If the domain of the function is then the value of is equal to, JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles-2024

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Solution

Q.
If the domain of the function $f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β],$ then the value of $5 β - 4 α$ is equal to [2024]

1 12

2 10

3 11

4 9

Ans.
(1)

Given function,

$f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$

Here, $\log (\frac{2 x + 3}{4 x^{2} + x - 3})$ will be defined, if

$\frac{2 x + 3}{4 x^{2} + x - 3} > 0 \Rightarrow \frac{2 x + 3}{(4 x - 3) (x + 1)} > 0$

$∴$ $x \in (\frac{- 3}{2}, - 1) \cup (\frac{3}{4}, \infty)$ ...(i)

Also, $\cos^{- 1} (\frac{2 x - 1}{x + 2})$ will be defined, if $- 1 \leq \frac{2 x - 1}{x + 2} \leq 1 \Rightarrow 0 \leq \frac{3 x + 1}{x + 2}$

$∴$ $x \in ([- \infty, - 2] \cup [\frac{- 1}{3}, \infty])$ ...(ii)

Also, $\frac{x - 3}{x + 2} \leq 0$

$∴$ $x \in [- 2, 3]$ ...(iii)

From (i), (ii) and (iii), we get

$β = 3$ and $α = \frac{3}{4} \Rightarrow 5 β - 4 α = 12$

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