Functions – Domain – JEE Main 22 Jan 2026 Shift 1

Solution:

The function is defined if both terms are defined. Let $f(x) = f_1(x) + f_2(x)$.

**For the first term** $f_1(x) = \sin^{-1}\left(\frac{5-x}{3+2x}\right)$:
The domain of $\sin^{-1}(t)$ is $[-1, 1]$.
$$ -1 \le \frac{5-x}{3+2x} \le 1 $$

This inequality splits into two parts:
**Part 1:** $\frac{5-x}{3+2x} \le 1$
$$ \frac{5-x}{3+2x} – 1 \le 0 $$
$$ \frac{5-x – (3+2x)}{3+2x} \le 0 $$
$$ \frac{2-3x}{3+2x} \le 0 $$
Multiplying by -1 reverses the inequality:
$$ \frac{3x-2}{2x+3} \ge 0 $$
Critical points are $x = -\frac{3}{2}$ and $x = \frac{2}{3}$.
Solution set $S_1 = \left(-\infty, -\frac{3}{2}\right) \cup \left[\frac{2}{3}, \infty\right)$.

**Part 2:** $\frac{5-x}{3+2x} \ge -1$
$$ \frac{5-x}{3+2x} + 1 \ge 0 $$
$$ \frac{5-x + (3+2x)}{3+2x} \ge 0 $$
$$ \frac{x+8}{2x+3} \ge 0 $$
Critical points are $x = -8$ and $x = -\frac{3}{2}$.
Solution set $S_2 = (-\infty, -8] \cup \left(-\frac{3}{2}, \infty\right)$.

**Domain of** $f_1(x) = S_1 \cap S_2$:
Intersection of $S_1$ and $S_2$ is:
$$ D_1 = (-\infty, -8] \cup \left[\frac{2}{3}, \infty\right) $$

**For the second term** $f_2(x) = \frac{1}{\log_{10}(10-x)}$:
The argument of the logarithm must be positive, and the denominator must not be zero.
1. $10 – x > 0 \Rightarrow x < 10$ 2. $\log_{10}(10-x) \ne 0 \Rightarrow 10-x \ne 1 \Rightarrow x \ne 9$
**Domain of** $f_2(x) = D_2 = (-\infty, 10) – \{9\}$.

**Total Domain** $= D_1 \cap D_2$:
$$ \left( (-\infty, -8] \cup \left[\frac{2}{3}, \infty\right) \right) \cap \left( (-\infty, 10) – \{9\} \right) $$
$$ = (-\infty, -8] \cup \left[\frac{2}{3}, 10\right) – \{9\} $$

Comparing this with the given form $(-\infty, \alpha] \cup [\beta,\gamma)-\{\delta\}$:
$$ \alpha = -8, \quad \beta = \frac{2}{3}, \quad \gamma = 10, \quad \delta = 9 $$

We need to find $6(\alpha+\beta+\gamma+\delta)$:
$$ \text{Sum} = \alpha+\beta+\gamma+\delta = -8 + \frac{2}{3} + 10 + 9 $$
$$ = 11 + \frac{2}{3} = \frac{33+2}{3} = \frac{35}{3} $$

$$ 6(\text{Sum}) = 6 \times \frac{35}{3} = 2 \times 35 = 70 $$

Ans. (1)