If the domain of the function $f(x)=\log_{e}\left(\frac{2x-3}{5+4x}\right)+\sin^{-1}\left(\frac{4+3x}{2-x}\right)$ is $[\alpha, \beta)$, then $\alpha^{2}+4\beta$ is equal to
- (1) 5
- (2) 4
- (3) 3
- (4) 7
Solution:
$$f(x) = \log_{e}\left(\frac{2x-3}{5+4x}\right) + \sin^{-1}\left(\frac{4+3x}{2-x}\right)$$
$$ \frac{2x-3}{5+4x} > 0 $$
$$ x \in \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right) $$
$$ -1 \le \frac{4+3x}{2-x} \le 1 $$
$$ \frac{4+3x}{2-x} + 1 \ge 0 \Rightarrow \frac{2x+6}{2-x} \ge 0 \Rightarrow \frac{x+3}{x-2} \le 0 $$
$$ x \in [-3, 2) $$
$$ \frac{4+3x}{2-x} – 1 \le 0 \Rightarrow \frac{4x+2}{2-x} \le 0 \Rightarrow \frac{2x+1}{x-2} \ge 0 $$
$$ x \in \left(-\infty, -\frac{1}{2}\right] \cup (2, \infty) $$
$$ x \in [-3, 2) \cap \left( \left(-\infty, -\frac{1}{2}\right] \cup (2, \infty) \right) $$
$$ x \in \left[-3, -\frac{1}{2}\right] $$
$$ x \in \left( \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right) \right) \cap \left[-3, -\frac{1}{2}\right] $$
$$ x \in \left[-3, -\frac{5}{4}\right) $$
$$ [\alpha, \beta) = \left[-3, -\frac{5}{4}\right) $$
$$ \alpha = -3, \quad \beta = -\frac{5}{4} $$
$$ \alpha^2 + 4\beta = (-3)^2 + 4\left(-\frac{5}{4}\right) $$
$$ = 9 – 5 = 4 $$
Ans. (2)