JEE Mathematics 2025 Question with Solution

Given: $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$

Find: The value of $\alpha + \beta + \gamma + \delta$ when the domain is $(\alpha, \beta) \cup (\gamma, \delta)$.

For the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ to be defined, we need two conditions to be satisfied.

The argument of the outer logarithm must be positive:

$$1 - \log_4(x^2 - 9x + 18) > 0$$ $$1 > \log_4(x^2 - 9x + 18)$$ $$4^1 > x^2 - 9x + 18$$ $$4 > x^2 - 9x + 18$$ $$0 > x^2 - 9x + 14$$ $$x^2 - 9x + 14 < 0$$

Factoring the quadratic:

$$(x - 2)(x - 7) < 0$$

This inequality holds for $2 < x < 7$. So, $x \in (2, 7)$.

The argument of the inner logarithm must be positive:

$$x^2 - 9x + 18 > 0$$

Factoring the quadratic:

$$(x - 3)(x - 6) > 0$$

This inequality holds for $x < 3$ or $x > 6$. So, $x \in (-\infty, 3) \cup (6, \infty)$.

The domain of the function is the intersection of the intervals obtained from these two conditions.

Intersection of $(-\infty, 3)$ and $(2, 7)$ is $(2, 3)$.

Intersection of $(6, \infty)$ and $(2, 7)$ is $(6, 7)$.

Therefore, the domain of the function is $(2, 3) \cup (6, 7)$.

Given that the domain is $(\alpha, \beta) \cup (\gamma, \delta)$, we have:

$$\alpha = 2, \quad \beta = 3, \quad \gamma = 6, \quad \delta = 7$$

Hence,

$$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$$

Therefore, the correct option is A.

Given: $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$

Find: The sum $\alpha + \beta + \gamma + \delta$.

To determine the domain of the function, we must ensure that the arguments of all logarithmic functions are positive.

First, ensure the argument of the inner logarithm is positive:

$$x^2 - 9x + 18 > 0$$ $$x^2 - 9x + 18 = (x - 3)(x - 6)$$

So,

$$x \in (-\infty, 3) \cup (6, \infty)$$

Next, the expression for the outer logarithm's argument must be positive:

$$1 - \log_4(x^2 - 9x + 18) > 0$$ $$\log_4(x^2 - 9x + 18) < 1$$ $$x^2 - 9x + 18 < 4$$ $$x^2 - 9x + 14 < 0$$

Factor as:

$$(x - 7)(x - 2) < 0$$

So,

$$x \in (2, 7)$$

The combined solution requires both conditions to be satisfied simultaneously. Hence we take the intersection:

$$\left[(-\infty, 3) \cup (6, \infty)\right] \cap (2, 7) = (2, 3) \cup (6, 7)$$

Thus,

$$\alpha = 2, \quad \beta = 3, \quad \gamma = 6, \quad \delta = 7$$

Therefore,

$$\alpha + \beta + \gamma + \delta = 18$$

So, the answer is $18$ and the correct option is A.