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QJEEM24Maf69f: JEE Main 2024 Mathematics - Class 11 JEEM24Maf69f Solution: If the domain of the function is , then the value of is equal to

If the domain of the function f(x)=log⁡e(2x+34x2+x−3)+cos⁡−1(2x−1x+2)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 (α,β](\alpha, \beta] , then the value of 5β−4α5 \beta-4 \alpha is equal to

Multiple Choice Options

  • A 9
  • B 12
  • C 11
  • D 10

Step-by-Step Verified Solution

amp;2x+34x2+x−3gt;0 and −1≤2x−1x+2≤1amp;2x+3(4x−3)(x+1)gt;03x+1x+2≥0&x−3x+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}

JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Functions Question 24 English Explanation

(−∞,−2)∪[−13,∞)(-\infty,-2) \cup\left[\frac{-1}{3}, \infty\right) ..... (1)

(−2,3](-2,3] ..... (2)

[−13,3]\left[\frac{-1}{3}, 3\right] ..... (3) (1)∩(2)∩(3)\quad (1) \cap(2) \cap(3)

amp;(34,3]amp;α=34β=3amp;5β−4α=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}

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