Question

Let $f\left(n\right)$ be the number of regions in which $n$ coplanar circles can divide the plane. If it is known that each pair of circles intersect in two different points and no three of them have common point of intersection, then

• A. $f\left(15\right)=212$
• B. $f^{-1}\left(134\right)=12$
• C. $f\left(n\right)$ is always an even number
• D. $f\left(n\right)$ is a perfect square

Answer 1

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

The correct options are
A $f\left(15\right)=212$
B $f^{-1}\left(134\right)=12$
C $f\left(n\right)$ is always an even number

Let the number of regions for $n$ circles be $f\left(n\right)$.
$\Rightarrow f\left(1\right)=2$
Now,
$f\left(n\right)=f(n-1)+2(n-1),\forall n\ge 2$
$\Rightarrow f\left(n\right)-f(n-1)=2(n-1)$
Putting $n=2,3,...,n$, we get
$f\left(n\right)-f\left(1\right)\Rightarrow f\left(n\right)-f\left(1\right)=2(1+2+3+...+n-1)\Rightarrow f\left(n\right)-f\left(1\right)=(n-1)n$
$\Rightarrow f\left(n\right)=n(n-1)+2$
$\Rightarrow f\left(n\right)=n^2-n+2$
$f\left(n\right)$ is always even.

$f\left(15\right)=212$

$n^2-n+2=134$
$\Rightarrow n^2-n-132=0$
$\Rightarrow n=12$ (since, $n$ can't be negative)