JEE Main 2025 Apr 3 Shift 2, Mathematics Q12: Introduction Of Functions

JEE Main2025Apr 3, Shift 2Mathematics

If the domain of the function $f (x) = lo g_{7} (1 - lo g_{4} (x^{2} - 9 x + 18))$ is $(α, β) \cup (γ, δ)$ , then $α + β + γ + δ$ is equal to

Solution

For the outer $lo g_{7}$ to be defined, the argument must be positive:

$1 - lo g_{4} (x^{2} - 9 x + 18) > 0 \Rightarrow lo g_{4} (x^{2} - 9 x + 18) < 1 \Rightarrow x^{2} - 9 x + 18 < 4$ .

For the inner $lo g_{4}$ to be defined, $x^{2} - 9 x + 18 > 0$ .

Factor: $x^{2} - 9 x + 18 = (x - 3) (x - 6) > 0 \Rightarrow x < 3$ or $x > 6$ . ( $*$ )

From the upper bound: $x^{2} - 9 x + 14 < 0 \Rightarrow (x - 2) (x - 7) < 0 \Rightarrow 2 < x < 7$ . ( $* *$ )

Intersect ( $*$ ) and ( $* *$ ): $x \in (2, 3) \cup (6, 7)$ .

So $α = 2, β = 3, γ = 6, δ = 7$ and $α + β + γ + δ = 18$ .

Concept behind this question

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