(d) Null set
\(\frac{x}{2x+1}\) > \(\frac14\) and \(\frac{6x}{4x-1}\)< \(\frac12\)
⇒ \(\frac{x}{2x+1}\) - \(\frac14\) > 0 and \(\frac{6x}{4x-1}\) - \(\frac12\) < 0
⇒ \(\frac{4x-2x-1}{4(2x+1)}\) > 0 and \(\frac{12x-4x+1}{2(4x-1)}\) < 0
⇒ \(\frac{2x-1}{2x+1}\) > 0 and \(\frac{8x+1}{4x-1}\) < 0
Now for \(\frac{2x-1}{2x+1}\) > 0 \(x\) ≠ \(\frac12\)
The critical points are \(\frac{-1}2\) and \(\frac12\) which are plotted on the real number line as shown below:
Examining the expression for the three parts of the number line, i.e, x < \(-\frac12\), \(\frac{-1}2\) < x < \(\frac12\), x > \(\frac12\), we see that
\(\frac{2x-1}{2x+1}\) > 0 when x < \(-\frac12\) or x > \(\frac12\)
∴ x ∈ \(\big(-∞,-\frac12\big)∪\big[\frac12,∞\big)\) ....(i)
Now for \(\frac{8x+1}{4x-1}<0,\) \(x\) ≠ \(\frac14\)
The critical points are \(-\frac18,\frac14\) which are plotted on the real number line as shown below:
The expression \(\frac{8x+1}{4x-1}<0,\) when x ∈ \(\big(\frac{-1}8,\frac14\big)\) ....(ii)
∴ It is clear from (i) and (ii) that the intersection of the solution sets given by (i) and (ii), i.e.,
\(\bigg\{\big(-∞,-\frac12\big)∪\bigg[\frac12,∞\big)\bigg]∩\bigg(-\frac18,\frac14\bigg)\) = ϕ.
