Let f:(0, π) → R be a function given by f(x)={(8/7)^((tan 8 x)/(tan 7 x)),, 0< x<π/2; a-8,, x=π/2; (1+| cot x)^(b/a|tan x|),, π/2 < x < π
where a, b ∈ Z. If f is continuous at x=π/2, then a²+b² is equal to _________.

Solution

Maths sir · 4 steps

Near π/2 the exponent tends to 0, not 8/7, so the left limit is just 1.

1. Shift to h. For the left-hand limit put x = π/2 - h with h → 0⁺. Then tan 8x = tan(4π - 8h) = -tan 8h and tan 7x = tan(7π/2 - 7h) = cot 7h.

   (tan 8x)/(tan 7x) = (-tan 8h)/(cot 7h) = -tan 8h tan 7h → 0

2. Left limit gives a. The left limit is (8/7)⁰ = 1. Continuity needs f(π/2) = a - 8 = 1, so a = 9.
3. Right limit gives b. As x → π/2⁺, let t = |cot x| → 0, so |tan x| = 1/t. Then (1+t)^(b/a·1/t) → e^(b/a) = e^(b/9). This must equal 1, so b = 0.
4. So. a² + b² = 81 + 0 = 81.

The idea to keep. Away from x = 0, do not use tan kx ≈ kx. Shift to x = c - h first. For a 1^∞ form, the limit is e raised to the limit of (base - 1) × exponent.

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