If α=lim x → 0⁺((e^(√(tan x))-e^(√x))/(√(tan x)-√x)) and β=lim x → 0(1+sin x)^(1/2 cot x) are the roots of the quadratic equation a x²+b x-√e=0, then 12 log ₑ(a+b) is equal to _________.

Solution

Maths sir · 4 steps

Find α and β first, then let the product and sum of roots give you a and b.

1. Find alpha. Put u = √(tan x) and v = √x. Both tend to 0, and (eᵘ - eᵛ)/(u - v) = eᵛ · (eᵘ⁻ᵛ - 1)/(u - v). As u - v → 0 the fraction tends to 1 and eᵛ → 1, so α = 1.
2. Find beta. This is a 1^∞ form, so β = e^L where L is the limit of (base - 1) times the exponent.

   L = lim(x → 0) sin x · 1/2 cot x = lim(x → 0) 1/2 cos x = 1/2 ⇒ β = e^(1/2) = √e

3. Use the roots. For ax² + bx - √e = 0, the product of roots is -√e/a = 1 × √e, so a = -1. The sum of roots is -b/a = b = 1 + √e.
4. So. a + b = -1 + 1 + √e = √e, and 12 logₑ √e = 12 × 1/2 = 6.

The idea to keep. For a 1^∞ limit, the answer is e raised to the limit of (base minus 1) times exponent. And when the roots are known, find a from the product first.

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