Tamil Nadu Board Class 12 Mathematics — Flashcards

A: Step 1: For a 2×2 matrix [[a, b], [c, d]], adjoint = [[d, -b], [-c, a]] Step 2: Replace a=2, b=-1, c=3, d=4 Step 3: adj(A) = [[4, 1], [-3, 2]] Answer: adj(A) = [[4, 1], [-3, 2]] Note: Adjoint is the

A: Step 1: Calculate determinant |A| = 2(4) - (-1)(3) = 8 + 3 = 11 Step 2: Find adj(A) = [[4, 1], [-3, 2]] (from previous problem) Step 3: Apply formula A⁻¹ = (1/11) × [[4, 1], [-3, 2]] Step 4: A⁻¹ = [[4

A: Step 1: Write in matrix form AX = B where A = [[2, 3], [1, 2]], X = [[x], [y]], B = [[8], [5]] Step 2: Find |A| = 2(2) - 3(1) = 4 - 3 = 1 ≠ 0, so A⁻¹ exists Step 3: Find adj(A) = [[2, -3], [-1, 2]] St

A: Step 1: Find the derivative: f'(x) = 6x + 2 Step 2: Evaluate at x = 1: f'(1) = 6(1) + 2 = 8 Step 3: The instantaneous rate of change at x = 1 is 8 units. Explanation: The derivative gives the slope of

A: Step 1: Velocity v(t) = ds/dt = 3t² - 12t + 9 Step 2: Set v(t) = 0: 3t² - 12t + 9 = 0 Step 3: Factor: 3(t² - 4t + 3) = 0 → 3(t - 1)(t - 3) = 0 Step 4: t = 1 or t = 3 Answer: The particle is at rest at

A: Use related rates when: - Two or more quantities are changing with respect to time - You know the rate of change of one quantity and need to find the rate of change of another - The quantities are rel

A: Step 1: Recognize that powers of i repeat every 4 terms: i^1=i, i^2=-1, i^3=-i, i^4=1 Step 2: Divide the exponent by 4: 7 ÷ 4 = 1 remainder 3 Step 3: Therefore i^7 = i^3 = -i Answer: -i Key insight:

A: Step 1: For i^1729: 1729 ÷ 4 = 432 remainder 1, so i^1729 = i^1 = i Step 2: For i^1950: 1950 ÷ 4 = 487 remainder 2, so i^1950 = i^2 = -1 Step 3: Add the results: i + (-1) = -1 + i Answer: -1 + i Meth

A: Step 1: Calculate individual terms: i + (-1) + (-i) + 1 + i + (-1) + (-i) + 1 Step 2: Group first 4 terms: (i - 1 - i + 1) = 0 Step 3: Group next 4 terms: (i - 1 - i + 1) = 0 Step 4: Total sum = 0 + 0