JEE Main 2025 Apr 4 Shift 2, Mathematics Q8: Introduction Of Functions

JEE Main2025Apr 4, Shift 2Mathematics

Let the domains of the functions $f (x) = lo g_{4} lo g_{3} lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5))$ and $g (x) = sin^{- 1} (\frac{7 x + 10}{x - 2})$ be $(α, β)$ and $[γ, δ]$ , respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to :-

A
15
B
13
C
16
D
14

Solution

Domain of $f$ . We need $lo g_{3} lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5)) > 0$ ,
i.e. $lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5)) > 1$ ,
i.e. $8 - lo g_{2} (x^{2} + 4 x + 5) > 7$ ,
so $lo g_{2} (x^{2} + 4 x + 5) < 1$ ,
hence $x^{2} + 4 x + 5 < 2$ ,
or $x^{2} + 4 x + 3 < 0$ , giving $x \in (- 3, - 1)$ .

So $α = - 3, β = - 1$ .

Domain of $g$ . Need $- 1 \leq \frac{7 x + 10}{x - 2} \leq 1$ . Solving the two inequalities (with care about the sign of $x - 2$ ) gives $x \in [- 2, - 1]$ .

So $γ = - 2, δ = - 1$ .

$∴ α^{2} + β^{2} + γ^{2} + δ^{2} = 9 + 1 + 4 + 1 = 15$ .

Concept behind this question

Introduction Of FunctionsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 4 Shift 2, Mathematics Q8: Introduction Of Functions

Concept behind this question

More previous year questions on Introduction Of Functions

Practice more Mathematics