Question

If $P=\{x\in N:\frac{14x}{x+1}-\left(\frac{9x-30}{x-4}\right)\le 0\},$
$Q=\{x\in Z:|x-1|\le 5\text{and}|x-1|\ge 2\}$
and $R=\{x\in R:{\log }_{6}x+\frac{2}{{\log }_{6}x}\}=3,$ then which of the following options is (are) CORRECT?

• A. $n(P\cup Q\cup R)=10$
• B. $n(P\cap Q\cap R)=1$
• C. $n(P\cap Q^{\prime }\cap R^{\prime })=1$
• D. $n(P\cap Q\cap R^{\prime })=1$

Answer 1

Answer: (A), (B), (C), (D)

$n(P\cap Q\cap R^{\prime })=1$

For set $P$
$\frac{14x}{x+1}-\left(\frac{9x-30}{x-4}\right)\le 0$
For the inequality to be defined, $x\ne -1,4$
$\frac{(14x^2-56x)-(9x^2-21x-30)}{(x+1)(x-4)}\le 0$
$\Rightarrow \frac{5x^2-35x+30}{(x+1)(x-4)}\le 0$
$\Rightarrow \frac{5(x-1)(x-6)}{(x+1)(x-4)}\le 0$
$\Rightarrow x\in (-1,1]\cup (4,6]$
Since $x\in N,P=\{1,5,6\}$

For set $Q$
$|x-1|\le 5$ and $|x-1|\ge 2$
$\Rightarrow 2\le |x-1|\le 5$
$\Rightarrow -5\le x-1\le -2$ or $2\le x-1\le 5$
$\Rightarrow -4\le x\le -1$ or $3\le x\le 6$
$\Rightarrow x\in [-4,-1]\cup [3,6]$
Since $x\in Z,Q=\{-4,-3,-2,-1,3,4,5,6\}$

For set $R$
Clearly, $x>0$ and $x\ne 1$
${\log }_{6}x+\frac{2}{{\log }_{6}x}=3$
$\Rightarrow ({\log }_{6}x{)}^2-3{\log }_{6}x+2=0$
$\Rightarrow ({\log }_{6}x-1)({\log }_{6}x-2)=0$
$\Rightarrow {\log }_{6}x=1$ or ${\log }_{6}x=2$
$\Rightarrow x=6,36$
$\therefore R=\{6,36\}$

$\therefore P\cup Q\cup R=\{1,5,6,-4,-3,-2,-1,3,4,36\}$
$P\cap Q\cap R=\left\{6\right\}$
$P\cap Q^{\prime }=\left\{1\right\},\therefore P\cap Q^{\prime }\cap R^{\prime }=\left\{1\right\}$
$P\cap Q=\{5,6\},\therefore P\cap Q\cap R^{\prime }=\left\{5\right\}$

So, $n(P\cup Q\cup R)=10$
$n(P\cap Q\cap R)=1$
$n(P\cap Q^{\prime }\cap R^{\prime })=1$
$n(P\cap Q\cap R^{\prime })=1$