Suppose lim _(x → 0) F(x)/x³ exists and is equal to L, where
F(x) = | a + sin x/2, - b cos x, 0; - b cos x, 0, a + sin x/2; 0, a + sin x/2, - b cos x |.
Then, -112 L is equal to ___________.

Solution

Maths sir · 5 steps

Call the two repeated entries p and q, and the whole determinant collapses to -(p³+q³).

1. Name the entries. Let p = a + sin x/2 and q = -b cos x. The rows are (p, q, 0), (q, 0, p) and (0, p, q). Expanding along the first row gives p(0 - p²) - q(q² - 0).

   F(x) = -(p³ + q³) = -(p+q)(p² - pq + q²)

2. Kill the constant. At x = 0, F(0) = -(a³ - b³). Dividing by x³ needs F(0) = 0, so a³ = b³, which gives a = b for real numbers.
3. Kill the x term. With a = b, p + q = sin x/2 + a(1 - cos x) ≈ x/2, and p² - pq + q² → 3a². So F ≈ -3a²/2x, which blows up after dividing by x³ unless a = 0. Hence a = b = 0.
4. Find L. Now q = 0 and F(x) = -sin³ x/2. Since sin x/2 ≈ x/2, L = lim(x → 0) -sin³(x/2)/x³ = -1/8.
5. So. -112L = -112 × (-1/8) = 14.

The common slip: -14. Expanding the determinant with the wrong sign gives F = p³ + q³, so L = 1/8 and the answer comes out as -14. Check the minus sign on the middle cofactor.

The idea to keep. For lim F(x)/xⁿ to exist, every term of F below xⁿ must vanish. Set the constant term to zero first, then the x term, and so on. That fixes the unknown constants.

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