If the function f(x)=(sin 3 x+α sin x-β cos 3 x)/x³, x ∈ R, is continuous at x=0, then f(0) is equal to :

Solution

Maths sir · 4 steps

Fix β and α so the numerator vanishes like x³, then the triple-angle identity finishes it.

1. Kill the constant. For f to be continuous at 0, lim(x → 0) f(x) must be finite. The denominator x³ goes to 0, so the numerator must also go to 0. At x = 0 the numerator is 0 + 0 - β cos 0 = -β, so β = 0.
2. Kill the x term. Near 0, sin 3x + α sin x ≈ 3x + α x = (3 + α)x. Dividing this by x³ would blow up unless 3 + α = 0, so α = -3.

   sin 3x - 3 sin x = (3 sin x - 4 sin³ x) - 3 sin x = -4 sin³ x

3. Take the limit. f(0) = lim(x → 0) (-4 sin³ x)/x³ = -4(lim(x → 0)(sin x)/x)³ = -4 × 1 = -4.
4. So. With β = 0 and α = -3, continuity forces f(0) = -4, which is option (c).

The idea to keep. When a limit over xⁿ must be finite, set each lower power in the numerator to zero one by one. The surviving xⁿ coefficient is the limit.

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