Question

Requesting to solve this SUM ​

Answer 1

Step-by-step explanation:

We have,

$\frac{ |2 {x}^{2} + x - 1 |(2x - 1) }{( {x}^{2} - 1)(2x - 1) ^{2} } \leqslant 0$

$\implies \: \frac{ |2 {x}^{2} +2 x - x - 1 | }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

$\implies \: \frac{ |2 x(x + 1) -1( x + 1)| }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

$\implies \: \frac{ |(2 x - 1)(x + 1) | }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

Now, consider the following cases,

CASE 1: When x < -1:

then (2x - 1) <0 and (x + 1) <0, so,

$\implies \: \frac{ (2 x - 1)(x + 1) }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

$\implies \: \frac{ 1 }{( x- 1) } \leqslant 0$

$\implies \: x \leqslant 1$

From this case, we canconclude that the value of x ≤ -1.

CASE 2: When $-1 ≤ x ≤ \frac{1}{2}$:

then, (x + 1) ≥0 and (2x - 1) <0

so,

$\implies \: \frac{ - (2 x - 1)(x + 1) }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

$\implies \: \frac{ (2 x - 1)(x + 1) }{( x- 1)(x + 1)(2x - 1) } \geqslant 0$

$\implies \: \frac{ 1 }{( x- 1) } \geqslant 0$

$\implies \: x \geqslant 1$

Previously we considered $x \in [-1, \frac{1}{2})$ but we got x ≥ 1, so in this case $x \in \phi$

CASE 3: When $x \geqslant \frac{1}{2}$:

then, (2x - 1) ≥ 0 and (x + 1) > 0

so,

$\implies \: \frac{ (2 x - 1)(x + 1) }{( x- 1)(x + 1)(2x - 1) } \leqslant 0$

$\implies \: \frac{ 1 }{( x- 1)} \leqslant 0$

$\implies \: x \leqslant 1$

so, in this case $x \in [\frac{1}{2}, 1)$

So, from these cases we get,

$x \in( - \infty ,- 1] \cup [\frac{1}{2}, 1)$