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Functions Part 2

Section: Mathematics

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Question : 24 of 73

Marks: +1, -0

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)

[\alpha, \beta)

\alpha^2+4 \beta

[3 Apr 2025 Shift 1]

3
7
4
5

Solution:

Domain conditions:

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

Thus

x\in\left(-\infty,-\frac54\right)\cup\left(\frac32,\infty\right)

.

Also,

-1\le\frac{4+3x}{2-x}\le1

.

First,

\frac{4+3x}{2-x}\le1

\Rightarrow \frac{4+3x}{2-x}-1\le0

\Rightarrow \frac{4+3x-2+x}{2-x}\le0

\Rightarrow \frac{4x+2}{2-x}\le0

\Rightarrow \frac{4x+2}{x-2}\ge0

\Rightarrow x\in\left(-\infty,-\frac12\right]\cup(2,\infty)

.

Next,

\frac{4+3x}{2-x}\ge-1

\Rightarrow \frac{4+3x}{2-x}+1\ge0

\Rightarrow \frac{4+3x+2-x}{2-x}\ge0

\Rightarrow \frac{2x+6}{2-x}\ge0

\Rightarrow \frac{2x+6}{x-2}\le0

\Rightarrow x\in[-3,2)

.

Taking the intersection of all conditions:

x\in\left[-3,-\frac54\right)

.

So

\alpha=-3

and

\beta=-\frac54

.

\alpha^2+4\beta=(-3)^2+4\left(-\frac54\right)=9-5=4

.

Hence the required value is

4

.

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