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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Functions

If the domain of the function f(x)= _e((2 x-3)/(5+4 x))+ ^ -1 ((4+3...

If the domain of the functionf(x)=log⁡e(2x−35+4x)+sin⁡−1(4+3x2−x) is [α,β)f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right) \quad \text{ is } \quad[\alpha, \beta)then α2+4β\alpha^2+4 \beta is equal to

Options

  1. A55
  2. B44
  3. C33
  4. D77

Correct answer

B. 44

Step-by-step solution

Given function is
f(x)=log⁡e(2x−35+4x)+sin⁡−1(4+3x2−x)f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right)
For domain, the conditions are
2x−35+4x>0 and ∣4+3x2−x∣≤1\frac{2 x-3}{5+4 x} \gt 0 \text{ and }\left|\frac{4+3 x}{2-x}\right| \leq 1
Now, 2x−35+4x>0⇒x∈(−∞,−54)∪[32,∞)\frac{2 \mathrm{x}-3}{5+4 \mathrm{x}} \gt 0 \Rightarrow \mathrm{x} \in\left(-\infty,-\frac{5}{4}\right) \cup\left[\frac{3}{2}, \infty\right)
and −1≤4+3x2−x≤1\quad-1 \leq \frac{4+3 x}{2-x} \leq 1
⇒(−1≤4+3x2−x)∩(4+3x2−x≤1)\Rightarrow\left(-1 \leq \frac{4+3 x}{2-x}\right) \cap\left(\frac{4+3 x}{2-x} \leq 1\right)
⇒(6+2x2−x≥0)∩(2+4x2−x≤0)\Rightarrow\left(\frac{6+2 \mathrm{x}}{2-\mathrm{x}} \geq 0\right) \cap\left(\frac{2+4 \mathrm{x}}{2-\mathrm{x}} \leq 0\right)
⇒6+2x2−x⋅2+4x2−x≤0⇒x∈[−3,−12]\begin{aligned} & \Rightarrow \frac{6+2 x}{2-x} \cdot \frac{2+4 x}{2-x} \leq 0 \\ & \Rightarrow x \in\left[-3,-\frac{1}{2}\right]\end{aligned}
Hence, we get the domain of f as x∈[−3,−54)\mathrm{x} \in\left[-3,-\frac{5}{4}\right) This means that α=−3,β=−54\alpha=-3, \beta=-\frac{5}{4} Thus, α2+4β=9−5=4\alpha^2+4 \beta=9-5=4
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