Let f:( - π/4,π/4 ) → R be defined as f(x) = { (1 + |sin x|)^(3a/(|sin x|)), ,, - π/4 < x < 0; b, ,, x = 0; e^(cot 4x/cot 2x), ,, 0 < x < π/4
If f is continuous at x = 0, then the value of 6a + b² is equal to :

Solution

Maths sir · 4 steps

Both one-sided limits must equal b: the left side is a 1^∞ form, the right is a plain ratio.

1. Left-hand limit. Put t = |sin x|. As x → 0⁻, t → 0⁺, so the expression is (1+t)^(3a/t). This is a 1^∞ form, and since (1+t)^(1/t) → e, the limit is e³ᵃ.
2. Right-hand limit. Write the exponent as (cot 4x)/(cot 2x) = (cos 4x sin 2x)/(sin 4x cos 2x). Using sin 4x = 2 sin 2x cos 2x, the sin 2x cancels and the exponent becomes (cos 4x)/(2 cos² 2x) → 1/2. So the limit is e^(1/2).

   e³ᵃ = b = e^(1/2)

3. Match them. Continuity needs both limits equal to f(0) = b. So 3a = 1/2, giving a = 1/6, and b = √e.
4. So. 6a + b² = 1 + e, which is option (c).

The idea to keep. For continuity, compute each one-sided limit separately and set both equal to f(0). Whenever you see (1+t)^(k/t) with t → 0, it goes to eᵏ.

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