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Question

If $${x}$$ is real and $$\displaystyle \frac{x-1}{4x+5}<\frac{x-3}{4x-3}$$, then $${x}$$ lies in the interval


A
(34,54)
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B
(54,34)
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C
(54,34)
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D
(34,75)
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Solution

The correct option is B $$\left(-\displaystyle \frac{5}{4}, \displaystyle \frac{3}{4}\right)$$
$$ x\in R$$
The inequality is $$\displaystyle \frac { x-1 }{ 4x+5 } <\frac { x-3 }{ 4x-3 }$$
$$ \Rightarrow  \displaystyle \frac { (x-1)(4x-3)-(x-3)(4x+5) }{ (4x+5)(4x-3) } <0$$
$$ \Rightarrow \quad \displaystyle \frac {18 }{ (4x+5)(4x-3) } <0 $$
$$ \Rightarrow \quad (4x+5)(4x-3)<0 $$
$$ \therefore \quad x=\left( -\displaystyle \frac { 5 }{ 4 } ,\displaystyle \frac { 3 }{ 4 }  \right) $$
Hence, option C is correct.

Mathematics

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