Let the domain of the function f ( x )=log 3 log 5 log 7 (9 x-x 2 -13 ) be the interval (m, n) . Let the hyperbola frac x 2 a 2 -frac y 2 b 2 =1 have eccentricity frac n 3 and the length of the latus rectum 8 m/3 . Then b 2 -a 2 is equal to :

Mathematics · Functions

JEE Main 2026 — 23 January, Morning Shift — Question 1

Let the domain of the function $\mathrm{f}(\mathrm{x})=\log _{3} \log _{5} \log _{7} \left(9 x-x^{2}-13\right)$ be the interval $(m, n)$ . Let the hyperbola $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 m}{3}$ . Then $b^{2}-a^{2}$ is equal to :

Option A:
5
Option B:
11
Option C:
9
Option D:
7
Correct

Answer: D

Step-by-step solution

\log_{5}\!\left(\log_{7}(9x - x^{2} - 13)\right) > 0

\Rightarrow 9x - x^{2} - 13 > 7

x^{2} - 9x + 20 < 0 \Rightarrow 4 < x < 5

m = 4, \quad n = 5

\Rightarrow e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \frac{5}{3}

\Rightarrow \frac{b^{2}}{a^{2}} = \frac{25}{9} - 1 = \frac{16}{9}

\Rightarrow \frac{b}{a} = \frac{4}{3}

\Rightarrow \frac{2b^{2}}{a} = \frac{8m}{3} \Rightarrow \frac{2b^{2}}{a} = \frac{32}{3}

\Rightarrow 2b^{2} = \frac{32}{3} \cdot \frac{3b}{4} \Rightarrow b = 4, \quad a = 3

b^{2} - a^{2} = 16 - 9 = 7

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Functions
Topic: Domain & range of functions