Exams Joint Entrance Examination Mathematics Functions Question 145 Question 145 of 162 Medium MCQ
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
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\[ 2 x + 3 4 x 2 + x − 3 > 0 and − 1 ≤ 2 x − 1 x + 2 ≤ 1 2 x + 3 ( 4 x − 3 ) ( x + 1 ) > 0 3 x + 1 x + 2 ≥ 0 x − 3 x + 2 ≤ 0 \begin{aligned}
& \frac{2 x+3}{4 x^2+x-3}>0 \text { and }-1 \leq \frac{2 x-1}{x+2} \leq 1 \\
& \frac{2 x+3}{(4 x-3)(x+1)}>0 \quad \frac{3 x+1}{x+2} \geq 0 & \frac{x-3}{x+2} \leq 0
\end{aligned} 4 x 2 + x − 3 2 x + 3 > 0 and − 1 ≤ x + 2 2 x − 1 ≤ 1 ( 4 x − 3 ) ( x + 1 ) 2 x + 3 > 0 x + 2 3 x + 1 ≥ 0 x + 2 x − 3 ≤ 0 \]
\[(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right)\] ..... (1)
\[(-2,3]\] ..... (2)
\[\left[\frac{-1}{3}, 3\right]\] ..... (3) \[\quad (1) \cap(2) \cap(3)\]
\[ ( 3 4 , 3 ] α = 3 4 β = 3 5 β − 4 α = 15 − 3 = 12 \begin{aligned}
& \left(\frac{3}{4}, 3\right] \\
& \alpha=\frac{3}{4} \beta=3 \\
& 5 \beta-4 \alpha=15-3=12
\end{aligned} ( 4 3 , 3 ] α = 4 3 β = 3 5 β − 4 α = 15 − 3 = 12 \] Answer the question to unlock