JEE Mathematics 2026 Question with Solution

Given: $f(x) = \log_3 \log_3 \log_7 (9x - x^2 - 13)$ has domain $(m,n)$.

Find: $b^2-a^2$ for the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with eccentricity $\frac{n}{3}$ and latus rectum $\frac{8m}{3}$.

For the domain, the outer logarithm requires

$$\log_3 \log_7 (9x-x^2-13) > 1$$

so

$$\log_7 (9x-x^2-13) > 3$$

From the extracted solution working, this simplifies as

$$\log_3 \log_7 (9x - x^2 - 13)>0 \implies \log_7 (9x - x^2 - 13)>3^0 = 1$$

and then

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

Therefore,

$$x^2 - 9x + 20 < 0$$ $$(x-4)(x-5)<0$$

Hence the domain is

$$(4,5)$$

So, $m=4$ and $n=5$.

Now for the hyperbola, eccentricity is

$$e=\frac{n}{3}=\frac{5}{3}$$

and latus rectum is

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

Thus,

$$b^2=\frac{16a}{3}$$

Also, for the hyperbola,

$$b^2=a^2(e^2-1)$$

Substituting $e=\frac{5}{3}$,

$$\frac{16a}{3}=a^2\left(\frac{25}{9}-1\right)=a^2\left(\frac{16}{9}\right)$$

So,

$$\frac{16a}{3}=\frac{16a^2}{9}$$

which gives

$$a=3$$

Then,

$$b^2=\frac{16(3)}{3}=16$$

Therefore,

$$b^2-a^2=16-9=7$$

So the correct option is D.

Given: domain is $(m,n)$ and the hyperbola has $e=\frac{n}{3}$, latus rectum $\frac{8m}{3}$.

Find: $b^2-a^2$.

From the logarithmic domain condition shown in the solution,

$$(x-4)(x-5)<0$$

So directly,

$$(m,n)=(4,5)$$

Hence,

$$e=\frac{5}{3}, \qquad \frac{2b^2}{a}=\frac{32}{3}$$

Using

$$b^2=a^2(e^2-1)=a^2\left(\frac{25}{9}-1\right)=\frac{16a^2}{9}$$

and also

$$b^2=\frac{16a}{3}$$

Equating,

$$\frac{16a}{3}=\frac{16a^2}{9}$$

which gives

$$a=3$$

Then,

$$b^2=16$$

Therefore,

$$b^2-a^2=16-9=7$$

Thus, the correct option is D.