Sets, Relations and Functions – Domain of Function – JEE Main 24 Jan 2026 Shift 1

Solution:

For the function to be defined, we need to satisfy the conditions for the logarithm base and argument.

**Condition 1: Argument > 0**
$$18x^2 – 11x + 1 > 0$$
$$(2x – 1)(9x – 1) > 0$$
$$x \in \left(-\infty, \frac{1}{9}\right) \cup \left(\frac{1}{2}, \infty\right)$$

**Condition 2: Base > 0**
$$10x^2 – 17x + 7 > 0$$
$$(10x – 7)(x – 1) > 0$$
$$x \in \left(-\infty, \frac{7}{10}\right) \cup (1, \infty)$$

**Condition 3: Base $\neq$ 1**
$$10x^2 – 17x + 7 \neq 1$$
$$10x^2 – 17x + 6 \neq 0$$
$$(5x – 6)(2x – 1) \neq 0$$
$$x \neq \frac{6}{5}, \quad x \neq \frac{1}{2}$$

**Combining all intervals:**
Intersection of Condition 1 and Condition 2:
Intervals: $\left(-\infty, \frac{1}{9}\right)$ overlaps with $\left(-\infty, \frac{7}{10}\right)$ to give $\left(-\infty, \frac{1}{9}\right)$.
Intervals: $\left(\frac{1}{2}, \infty\right)$ overlaps with $\left(-\infty, \frac{7}{10}\right)$ to give $\left(\frac{1}{2}, \frac{7}{10}\right)$.
Intervals: $(1, \infty)$ overlaps with $\left(\frac{1}{2}, \infty\right)$ to give $(1, \infty)$.

So, the combined domain is:
$$x \in \left(-\infty, \frac{1}{9}\right) \cup \left(\frac{1}{2}, \frac{7}{10}\right) \cup (1, \infty) – \left\{\frac{6}{5}\right\}$$
Note: $x \neq \frac{1}{2}$ is already excluded by the open brackets.

Comparing with $(-\infty, a) \cup (b, c) \cup (d, \infty) – \{e\}$:
$a = \frac{1}{9}, \quad b = \frac{1}{2}, \quad c = \frac{7}{10}, \quad d = 1, \quad e = \frac{6}{5}$

We need to find $90(a+b+c+d+e)$:
$$90\left(\frac{1}{9} + \frac{1}{2} + \frac{7}{10} + 1 + \frac{6}{5}\right)$$
$$= 90 \times \frac{1}{9} + 90 \times \frac{1}{2} + 90 \times \frac{7}{10} + 90 \times 1 + 90 \times \frac{6}{5}$$
$$= 10 + 45 + 63 + 90 + 108$$
$$= 316$$

Ans. (4)