Question

Solve the following inequality. $\frac{2}{p-1}\ge \frac{3}{4}$

• A. $1<p\le \frac{11}{3}$
• B. $1<p\le \frac{13}{3}$
• C. $2<p\le \frac{11}{3}$
• D. None of these

Answer 1

The correct option is A $1<p\le \frac{11}{3}$

The given inequality is,

$\frac{2}{p-1}\ge \frac{3}{4}$

Subtract $\frac{3}{4}$ from both the sides, we get

$\frac{2}{p-1}-\frac{3}{4}\ge \frac{3}{4}-\frac{3}{4}$

$\therefore \frac{2}{p-1}-\frac{3}{4}\ge 0$

Performing cross multiplication, we get,

$\frac{(2\times 4)-3(p-1)}{4(p-1)}\ge 0$

$\therefore \frac{8-3p+3}{4p-4}\ge 0$

$\therefore \frac{11-3p}{4p-4}\ge 0$

There are following two possibilities for this inequality-

Case (i) $\therefore 11-3p\ge 0$ and $4p>4$

$\therefore 11\ge 3p$ and $p>1$

$\therefore p\le \frac{11}{3}$ and $p>1$

Thus, range of values of p is, $1<p\le \frac{11}{3}$

Case (ii) $11-3p\le 0$ and $4p-4<0$

$\therefore 11\le 3p$ and $4p<4$

$\therefore p\ge \frac{11}{3}$ and $p<1$

Satisfying these two conditions simultaneously is not possible. Thus, solution of the given inequality is $1<p\le \frac{11}{3}$

Thus, Answer is option (A)