If the domain of the function is equal to [2025]

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Solution

Q.
If the domain of the function

$f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x}) is [α, β), then α^{2} + 4 β$

is equal to [2025]

1 7

2 5

3 3

4 4

Ans.
(4)

We have, $f (x) = \log_{e} (\frac{2 x - 3}{5 + 4 x}) + \sin^{- 1} (\frac{4 + 3 x}{2 - x})$

For f(x) to be defined we have,

$\frac{2 x - 3}{5 + 4 x} > 0 and | \frac{4 + 3 x}{2 - x} | \leq 1$

Now, $\frac{2 x - 3}{5 + 4 x} > 0$

Case I : 2x – 3 > 0 and 5 + 4x > 0

$\Rightarrow$ x > 3/2 and x > – 5/4

$\Rightarrow x \in (3 / 2, \infty)$ ... (i)

Case II : 2x – 3 < 0 and 5 + 4x < 0

$\Rightarrow$ x < 3/2 and x < – 5/4

$\Rightarrow x \in (- \infty, - 5 / 4)$ ... (ii)

From (i) and (ii), we get

$x \in (- \infty, - \frac{5}{4}) \cup (\frac{3}{2}, \infty)$ ... (iii)

Also, $| \frac{4 + 3 x}{2 - x} | \leq 1$

$\Rightarrow - 1 \leq \frac{4 + 3 x}{2 - x} \leq 1 \Rightarrow - 1 \leq \frac{4 + 3 x}{2 - x} and \frac{4 + 3 x}{2 - x} \leq 1$

$\Rightarrow 0 \leq \frac{4 + 3 x}{2 - x} + 1 and \frac{4 + 3 x}{2 - x} - 1 \leq 0$

$\Rightarrow 0 \leq \frac{6 + 2 x}{2 - x} and \frac{2 + 4 x}{2 - x} \leq 0$

$\Rightarrow - 3 \leq x and x \leq - \frac{1}{2}, x \neq 2$

$\Rightarrow x \in [- 3, - \frac{1}{2}]$ ... (iv)

From (iii) and (iv), we get

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

$∴ α = - 3 and β = - \frac{5}{4}$

Thus, $α^{2} + 4 β = 9 - 5 = 4 .$ .

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