Let the domain of the function f(x)=log ₂ log ₄ log ₆(3+4 x-x²) be (a, b). If ∫₀ᵇ⁻ᵃ[x²] d x=p-√q-√r, p, q, r ∈ ℕ, gcd(p, q, r)=1, where [·] is the greatest integer function, then p+q+r is equal to

Solution

Maths sir · 4 steps

Peel the logs from the outside in to get the domain, then split the integral where x² crosses an integer.

1. Outer log first. For log₂ to exist, its input must be positive: log₄ log₆(3+4x-x²) > 0. That means log₆(3+4x-x²) > 1, so 3+4x-x² > 6. This one condition also makes the inner logs defined.
2. Find the domain. 3+4x-x² > 6 gives x²-4x+3 < 0, that is (x-1)(x-3) < 0. So the domain is (1, 3), with a = 1, b = 3 and b - a = 2.
3. Split the integral. On [0, 2], x² runs from 0 to 4. So [x²] changes value at x = 1, √2 and √3. It equals 0, 1, 2 and 3 on the four pieces.

   ∫₀² [x²] dx = 0(1-0) + 1(√2-1) + 2(√3-√2) + 3(2-√3) = 5 - √2 - √3

4. So. p = 5, q = 2, r = 3, and their gcd is 1. So p + q + r = 10, which is option (a).

The idea to keep. For nested logs, the outermost log decides the domain: keep making the inside bigger until you reach the polynomial. For ∫ [x²] dx, break the range at x = √n.

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