JEE Main 2025 Apr 3 Shift 1, Mathematics Q8: Some Important Properties Of Functions

JEE Main2025Apr 3, Shift 1Mathematics

If the domain of the function $f (x) = lo g_{e} (\frac{2 x - 3}{5 + 4 x}) + sin^{- 1} (\frac{4 + 3 x}{2 - x})$ is $[α, β)$ , then $α^{2} + 4 β$ is equal to

Solution

Two conditions are needed.

Logarithm: $\frac{2 x - 3}{5 + 4 x} > 0 \Rightarrow x \in (- \infty, - \frac{5}{4}) \cup (\frac{3}{2}, \infty) .$

Arcsine: $- 1 \leq \frac{4 + 3 x}{2 - x} \leq 1.$ Rearrange each side:

Upper: $\frac{4 + 3 x}{2 - x} - 1 \leq 0 \Rightarrow \frac{2 + 4 x}{2 - x} \leq 0.$

Lower: $\frac{4 + 3 x}{2 - x} + 1 \geq 0 \Rightarrow \frac{6 + 2 x}{2 - x} \geq 0.$

Multiplying these (carefully tracking sign) gives $\frac{( 6 + 2 x ) ( 2 + 4 x )}{( 2 - x ) ^{2}} \leq 0$ , i.e. $(3 + x) (1 + 2 x) \leq 0 \Rightarrow x \in [- 3, - \frac{1}{2}] .$

Intersection with the log domain ( $x < - \frac{5}{4}$ or $x > \frac{3}{2}$ ): only the left branch survives, giving

$x \in [- 3, - \frac{5}{4}) .$

So $α = - 3, β = - \frac{5}{4}$ , and $α^{2} + 4 β = 9 - 5 = 4.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q8: Some Important Properties Of Functions

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