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
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
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The Correct Option is A
Approach Solution - 1
The problem requires us to first determine the domain (a, b) of the function \( f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) \). Then, we need to evaluate the integral \( \int_0^{b-a} [x^2] dx \), where [.] denotes the greatest integer function. Finally, by comparing the result with the given form \( p - \sqrt{q} - \sqrt{r} \), we need to find the value of \( p + q + r \).
Concept Used:
1. Domain of Nested Logarithmic Functions: For a function of the form \( \log_{b_1}(\log_{b_2}(\ldots \log_{b_n}(y)\ldots)) \) to be defined, where all bases \( b_i > 1 \), the argument of each logarithm must be greater than 1. This means we must satisfy the condition \( \log_{b_{n}}(y) > 1 \), which implies \( y > b_n \).
2. Integral of the Greatest Integer Function: To evaluate an integral of the form \( \int [g(x)] dx \), the interval of integration must be split into subintervals where the value of \( [g(x)] \) remains constant. The split points are the values of x for which \( g(x) \) is an integer.
Step-by-Step Solution:
Step 1: Determine the domain of the function \( f(x) \).
For \( f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2) \) to be defined, the argument of the innermost logarithm must be positive, and subsequently, the argument of each outer logarithm must also be positive. For a nested logarithm with bases greater than 1, we require the argument of the second logarithm to be greater than 1.
\[ \log_4 (\log_6 (3 + 4x - x^2)) > 0 \]
Since the base is 4 (> 1), this simplifies to:
\[ \log_6 (3 + 4x - x^2) > 4^0 = 1 \]
Again, since the base is 6 (> 1), this gives:
\[ 3 + 4x - x^2 > 6^1 = 6 \]
We solve this quadratic inequality:
\[ -x^2 + 4x - 3 > 0 \]
Multiplying by -1 reverses the inequality sign:
\[ x^2 - 4x + 3 < 0 \]
Factoring the quadratic expression, we get:
\[ (x - 1)(x - 3) < 0 \]
This inequality holds true for \( 1 < x < 3 \). Thus, the domain of the function is (1, 3). Comparing this with the given domain (a, b), we have \( a = 1 \) and \( b = 3 \).
Step 2: Determine the limits of the integral.
The upper limit of the integral is \( b - a \). Substituting the values of a and b:
\[ b - a = 3 - 1 = 2 \]
So, we need to evaluate the integral \( I = \int_0^2 [x^2] dx \).
Step 3: Evaluate the integral \( \int_0^2 [x^2] dx \).
The value of \( [x^2] \) changes when \( x^2 \) becomes an integer. We split the integration interval [0, 2] at points where \( x^2 \) is an integer, i.e., at \( x = \sqrt{1}, \sqrt{2}, \sqrt{3} \).
\[ I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx \]
Now, we substitute the value of \( [x^2] \) in each subinterval:
- For \( 0 \le x < 1 \), \( 0 \le x^2 < 1 \), so \( [x^2] = 0 \).
- For \( 1 \le x < \sqrt{2} \), \( 1 \le x^2 < 2 \), so \( [x^2] = 1 \).
- For \( \sqrt{2} \le x < \sqrt{3} \), \( 2 \le x^2 < 3 \), so \( [x^2] = 2 \).
- For \( \sqrt{3} \le x < 2 \), \( 3 \le x^2 < 4 \), so \( [x^2] = 3 \).
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 part:
\[ I = 0 + [x]_1^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^2 \] \[ I = (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (3(2) - 3\sqrt{3}) \] \[ I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} \] \[ I = (6 - 1) + (\sqrt{2} - 2\sqrt{2}) + (2\sqrt{3} - 3\sqrt{3}) \] \[ I = 5 - \sqrt{2} - \sqrt{3} \]
Final Computation & Result:
We are given that the value of the integral is \( p - \sqrt{q} - \sqrt{r} \).
Comparing our result \( 5 - \sqrt{2} - \sqrt{3} \) with the given form, we can identify:
\[ p = 5, \quad q = 2, \quad r = 3 \]
We verify the conditions: \( p, q, r \in \mathbb{N} \) and \( \text{gcd}(p, q, r) = \text{gcd}(5, 2, 3) = 1 \). The conditions are satisfied.
The question asks for the value of \( p + q + r \).
\[ p + q + r = 5 + 2 + 3 = 10 \]
The value of \( p + q + r \) is 10.
Approach Solution -2
\( \log_6(3 + 4x - x^2)>1 \)
\( 3 + 4x - x^2>6 \) \( x^2 - 4x + 3<0 \)
\( (x-1)(x-3)<0 \) \( x \in (1, 3) \) So \( a = 1 \) and \( b = 3 \)
\( \Rightarrow \int_0^{2} [x^2] dx = ? \) \( I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^{2} [x^2] dx \)
\( = 0 + |x|_1^{\sqrt{2}} + 2|x|_{\sqrt{2}}^{\sqrt{3}} + 3|x|_{\sqrt{3}}^{2} \) \( = (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) \) \( = 5 - \sqrt{2} - \sqrt{3} \) \( \Rightarrow p + q + r = 10 \)
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