Sets, Relations and Functions – Domain – JEE Main 02 April 2025 Shift 2

Solution:

For the function to be defined, the expressions inside the square roots in the denominator must be strictly positive.

$$10+3x-x^{2} > 0$$
$$x^{2}-3x-10 < 0$$
$$(x-5)(x+2) < 0$$
$$x \in (-2, 5)$$

$$x+|x| > 0$$
$$|x| > -x$$

This inequality holds true only for strictly positive real numbers.
$$x \in (0, \infty)$$

The domain of the function is the intersection of the intervals from both conditions:
$$x \in (-2, 5) \cap (0, \infty)$$
$$x \in (0, 5)$$

Comparing this with the given domain $(a, b)$:
$$a = 0, \quad b = 5$$

Calculate the required value:
$$(1+a)^{2} + b^{2} = (1+0)^{2} + (5)^{2}$$
$$(1+a)^{2} + b^{2} = 1 + 25 = 26$$

Ans. (1)