JEE Main 2025 Apr 2 Shift 2, Mathematics Q12: Introduction Of Functions

JEE Main2025Apr 2, Shift 2Mathematics

If the domain of the function $f (x) = \frac{1}{10 + 3 x - x ^{2}} + \frac{1}{x + ∣ x ∣}$ is $(a, b)$ , then $(1 + a)^{2} + b^{2}$ is equal to :

Solution

Two strict positivity conditions must hold.

From $x + ∣ x ∣ > 0$ : for $x \leq 0$ the expression is $0$ ; only $x > 0$ works.

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

Intersection: $x \in (0, 5)$ . So $a = 0, b = 5$ and $(1 + a)^{2} + b^{2} = 1 + 25 = 26$ .

Concept behind this question

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