(c) \(\big(-\frac14,\frac56\big)\)
\(\frac{6x-5}{4x+1}\) < 0 on putting (6x – 5) and (4x + 1) equal to zero, we get the critical points as \(x=\frac56\) and \(x=-\frac14\). On the real number line they can be shown as:
• When \(x<-\frac14\), both numerator and denominator of the expression \(\frac{6x-5}{4x+1}\) are negative making the expression positive.
• When \(x\) lies between \(-\frac14\) and \(\frac56,\) the expression \(\frac{6x-5}{4x+1}\) becomes negative, hence less than zero.
• When \(x>\frac56\), both (6x – 5) and (4x + 1) are positive, making the expression \(\frac{6x-5}{4x+1}\) positive and hence greater than zero.
∴ For \(\frac{6x-5}{4x+1}\) < 0, \(x\) ∈ \(\big(-\frac14,\frac56\big)\).
