Question

The number of elements in the set \left\\{ \left. (a,b):\,2{{a}^{2}}+3{{b}^{2}}=35,\,\,\,a,\,\,b\,\in \,\,Z \right\\} \right. , where $\,\,\,Z$ is the set of all integers, is
A. 2
B. 4
C. 8
D. 12

Answer 1

we have to find the number of elements from the given set. In set equation is given that $2{{a}^{2}}+3{{b}^{2}}=35$ we have to take the value of $a$ and $\,b$ in such a way that it satisfies the equation.
In set $a,\,\,b\,\in \,\,Z$ where $Z$ is the set of all integers.

Complete step by step answer:
The equation which is given by:
$\,2{{a}^{2}}+3{{b}^{2}}=35--(1)$
We have to check the equation by substitute the value of $a$ and $\,b$
Substitute $a=2$ and $\,b=3$ in equation $(1)$
$\,=2{{(2)}^{2}}+3{{(3)}^{2}}$
After simplifying further we get:
$\,=8+27$
$=35$
Substitute $a=-2$ and $b=-3$ in equation $(1)$
$\,=2{{(-2)}^{2}}+3{{(-3)}^{2}}$
After simplifying further we get:
$\,=8+27=35$
Substitute $a=-2$ and $\,b=3$ in equation $(1)$
$\,=2{{(-2)}^{2}}+3{{(3)}^{2}}$
After simplifying further we get:
$\,=8+27=35$
Substitute $a=2$ and $b=-3$ in equation $(1)$
$\,=2{{(2)}^{2}}+3{{(-3)}^{2}}$
After simplifying further we get:
$\,=8+27=35$
Substitute $a=4$ and $b=1$ in equation $(1)$
$\,=2{{(4)}^{2}}+3{{(1)}^{2}}$
After simplifying further we get:
$\,=32+3=35$
Substitute $a=-4$ and $b=-1$ in equation $(1)$
$\,=2{{(-4)}^{2}}+3{{(-1)}^{2}}$
After simplifying further we get:
$\,=32+3=35$
Substitute $a=-4$ and $b=1$ in equation $(1)$
$\,=2{{(-4)}^{2}}+3{{(1)}^{2}}$
After simplifying further we get:
$\,=32+3=35$
Substitute $a=4$ and $b=-1$ in equation $(1)$
$\,=2{{(4)}^{2}}+3{{(-1)}^{2}}$
After simplifying further we get:
$\,=32+3=35$
Elements which is present in the sets are:
$\\{(2,3),(-2,-3),(-2,3),(2,-3),(4,1),(-4,-1),(-4,1),(4,-1)\\}$
Therefore, Total number of elements are $8$

So, the correct answer is “Option C”.

Note: According to the question which is represented as set in that value of $a$ and $\,b$ are taken so that it’s satisfies the equation that is in the above solution you can see we have substitute the values to satisfies the equation. Hence we found that after substituting there are 8 elements which are represented as sets. So, in this way we can solve similar problems.