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1.Relation and Function
hard
If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to
A
$10$
B
$12$
C
$11$
D
$9$
(JEE MAIN-2024)
Solution
$\frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 $
$\frac{2 \times+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \quad \& \quad \frac{x-3}{x+2} \leq 0 $
$(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right) \quad \ldots . .(1) $
$(-2,3] \quad \ldots . .(2)$
${\left[\frac{-1}{3}, 3\right] \ldots \ldots(3) \quad(1) \cap(2) \cap(3)} $
$ \left(\frac{3}{4}, 3\right] $
$ \alpha=\frac{3}{4} \beta=3 $
$ 5 \beta-4 \alpha=15-3=12$
Standard 12
Mathematics
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