JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

Let the domain of the function $f (x) = lo g_{3} lo g_{5} lo g_{7} (9 x - x^{2} - 13)$ 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 $\frac{8 m}{3}$ . Then $b^{2} - a^{2}$ is equal to :

A
$5$
B
$11$
C
$9$
D
$7$

Solution

Step 1 — Domain. For $lo g_{3}$ to be defined we need $lo g_{5} lo g_{7} (9 x - x^{2} - 13) > 0$ , hence $lo g_{7} (9 x - x^{2} - 13) > 1$ , i.e. $9 x - x^{2} - 13 > 7$ . This gives $x^{2} - 9 x + 20 < 0$ , so $4 < x < 5$ . Thus $m = 4$ , $n = 5$ .

Step 2 — Hyperbola parameters. Eccentricity $e = n /3 = 5/3$ , so $\frac{b ^{2}}{a ^{2}} = e^{2} - 1 = \frac{16}{9}$ , giving $\frac{b}{a} = \frac{4}{3}$ .

Step 3 — Latus rectum: $\frac{2 b ^{2}}{a} = \frac{8 m}{3} = \frac{32}{3}$ , so $b^{2} = \frac{16 a}{3}$ .

Step 4 — Solve. Combining $b^{2} = \frac{16 a ^{2}}{9}$ with $b^{2} = \frac{16 a}{3}$ gives $a = 3$ and $b = 4$ .

$∴ b^{2} - a^{2} = 16 - 9 = 7$ .

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