If the function f(x) = { (1 + |cos x|)^(λ / |cos x|), ,, 0 < x < π/2; μ, ,, x = π/2; e^((cot 6x)/(cot 4x)), ,, π/2 < x < π
is continuous at x = π/2, then 9λ + 6log ₑμ + μ ⁶ - e^(6λ) is equal to

Solution

Maths sir · 4 steps

Both one-sided limits must equal μ: the left one fixes λ, the right one fixes μ.

1. Left-hand limit. As x → π/2⁻, put t = |cos x| → 0⁺. Then (1+t)^(λ/t) = [(1+t)^(1/t)]^λ → e^λ, using the standard limit (1+t)^(1/t) → e.
2. Right-hand limit. Put x = π/2 + h with h → 0⁺. Then sin 6x = -sin 6h, cos 6x = -cos 6h, sin 4x = sin 4h and cos 4x = cos 4h.

   (cot 6x)/(cot 4x) = (cos 6x)/(cos 4x) × (sin 4x)/(sin 6x) = (cos 6h)/(cos 4h) × (sin 4h)/(sin 6h) → 1 × 4/6 = 2/3

3. Match the limits. Continuity at π/2 needs e^λ = μ = e^(2/3). So λ = 2/3 and μ = e^(2/3).
4. So. 9λ = 6, 6 logₑ μ = 4, μ⁶ = e⁴ and e^(6λ) = e⁴. The expression is 6 + 4 + e⁴ - e⁴ = 10, option (b).

The idea to keep. For (1+t)^(k/t) with t → 0, the limit is eᵏ. For a ratio of cotangents at a point where both blow up, shift to x = a + h and use (sin mh)/(sin nh) → m/n.

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