Question

Let the domain of \( f(x) = \log_3 \log_3 \log_7 (9x - x^2 - 13) \) be (m, n). Let the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) have eccentricity \( \frac{n}{3} \) and latus rectum \( \frac{8m}{3} \). Then \( b^2 - a^2 \) is equal to :

• A. 9
• B. 11
• C. 5
• D. 7

Hint:

For \( \log_a(\log_b(x))>0 \), remember it simplifies to \( \log_b(x)>a^0 \), which is \( \log_b(x)>1 \), and finally \( x>b^1 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The domain of a log function \( \log(u) \) requires \( u>0 \). We apply this nested condition three times to find the values of \( m \) and \( n \).
Step 2: Detailed Explanation:
For the domain: 1. \( \log_3 \log_7 (9x - x^2 - 13)>0 \implies \log_7 (9x - x^2 - 13)>3^0 = 1 \). 2. \( 9x - x^2 - 13>7^1 \implies x^2 - 9x + 20<0 \implies (x-4)(x-5)<0 \). Thus, domain is \( (4, 5) \), so \( m = 4 \) and \( n = 5 \). Hyperbola: Eccentricity \( e = 5/3 \). Latus Rectum \( \frac{2b^2}{a} = \frac{8(4)}{3} = \frac{32}{3} \implies b^2 = \frac{16a}{3} \). Using \( b^2 = a^2(e^2 - 1) \): \[ \frac{16a}{3} = a^2 \left(\frac{25}{9} - 1\right) = a^2 \left(\frac{16}{9}\right) \] \[ \frac{16a}{3} = \frac{16a^2}{9} \implies a = 3 \] Then \( b^2 = \frac{16(3)}{3} = 16 \). \( b^2 - a^2 = 16 - 9 = 7 \).
Step 3: Final Answer:
The value is 7.