JEE Mathematics 2025 Question with Solution

Given: The domain of $\log_5(18x - x^2 - 77)$ is $(\alpha, \beta)$ and the domain of $\log_{(x-1)}\left(\frac{2x^2 + 3x - 2}{x^2 - 3x - 4}\right)$ is $(\gamma, \delta)$.

Find: The value of $\alpha^2 + \beta^2 + \gamma^2$.

For the first logarithmic function, the argument must be positive:

$$18x - x^2 - 77 > 0$$

Rearranging,

$$x^2 - 18x + 77 < 0$$

Factoring,

$$(x - 7)(x - 11) < 0$$

Hence,

$$x \in (7, 11)$$

So, $\alpha = 7$ and $\beta = 11$.

For the second logarithmic function, both the base conditions and the argument condition must hold.

Base condition:

$$x - 1 > 0$$

so $x > 1$. Also, the base cannot be $1$, so $x \ne 2$.

Argument condition:

$$\frac{2x^2 + 3x - 2}{x^2 - 3x - 4} > 0$$

Factoring,

$$\frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)} > 0$$

Using the sign chart method, the valid range is

$$x \in (4, \infty)$$

Thus, $\gamma = 4$.

Now compute:

$$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + 11^2 + 4^2 = 49 + 121 + 16 = 186$$

Therefore, the correct option is C.

Given: Two logarithmic expressions with domains $(\alpha, \beta)$ and $(\gamma, \delta)$.

Find: $\alpha^2 + \beta^2 + \gamma^2$.

For $\log_5(18x - x^2 - 77)$, only the argument condition is needed because the base $5$ is already valid.

$$18x - x^2 - 77 > 0$$ $$-x^2 + 18x - 77 > 0$$ $$x^2 - 18x + 77 < 0$$ $$(x - 7)(x - 11) < 0$$

So the first domain is

$$(\alpha, \beta) = (7, 11)$$

For $\log_{(x-1)}\left(\frac{2x^2 + 3x - 2}{x^2 - 3x - 4}\right)$, the logarithm exists only when:

1. the base is positive,
2. the base is not equal to $1$,
3. the argument is positive.

From the base,

$$x - 1 > 0 \Rightarrow x > 1$$

and

$$x - 1 \ne 1 \Rightarrow x \ne 2$$

Now solve

$$\frac{2x^2 + 3x - 2}{x^2 - 3x - 4} > 0$$

Factor both polynomials:

$$2x^2 + 3x - 2 = (2x - 1)(x + 2)$$ $$x^2 - 3x - 4 = (x - 4)(x + 1)$$

Thus,

$$\frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)} > 0$$

From the sign chart, the valid interval used in the solution is

$$x \in (4, \infty)$$

Hence,

$$\gamma = 4$$

Finally,

$$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + 11^2 + 4^2$$ $$= 49 + 121 + 16$$ $$= 186$$

Therefore, the correct option is C, and the required value is $186$.