|) = 1/(x + 1) Step 3 — Apply the solution formula: y · IF = ∫ Q · IF dx + C y · (1/(x + 1)) = ∫ (x + 1) · (1/(x + 1)) dx + C y/(x + 1) = ∫ 1 dx + C y/(x + 1) = x + C y = (x + 1)(x + C) y = x(x + 1) + C(x + 1) Result: The general solution is y = x(x + 1) + C(x + 1), or equivalently y = (x + 1)(x + C).

4. Find the shortest distance between the skew lines: L₁: r⃗ = (î + 2ĵ + 3k̂) + λ(2î + 3ĵ + k̂) and L₂: r⃗ = (4î + 6ĵ + 7k̂) + μ(2î + 3ĵ + k̂).

Step 1 — Check if lines are parallel or skew: b⃗₁ = 2î + 3ĵ + k̂ and b⃗₂ = 2î + 3ĵ + k̂. Since b⃗₁ = b⃗₂, these are parallel lines, not skew. For parallel lines, the shortest distance formula simplifies: d = |(a⃗₂ − a⃗₁) × b⃗| / |b⃗| Step 2 — Compute vectors: a⃗₂ − a⃗₁ = (4î + 6ĵ + 7k̂) − (î + 2ĵ + 3k̂) = 3î + 4ĵ + 4k̂ b⃗ = 2î + 3ĵ + k̂ Step 3 — Cross product: (a⃗₂ − a⃗₁) × b⃗ = |î ĵ k̂; 3 4 4; 2 3 1| = î(4 − 12) − ĵ(3 − 8) + k̂(9 − 8) = −8î + 5ĵ + k̂ |(a⃗₂ − a⃗₁) × b⃗| = √(64 + 25 + 1) = √90 = 3√10 |b⃗| = √(4 + 9 + 1) = √14 Step 4 — Compute distance: d = 3√10/√14 = 3√10/√14 = 3√(5/7) units Result: The shortest distance between the parallel lines is 3√(5/7) units ≈ 2.53 units.

5. Bag I contains 3 red and 4 black balls. Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II (using Bayes' theorem).
Answer

Step 1 — Define events: E₁ = Ball drawn from Bag I; E₂ = Ball drawn from Bag II A = Ball drawn is red Step 2 — Prior probabilities: P(E₁) = 1/2, P(E₂) = 1/2 (equal chance of choosing either bag) Step 3 — Conditional probabilities: P(A|E₁) = P(red from Bag I) = 3/7 P(A|E₂) = P(red from Bag II) = 5/11 Step 4 — Apply Bayes' theorem: P(E₂|A) = P(E₂) · P(A|E₂) / [P(E₁) · P(A|E₁) + P(E₂) · P(A|E₂)] = (1/2)(5/11) / [(1/2)(3/7) + (1/2)(5/11)] = (5/22) / [3/14 + 5/22] = (5/22) / [(33 + 35)/154] = (5/22) / (68/154) = (5/22) × (154/68) = (5 × 154) / (22 × 68) = 770/1496 = 35/68 Result: The probability that the red ball was drawn from Bag II is 35/68 ≈ 0.515.

6. Solve the following linear programming problem graphically: Maximise Z = 5x + 3y subject to x + y ≤ 6, 2x + 5y ≤ 20, x ≥ 0, y ≥ 0.

Step 1 — Find the feasible region: Constraint 1: x + y ≤ 6 → line passes through (6, 0) and (0, 6) Constraint 2: 2x + 5y ≤ 20 → line passes through (10, 0) and (0, 4) Step 2 — Find corner points of the feasible region: - O(0, 0): intersection of x = 0, y = 0 - A(6, 0): intersection of x + y = 6 with y = 0 - B(0, 4): intersection of 2x + 5y = 20 with x = 0 - C: intersection of x + y = 6 and 2x + 5y = 20 From x + y = 6 → x = 6 − y Substituting: 2(6 − y) + 5y = 20 → 12 − 2y + 5y = 20 → 3y = 8 → y = 8/3 x = 6 − 8/3 = 10/3 C(10/3, 8/3) Step 3 — Evaluate Z at each corner point: - Z at O(0, 0) = 0 - Z at A(6, 0) = 5(6) + 3(0) = 30 - Z at B(0, 4) = 5(0) + 3(4) = 12 - Z at C(10/3, 8/3) = 5(10/3) + 3(8/3) = 50/3 + 24/3 = 74/3 ≈ 24.67 Step 4 — Determine optimum: Maximum Z = 30 at point A(6, 0). Result: The maximum value of Z is 30, achieved at x = 6, y = 0.


Common Mistakes to Avoid

1. Forgetting the constant of integration (+C) in indefinite integrals. Every indefinite integral must include "+ C". Board examiners deduct marks for this omission.

2. Wrong ILATE choice in integration by parts. Remember the priority order: Inverse trig > Logarithmic > Algebraic > Trigonometric > Exponential. Choosing the wrong function as u leads to a more complicated integral instead of a simpler one.

3. Mixing up order and degree of differential equations. Order = highest derivative present; Degree = power of the highest order derivative (only when the DE is polynomial in derivatives). The degree of dy/dx + sin(dy/dx) = 0 is NOT defined.

4. Using wrong limits in area problems. Always find the correct points of intersection before setting up the definite integral. Sketching the curves helps identify which function is on top.

5. Confusing dot product and cross product properties. Dot product gives a scalar; cross product gives a vector. Dot product is zero for perpendicular vectors; cross product is zero for parallel vectors. Do not interchange these conditions.

6. Incorrect substitution in Bayes' theorem. Students often swap P(AE) and P(EA). Remember: Bayes' theorem gives P(EA), which requires computing P(AE) first. Always label events clearly and use a tree diagram when in doubt.

Exam Checklist

  • [ ] Indefinite integrals: Every answer ends with "+ C" — no exceptions.
  • [ ] Integration by parts: Applied ILATE rule correctly; verified by differentiating the result.
  • [ ] Definite integrals: Substituted both limits (upper and lower) properly; checked for even/odd function shortcuts.
  • [ ] Area problems: Drew a sketch, found points of intersection, identified the top function, and used correct limits.
  • [ ] Differential equations: Checked that the DE is in standard form before finding the integrating factor; included the constant of integration before multiplying by IF.
  • [ ] Vector/cross products: Wrote the determinant correctly (î, ĵ, k̂ in the first row); verified that a⃗ × b⃗ = −(b⃗ × a⃗).
  • [ ] 3D geometry: Used direction cosines (not direction ratios) in angle formulae; verified l² + m² + n² = 1.
  • [ ] Probability: Checked that all probabilities in a distribution sum to 1; used Bayes' theorem only when prior and conditional probabilities are both given.

Frequently Asked Questions

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