JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 18)
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
9
12
11
10
Explanation
$$\begin{aligned} & \frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 \\ & \frac{2 x+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 \& \frac{x-3}{x+2} \leq 0 \end{aligned}$$
_30th_January_Evening_Shift_en_18_1.png)
$$(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right)$$ ..... (1)
$$(-2,3]$$ ..... (2)
$$\left[\frac{-1}{3}, 3\right]$$ ..... (3) $$\quad (1) \cap(2) \cap(3)$$
$$\begin{aligned} & \left(\frac{3}{4}, 3\right] \\ & \alpha=\frac{3}{4} \beta=3 \\ & 5 \beta-4 \alpha=15-3=12 \end{aligned}$$
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