The number of elements in the set `{(a, b) : 2a^2 + 3b^2 = 35. a . b in Z}`,where `Z` is the set of all integers, is

Options

A. 2
B. 4
C. 8
D. 12

Answer

A

Explanation

Let
`{(a,b) : 2a ^ 2 + 3b ^ 2 = 35, a, b in Z} `
Now, ` 2a ^ 2 + 3b ^ 2 = 35 `
` rArr 2a ^ 2 = 35 - 3b ^ 2 `
` rArr a ^ 2 = (( 35 - 3b ^ 2)/( 2)) `
since, a, b ` in ` Z. Hence, it is only possible when ` b = pm 3 or b = pm 1 `
We have , ` a ^ 2 = ( 35 - 27 ) /( 2 ) = ( 8 ) /( 2 ) = 4 `
` rArr a ^ 2 = 4 rArr a = pm 2 `
or ` a ^ 2 = ( 35 - 2 ) /( 2 ) = ( 32 ) /( 2) = 16 `
` rArr a ^ 2 = 16 rArr a = pm 4 `
Hence, solution sets are ` { pm 4, pm1 } or { pm 2, pm 3} `