CBSE 12th Maths Question Paper 2018 (Set 1)

We know that |(AB)⁻¹| = 1/|AB| = 1/(|A||B|). Given |A| = 2 and |(AB)⁻¹| = 1/6, we have 1/(2|B|) = 1/6. Thus, 2|B| = 6, which implies |B| = 3.

Let u = x². The expression becomes sin²(u). Differentiating sin²(u) with respect to u, we get d/du(sin²u) = 2 sin u cos u = sin(2u). Substituting back, we get sin(2x²).

The order of a differential equation is the order of the highest derivative present. In the given equation, the highest derivative is d²y/dx², which is of order 2.

For the first part: The sum of squares of direction cosines is 1. So, (1/√3)² + (1/√6)² + n² = 1, which means 1/3 + 1/6 + n² = 1. Thus 1/2 + n² = 1, so n² = 1/2, n = 1/√2. The angle θ with z-axis is given by cos θ = n = 1/√2, so θ = 45° (or π/4 radians). For the OR part: The direction ratios are (2, 1, 2). The magnitude is √(2² + 1² + 2²) = √9 = 3. Dividing by magnitude, the direction cosines are (2/3, 1/3, 2/3).

Let (e₁, e₂) be the identity element. Then (a, b) * (e₁, e₂) = (a, b). This implies (ae₂ + be₁, be₂) = (a, b). From be₂ = b, we get e₂ = 1 (since b ≠ 0). From ae₂ + be₁ = a, putting e₂ = 1 gives a + be₁ = a, implying be₁ = 0, so e₁ = 0. Thus, the identity element is (0, 1).