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 at two different points and no three of them have a common point of intersection, then
A) $f\left( {20} \right) = 382$
B) $f\left( n \right)$ is always an even number.
C) ${f^{ - 1}}\left( {92} \right) = 10$
D) $f\left( n \right)$ can be odd.

Answer 1

Here, we have to find the condition for the number of regions in which the number of coplanar circles dividing the plane. We will find the number of regions for a number of circles by using the given information. Then by adding the number of regions of all the number of circles, we will find the condition for a finite number of circles.

Formula Used:
Sum of $n$ natural numbers is given by the formula $S = \dfrac{{n\left( {n + 1} \right)}}{2}$

Complete step by step solution:
We are given that $f\left(n \right)$ be the number of regions in which $n$ coplanar circles can divide the plane.
We are also given that each pair of circles intersect at two different points, so we get $f\left( 1 \right) = 2$
Now, we have $f\left( n \right) = f\left( {n - 1} \right) + 2\left( {n - 1} \right)$ for the number of regions in which the coplanar circles dividing the plane greater than 2. So,
$f\left( n \right) = f\left( {n - 1} \right) + 2\left( {n - 1} \right),\forall n \ge 2$
$\Rightarrow f\left( n \right) - f\left( {n - 1} \right) = 2\left( {n - 1} \right)$………………………………………………………………$\left( 1 \right)$
Substituting $n = 2$ in the equation $\left( 1 \right)$, we get
$\Rightarrow f\left( 2 \right) - f\left( {2 - 1} \right) = 2\left( {2 - 1} \right) \Rightarrow f\left( 2 \right) - f\left( 1 \right) = 2$
Substituting $n = 3$ in the equation $\left( 1 \right)$, we get
$\Rightarrow f\left( 3 \right) - f\left( {3 - 1} \right) = 2\left( {3 - 1} \right) \Rightarrow f\left( 3 \right) - f\left( 2 \right) = 2\left( 2 \right)$
Substituting $n = 4$ in the equation $\left( 1 \right)$, we get
$\Rightarrow f\left( 4 \right) - f\left( {4 - 1} \right) = 2\left( {4 - 1} \right) \Rightarrow f\left( 4 \right) - f\left( 3 \right) = 2\left( 3 \right)$
Substituting $n = n$ in the equation $\left( 1 \right)$, we get
$\Rightarrow f\left( n \right) - f\left( {n - 1} \right) = 2\left( {n - 1} \right) \Rightarrow f\left( n \right) - f\left( {n - 1} \right) = 2\left( {n - 1} \right)$
Now we will add all the equations together. Therefore, we get
$f\left( n \right) - f\left( {n - 1} \right) + f\left( {n - 1} \right) - f\left( {n - 2} \right) + ... + f\left( 2 \right) - f\left( 1 \right) = 2\left( {n - 1} \right) + 2\left( n \right) + ... + 2\left( 3 \right) + 2\left( 2 \right) + 2\left( 1 \right)$Since the consecutive terms are opposite in signs, so they will cancel each other. Therefore,
$\Rightarrow f\left( n \right) - f\left( 1 \right) = 2\left[ {1 + 2 + 3 + ......... + n + \left( {n - 1} \right)} \right]$
Sum of $n$natural numbers is given by the formula $S = \dfrac{{n\left( {n + 1} \right)}}{2}$
By using the sum of $n$natural numbers, we get
$\Rightarrow f\left( n \right) - f\left( 1 \right) = 2\left[ {\dfrac{{\left( {n - 1} \right)\left( {n - 1 + 1} \right)}}{2}} \right]$
$\Rightarrow f\left( n \right) - f\left( 1 \right) = \left( {n - 1} \right)n$
Now by substituting$f\left( 1 \right) = 2$, we get
$\Rightarrow f\left( n \right) - 2 = \left( {n - 1} \right)n$
Adding 2 on both sides, we get
$\Rightarrow f\left( n \right) = \left( {n - 1} \right)n + 2$
By multiplying the terms, we get
$\Rightarrow f\left( n \right) = {n^2} - n + 2$
Since the expression is added to $2$, thus $f\left( n \right)$ is always even.
Now, substituting $n = 20$ in the above equation, we get
$\Rightarrow f\left( {20} \right) = {20^2} - 20 + 2$
Applying the exponent on the terms, we get
$\Rightarrow f\left( {20} \right) = 400 - 20 + 2 = 382$
Also we know that ${n^2} - n + 2 = 92$.
By rewriting the equation, we get
$\Rightarrow {n^2} - n + 2 - 92 = 0$
Subtracting the like terms, we get
$\Rightarrow {n^2} - n - 90 = 0$
By factoring the given quadratic equation, we get
$\Rightarrow {n^2} - 10n + 9n - 90 = 0$
By taking out the common factors, we get
$\Rightarrow n\left( {n - 10} \right) + 9\left( {n - 10} \right) = 0$
Factoring out common terms, we get
$\Rightarrow \left( {n + 9} \right)\left( {n - 10} \right) = 0$
Now by zero product property, we get
$\Rightarrow \left( {n + 9} \right) = 0$
$\Rightarrow n = - 9$
Or
$\Rightarrow \left( {n - 10} \right) = 0$
$\Rightarrow n = 10$

Since $n$ cannot be negative, so we get $n = 10$
Therefore, $f\left( {20} \right) = 382$ , $f\left( n \right)$ is always an even number and ${f^{ - 1}}\left( {92} \right) = 10$.

Thus, option (A), (B), (C) are the correct answer.

Note:
We know that Coplanar circles are defined as the circles intersecting the plane at no point, either at one point or at two points. A coplanar circle intersecting only at one point is known as the Tangent circle. A coplanar circle with only one centre is called the concentric circle. Coplanar points are defined as three or more points lying on the same plane.