Question

Let $a,b\in N$ with $2\le a\le 2020$ and $2\le b\le 2020$.
$P,Q,R$ are three sets defined as
$P=\left\{\right(a,b):{\log }_{a}b+6{\log }_{b}a=5\}$
$Q=\left\{\right(a,b):b=a^2\}$
$R=\left\{\right(a,b):b=a^3\}.$
Then which of the following is (are) correct?

• A. $n\left(P\right)=54$
• B. $n(P\cap Q)=44$
• C. $n(P\cap R)=11$
• D. $n(P\cup Q\cup R)=54$

Answer 1

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

The correct options are
A $n\left(P\right)=54$
C $n(P\cap R)=11$
D $n(P\cup Q\cup R)=54$

Let ${\log }_{a}b=t$
Then, $t^2-5t+6=0$
$(t-2)(t-3)=0$
$\Rightarrow t=2,t=3$
$\Rightarrow b=a^2$ or $b=a^3$

Case$-1:$
If $b=a^2,$
then $a^2\le 2020$
$\Rightarrow a\le 44$
So, ordered pairs $(a,b)$ are of form $(a,a^2)$
$\left\{\right(2,2^2),(3,3^2),\dots ,(44,{44}^2\left)\right\}$
Number of ordered pairs $=43$

Case$-2:$
If $b=a^3,$
then $a^3\le 2020$
$\Rightarrow a\le 12$
$\left\{\right(2,2^3),(3,3^3)\cdots (12,{12}^3\left)\right\}$
Number of ordered pairs $=11$
$\therefore n\left(P\right)=43+11=54$

Since, $2\le b\le 2020$
$Q=\left\{\right(2,2^2),(3,3^2),\dots ,(44,{44}^2\left)\right\}$
$n(P\cap Q)=43$
and $R=\left\{\right(2,2^3),(3,3^3),\dots ,(12,{12}^3\left)\right\}$
$n(P\cap R)=11$

$n(P\cup Q\cup R)=\sum n\left(P\right)-\sum n(P\cap Q)+\sum n(P\cap Q\cap R)$
$=54+43+11-43-0-11+0$
$=54$