JEE Main 2025 Apr 8 Shift 2, Mathematics Q22: Invertible And Composite Functions

JEE Main2025Apr 8, Shift 2Mathematics

Let the domain of the function $f (x) = cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = lo g_{2} (2 - 6 lo g_{27} (2 x + 5))$ be $(γ, δ)$ . Then $∣7 (α + β) + 4 (γ + δ) ∣$ is equal to ______

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Solution

For $f (x)$ , need $- 1 \leq \frac{4 x + 5}{3 x - 7} \leq 1$ .

Right inequality: $\frac{4 x + 5}{3 x - 7} - 1 \leq 0$ , i.e. $\frac{x + 12}{3 x - 7} \leq 0$ , giving $x \in [- 12, \frac{7}{3})$ .

Left inequality: $\frac{4 x + 5}{3 x - 7} + 1 \geq 0$ , i.e. $\frac{7 x - 2}{3 x - 7} \geq 0$ , giving $x \in (- \infty, \frac{2}{7}] \cup (\frac{7}{3}, \infty)$ .

Intersecting: domain is $[- 12, \frac{2}{7}]$ , so $α = - 12$ , $β = \frac{2}{7}$ .

For $g (x)$ , need $2 - 6 lo g_{27} (2 x + 5) > 0$ and $2 x + 5 > 0$ .

First: $\log_{27}(2x+5) < \dfrac{1}{3}$, i.e. $2x+5 < 27^{1/3} = 3$, giving $x < -1$.

Second: $x > - \frac{5}{2}$ .

Domain of $g$ : $(- \frac{5}{2}, - 1)$ , so $γ = - \frac{5}{2}$ , $δ = - 1$ .

Compute: $7 (α + β) = 7 (- 12 + \frac{2}{7}) = - 84 + 2 = - 82$ ; $4 (γ + δ) = 4 (- \frac{5}{2} - 1) = 4 \cdot (- \frac{7}{2}) = - 14$ .

$∣ - 82 + (- 14) ∣ = ∣ - 96 ∣ = 96$ .

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q22: Invertible And Composite Functions

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