If lim x → 1⁺ ((x-1)(6+λ cos (x-1))+μ sin (1-x))/(x-1)³=-1, where λ, μ ∈ ℝ, then λ+μ is equal to

Solution

Maths sir · 5 steps

Put t = x - 1, expand cos and sin up to t³, and make the numerator start at t³.

1. Shift the variable. Let t = x - 1, so t → 0⁺. Since sin(1-x) = -sin t, the numerator becomes t(6 + λ cos t) - μ sin t and the denominator is t³.
2. Expand in powers. Use cos t = 1 - t²/2 + … and sin t = t - t³/6 + … and collect the terms by power of t.

   t(6+λ cos t) - μ sin t = (6 + λ - μ) t + (-λ/2 + μ/6)t³ + …

3. Match coefficients. The limit is finite, so the t term must vanish: 6 + λ - μ = 0. The t³ coefficient must equal the limit: -λ/2 + μ/6 = -1, that is, μ = 3λ - 6.
4. Solve. Substituting gives 6 + λ - 3λ + 6 = 0, so λ = 6 and μ = 12.
5. So. λ + μ = 6 + 12 = 18, which is option (c).

The idea to keep. When a limit over tⁿ is finite, every term in the numerator below tⁿ must cancel to zero. The tⁿ coefficient then equals the limit. Expand only as far as you need.

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