Question

Solve the following inequality

$\frac{1}{x - 1} - \frac{1}{x - 4} - \frac{4}{x - 2} + \frac{4}{x - 3} < \frac{1}{30}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\dfrac{1}{x - 1} - \dfrac{1}{x - 4} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} < \dfrac{1}{30}$

$\dfrac{x - 4 - x + 1}{(x - 1)(x - 4)}+ \dfrac{4x - 8 - 4x + 12}{(x - 3)(x - 2)} < \dfrac{1}{30}$

$\dfrac{ - 3}{(x - 1)(x - 4)}+ \dfrac{4}{(x - 3)(x - 2)} < \dfrac{1}{30}$

$\dfrac{ - 3}{ {x}^{2} - 5x + 4}+ \dfrac{4}{ {x}^{2} - 5x + 6} < \dfrac{1}{30}$

Let assume that

$\rm \: {x}^{2} - 5x + 4 = y$

So, above expression can be rewritten as

$\rm \: \dfrac{ - 3}{y} + \dfrac{4}{y + 2} < \dfrac{1}{30}$

$\rm \: \dfrac{ -3 y - 6 + 4y}{y(y + 2)} < \dfrac{1}{30}$

$\rm \: \dfrac{y - 6}{y(y + 2)} < \dfrac{1}{30}$

$\rm \: \dfrac{y - 6}{y(y + 2)} - \dfrac{1}{30} < 0$

$\rm \: \dfrac{30y - 180 - {y}^{2} - 2y }{30y(y + 2)} < 0$

$\rm \: \dfrac{ - {y}^{2} + 28y - 180}{30y(y + 2)} < 0$

$\rm \: \dfrac{{y}^{2} - 28y + 180}{y(y + 2)} > 0$

$\rm \: \dfrac{(y - 10)(y - 18)}{y(y + 2)} > 0$

Using number line method, we have

$\rm\implies \:y \in \: ( - \infty , - 2) \cup \: (0,10) \cup \: (18, \infty )$

Consider Case :- 1

$\rm \: y \in \: ( - \infty , - 2)$

$\rm \: y < - 2$

$\rm \: {x}^{2} - 5x + 4 < - 2$

$\rm \: {x}^{2} - 5x + 6 < 0$

$\rm \: (x - 2)(x - 3) < 0$

$\rm\implies \:x \: \in \: (2,3) \: \cdots \: (1)$

Consider, Case :- 2

$\rm \: y \: \in \: (0,10)$

$\rm \: y > 0 \: \:and \: \: y < 10$

$\rm \: {x}^{2} - 5x + 4 > 0 \: \:and \: \: {x}^{2} - 5x + 4 < 10$

$\rm \: (x - 1)(x - 4) > 0 \: \:and \: \: {x}^{2} - 5x - 6 < 0$

$\rm \: (x - 1)(x - 4) > 0 \: \:and \: \: (x - 6)(x + 1)< 0$

$\rm\implies \:x \: \in \: ( - 1,1) \: \cup \: (4,6) \: \cdots \: (2)$

Consider Case :- 3

$\rm \: y > 18$

$\rm \: {x}^{2} - 5x + 4 > 18$

$\rm \: {x}^{2} - 5x - 14 > 0$

$\rm \: (x - 7)(x + 2) > 0$

$\rm\implies \:x \: \in \: ( - \infty , - 2) \: \cup \: (7, \infty ) \: \cdots \: (3)$

So, Combining equation (1), (2) and (3), we get

$\rm\implies \:x \in ( - \infty , - 2) \cup( - 1,1)\cup (2,3) \cup(4,6) \cup (7, \infty )$