If α = lim _(x → π/4) (tan ³x - tan x)/(cos ( x + π/4 )) and β = lim _(x → 0) (cos x)ᶜᵒᵗˣ are the roots of the equation, ax² + bx - 4 = 0, then the ordered pair (a, b) is :

Solution

Maths sir · 4 steps

Find both limits first, then read a and b straight off the sum and product of roots.

1. Find alpha. At x = π/4 the top is 1 - 1 = 0 and the bottom is cos π/2 = 0, so this is 0/0. Use L'Hospital's rule. The top differentiates to (3 tan² x - 1)sec² x = (3 - 1)(2) = 4, and the bottom to -sin(x + π/4) = -1. So α = -4.
2. Find beta. As x → 0, cos x → 1 and cot x → ∞, a 1^∞ form. For this form, lim fᵍ = eˡⁱᵐᵍ⁽ᶠ⁻¹⁾.

   β = e^(lim(x → 0) cot x (cos x - 1)) = e^(lim(x → 0) (cos x - 1)/(tan x)) = e⁰ = 1

3. Match the roots. The roots are -4 and 1. For ax² + bx - 4 = 0, the product is -4/a = (-4)(1) = -4, so a = 1. The sum is -b/a = -4 + 1 = -3, so b = 3.
4. So. The equation is x² + 3x - 4 = 0, and (a, b) = (1, 3), which is option (d).

The idea to keep. For a 1^∞ limit, use eˡⁱᵐᵍ⁽ᶠ⁻¹⁾. For 0/0 in trigonometric forms, L'Hospital is usually the fastest route. Then use sum = -b/a and product = c/a.

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