For some a, b, let f(x)=|a+(sin x)/x, 1, b; a, 1+(sin x)/x, b; a, 1, b+(sin x)/x|, x ≠ 0, lim _(x → 0) f(x)=λ+μ a+ν b. Then (λ+μ+v)² is equal to :

Solution

Maths sir · 4 steps

Let x → 0 inside the determinant first, then two row subtractions make it easy to expand.

1. Limit goes inside. The determinant is a polynomial in its entries, so the limit can be taken entry by entry. Since lim(x → 0) (sin x)/x = 1, every (sin x)/x becomes 1.

   lim(x → 0) f(x) = |a+1, 1, b; a, 2, b; a, 1, b+1|

2. Clear the rows. Do R₁ → R₁ - R₂ and R₃ → R₃ - R₂. The rows become (1, -1, 0), (a, 2, b) and (0, -1, 1), and the determinant does not change.
3. Expand along R1. 1 × (2 × 1 - b × (-1)) - (-1) × (a × 1 - b × 0) = 2 + b + a. Comparing with λ + μ a + ν b gives λ = 2, μ = 1, ν = 1.
4. So. λ + μ + ν = 4, and (λ + μ + ν)² = 16, which is option (b).

The idea to keep. A determinant is continuous in its entries, so a limit can go straight inside. Then use row operations to create zeros before you expand.

Answer from the NTA final answer key. Solution by PrepWiser.