Question

If X = \left\\{ {p:{\text{where }}p = \dfrac{{\left( {n + 2} \right)\left( {2{n^2} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{{n^2} + 2n{\text{ and }}n,p \in {Z^ + }}}} \right\\} , then find the number of elements in the set X.
(a) 2
(b) 3
(c) 4
(d) 6

Answer 1

Hint : In this question a set is given and inside this set there is an expression in terms of two positive integer values is given and we have to find the number of the elements in the set X which satisfy all the necessary conditions. First, we simplify the expression then obtain a value in the expression that is undefined and using that value of the expression we find the numbers of the elements for which the necessary conditions of the set X are completely satisfied.

Complete step-by-step answer :
Given:
X = \left\\{ {p:{\text{where }}p = \dfrac{{\left( {n + 2} \right)\left( {2{n^2} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{{n^2} + 2n{\text{ and }}n,p \in {Z^ + }}}} \right\\}
Where, $n{\text{ and }}p$ are positive integer values.
We can solve for the value of $p$ in the following way –
$\Rightarrow {\text{ }}p = \dfrac{{\left( {n + 2} \right)\left( {2{n^5} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{{n^2} + 2n{\text{ }}}}\\\ = \dfrac{{\left( {n + 2} \right)\left( {2{n^5} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{n\left( {n + 2} \right)}}\\\ = \dfrac{{\left( {2{n^5} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{n}\\\ = \left( {2{n^4} + 3{n^3} + 4{n^2} + 5n} \right) + \left( {\dfrac{6}{n}} \right)$
From this expression we can say that for this first part of the expression $\left( {2{n^4} + 3{n^3} + 4{n^2} + 5n} \right)$ for each positive integer value of $n$ the value of this expression would also be an positive integer, so it means that this part satisfies the necessary condition. Now we have to find for the second part of the expression $\left( {\dfrac{6}{n}} \right)$ that for how many positive values of $n$ the value of the expression $\left( {\dfrac{6}{n}} \right)$ would also be a positive integer.
So, the possible numbers for which the value of the expression $\left( {\dfrac{6}{n}} \right)$ would be an integer are –
$\Rightarrow \left( {\dfrac{6}{n}} \right) \to - 6,-2,-3, - 1,1,2,3{\text{ and }}6$
Also, we know that these four numbers of elements in $p$ belong to the value of set X.
But we also know that $n,p \in {Z^ + }$ which means that $n$ and $p$ belongs to only positive integer values. So neglecting the negative values $- 6$, $-3$, $-2$ and $- 1$ , we have only four positive integer values $1$, $2$, $3$ and $6$ left.
Therefore, we can say that the number of positive elements in the set X $= 4$
So, the correct answer is “Option C”.

Note : A set of elements is always defined by the necessary conditions and the elements inside the set that fulfil that condition. Any element that does not satisfy the necessary condition of the set is not a part of the set, hence it is neglected. For example, in the set given in the question according to the condition given all the elements inside the set X have to be positive integer values and any negative integer value would be neglected.