Let the domain of the function \(f(\mathrm{x})=\log _{2} \log _{4} \log _{6}\left(3+4 \mathrm{x}-\mathrm{x}^{2}\right)\) be \((\mathrm{a}, \mathrm{b})\). If \(\int_{0}^{\mathrm{b}-\mathrm{a}}\left[\mathrm{x}^{2}\right] \mathrm{dx}=\mathrm{p}-\sqrt{\mathrm{q}}-\sqrt{\mathrm{r}}, \mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathbb{N}, \operatorname{gcd}(\mathrm{p}, \mathrm{q}, \mathrm{r})=1\), where \([\cdot]\) is the greatest integer function, then \(\mathrm{p}+\mathrm{q}+\mathrm{r}\) is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Sets Relations and Functions

Let the domain of the function

f (x) = lo g_{2} lo g_{4} lo g_{6} (3 + 4 x - x^{2})

(a, b)

. If

\int_{0}^{b - a} [x^{2}] dx = p - q - r, p, q, r \in N, gcd (p, q, r) = 1

,
where

[\cdot]

is the greatest integer function, then

p + q + r

is equal to

Answer: (A)

The outer logarithm requires

lo g_{4} lo g_{6} (3 + 4 x - x^{2}) > 0

, i.e.

lo g_{6} (3 + 4 x - x^{2}) > 1

, i.e.

3 + 4 x - x^{2} > 6

. This is

x^{2} - 4 x + 3 < 0

, so

(x - 1) (x - 3) < 0

and

x \in (1, 3)

; hence

a = 1, b = 3

and

b - a = 2

. On

[0, 2]

[x^{2}]

takes values

0, 1, 2, 3

[0, 1), [1, 2), [2, 3), [3, 2)

, so

\int_{0}^{2} [x^{2}] d x = (2 - 1) + 2 (3 - 2) + 3 (2 - 3) = 5 - 2 - 3 .

Thus

p = 5, q = 2, r = 3

and

p + q + r = 10

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option