Question

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

• A. 10
• B. 8
• C. 11
• D. 9

Hint:

Find the domain of the function by solving the inequalities and then evaluate the integral by splitting it into intervals where \( [x^2] \) is constant.

Answer 1

The Correct Option is A

The objective is to determine the domain (a, b) for the function \( f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) \). Subsequently, we must compute the integral \( \int_0^{b-a} [x^2] dx \), where [.] represents the greatest integer function. Finally, by equating this result to \( p - \sqrt{q} - \sqrt{r} \), we will find the value of \( p + q + r \).

Concept Used:

1. Domain of Nested Logarithms: For a nested logarithmic function \( \log_{b_1}(\log_{b_2}(\ldots \log_{b_n}(y)\ldots)) \) with bases \( b_i>1 \), each argument must be positive. Specifically, for the second logarithm, the condition is \( \log_{b_{n}}(y)>1 \), implying \( y>b_n \).

2. Integral of Greatest Integer Function: To evaluate \( \int [g(x)] dx \), divide the integration interval into subintervals where \( [g(x)] \) is constant. The endpoints of these subintervals are where \( g(x) \) is an integer.

Step-by-Step Solution:

Step 1: Determine the domain of \( f(x) \).

For \( f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) \) to be defined, and given the bases are greater than 1, we need:

\[ \log_4 (\log_6 (3 + 4x - x^2)) > 0 \]

This implies:

\[ \log_6 (3 + 4x - x^2) > 4^0 = 1 \]

This further implies:

\[ 3 + 4x - x^2 > 6^1 = 6 \]

Solving the quadratic inequality \( -x^2 + 4x - 3>0 \), which is equivalent to \( x^2 - 4x + 3<0 \), we factor it to \( (x - 1)(x - 3)<0 \). This inequality holds for \( 1<x<3 \). Therefore, the domain is (1, 3), meaning \( a = 1 \) and \( b = 3 \).

Step 2: Determine the integral limits.

The upper limit is \( b - a = 3 - 1 = 2 \). The integral to evaluate is \( I = \int_0^2 [x^2] dx \).

Step 3: Evaluate \( \int_0^2 [x^2] dx \).

We split the interval [0, 2] where \( x^2 \) is an integer: at \( x = \sqrt{1}, \sqrt{2}, \sqrt{3} \).

\[ I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx \]

The values of \( [x^2] \) in each subinterval are:

• \( [x^2] = 0 \) for \( 0 \le x < 1 \)
• \( [x^2] = 1 \) for \( 1 \le x < \sqrt{2} \)
• \( [x^2] = 2 \) for \( \sqrt{2} \le x < \sqrt{3} \)
• \( [x^2] = 3 \) for \( \sqrt{3} \le x < 2 \)

The integral becomes:

\[ I = \int_0^1 0 \,dx + \int_1^{\sqrt{2}} 1 \,dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \,dx + \int_{\sqrt{3}}^2 3 \,dx \]

Evaluating each integral:

\[ I = 0 + [x]_1^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^2 \] \[ I = (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) \] \[ I = 5 - \sqrt{2} - \sqrt{3} \]

Final Computation & Result:

The integral's value is \( 5 - \sqrt{2} - \sqrt{3} \), which matches the form \( p - \sqrt{q} - \sqrt{r} \). Thus, \( p = 5, q = 2, r = 3 \). These are natural numbers and their greatest common divisor is 1, satisfying the conditions.

The required value is \( p + q + r = 5 + 2 + 3 = 10 \).

The value of \( p + q + r \) is 10.