Let x=2 be a root of the equation x²+px+q=0 and f(x) = { (1 - cos (x² - 4px + q² + 8q + 16))/(x - 2p)⁴,, x ≠ 2p; 0,, x = 2p
Then lim _(x → 2p +) [f(x)], where [ . ] denotes greatest integer function, is

Solution

Maths sir · 4 steps

Use the root to turn the cosine's argument into a perfect square, then check which side of 1/2 f sits on.

1. Use the root. Since x = 2 is a root, 4 + 2p + q = 0, so q + 4 = -2p. Then q² + 8q + 16 = (q+4)² = 4p².
2. Clean up f. The argument of the cosine becomes x² - 4px + 4p² = (x - 2p)². Put u = (x-2p)², so the denominator (x-2p)⁴ is u².

   f(x) = (1-cos u)/u² = (2 sin²(u/2))/u² = 1/2(sin(u/2)/(u/2))²

3. Which side of 1/2. As x → 2p⁺, u → 0⁺, so f(x) → 1/2. Because 0 < sin θ < θ for small positive θ, the bracket is a little less than 1. So 0 < f(x) < 1/2 near x = 2p.
4. So. Near x = 2p the value of f(x) lies strictly between 0 and 1/2, so [f(x)] = 0 there. The limit is 0, option (c).

The idea to keep. With a greatest integer function outside a limit, finding the limit value is not enough. Check whether f approaches it from above or below. Here (sin θ)/θ < 1 keeps f below 1/2.

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