Functions – Domain & Conic Sections – JEE Main 23 Jan 2026 Shift 1

Solution:

For the function $f(x) = \log_{3}(\log_{5}(\log_{7}(9x-x^{2}-13)))$ to be defined, the argument of the outermost logarithm must be positive.
$$ \log_{5}(\log_{7}(9x-x^{2}-13)) > 0 $$

Using the definition of logarithms ($ \log_b a > 0 \Rightarrow a > b^0 $ for $b > 1$):
$$ \log_{7}(9x-x^{2}-13) > 5^0 $$

$$ \log_{7}(9x-x^{2}-13) > 1 $$

$$ 9x-x^{2}-13 > 7^1 $$

$$ 9x-x^{2}-13 > 7 $$

$$ x^{2}-9x+20 < 0 $$
Factorizing the quadratic expression:
$$ (x-4)(x-5) < 0 $$
$$ \Rightarrow x \in (4, 5) $$
Comparing this with the interval $(m, n)$, we get $m = 4$ and $n = 5$.

Now, consider the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Given:
Eccentricity $e = \frac{n}{3} = \frac{5}{3}$
Length of Latus Rectum $L.R. = \frac{8m}{3} = \frac{8(4)}{3} = \frac{32}{3}$

Formula for Latus Rectum:
$$ \frac{2b^2}{a} = \frac{32}{3} \Rightarrow b^2 = \frac{16a}{3} \quad \dots(1) $$

Formula for Eccentricity ($b^2 = a^2(e^2-1)$):
$$ b^2 = a^2\left(\left(\frac{5}{3}\right)^2 – 1\right) $$

$$ b^2 = a^2\left(\frac{25}{9} – 1\right) = a^2\left(\frac{16}{9}\right) \quad \dots(2) $$

Equating (1) and (2):
$$ \frac{16a}{3} = \frac{16a^2}{9} $$

$$ \frac{1}{3} = \frac{a}{9} \Rightarrow a = 3 $$

Substituting $a=3$ in (1):
$$ b^2 = \frac{16(3)}{3} = 16 $$

We need to find $b^2 – a^2$:
$$ b^2 – a^2 = 16 – 3^2 = 16 – 9 = 7 $$

Ans. (4)