If α is positive root of the equation, p(x) = x² - x - 2 = 0, then
lim _(x → α +) √(1 - cos ( p( x ) ))/(x + α - 4) is equal to :

Solution

Maths sir · 4 steps

Turn 1-cos into 2 sin², take the root, and let p(x) factor so that x-2 cancels.

1. Find alpha. x² - x - 2 = (x-2)(x+1), so the roots are 2 and -1. The positive root is α = 2, and the denominator becomes x + 2 - 4 = x - 2.
2. Remove the root. Use 1 - cos θ = 2 sin² θ/2, so √(1-cos p(x)) = √2|sin p(x)/2|. As x → 2⁺, p(x) = (x-2)(x+1) is a small positive number, so the sine is positive and the modulus can be dropped.

   lim(x → 2⁺) (√2 sin p(x)/2)/(p(x)/2) · (x-2)(x+1)/2(x-2) = √2 × 1 × 3/2

3. Cancel and evaluate. The first fraction tends to 1 because p(x)/2 → 0. In the second, x-2 cancels and (x+1)/2 → 3/2.
4. So. The limit is 3√2/2 = 3/√2, which is option (c).

The idea to keep. Whenever you see √(1-cos θ), write it as √2 |sin θ/2| and check the sign of θ from the side of approach. Then factor so that the zero term cancels.

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