JEE Main 2023 — Mathematics Algebra

Given,

$$A={1,2,3,\ldots ,10}B={0,1,2,4}$$A=1,2,3,…,10B=0,1,2,4

$(a,b)\in A\times A$(a,b)∈A×A such that

$$2{(a-b)}^{2}+3(a-b)-k=0$$2(a−b)2+3(a−b)−k=0

where $k\in {0,1,2,3,4}$k∈0,1,2,3,4

Now finding discriminant $D=9-4\times 2(-k)$D=9−4×2(−k)

And $9-4\times 2(-k)$9−4×2(−k) a perfect square for any possible $(a,b)$(a,b) as $(a-b)$(a−b) will be a integer,

So, $9+8k$9+8k is a perfect square

$\Rightarrow k=0$⇒k=0 or $k=2$k=2

Now for $k=0$k=0,

$$2{(a-b)}^{2}+3(a-b)=0$$2(a−b)2+3(a−b)=0

$$\Rightarrow (a-b)[2{(a-b)}^{2}+3]=0$$⇒(a−b)[2(a−b)2+3]=0

$$\Rightarrow a-b=0\Rightarrow (a,b)\in {(1,1),(2,2)\ldots .(10,10)}$$⇒a−b=0⇒(a,b)∈(1,1),(2,2)….(10,10)

$\Rightarrow$⇒Total $10$10 elements belonging to $R.$R.

And, $a-b=\frac{3}{2}$a−b=23​ is not possible

Now for $k=2$k=2,

$$2{(a-b)}^{2}+3(a-b)-2=0$$2(a−b)2+3(a−b)−2=0

$\Rightarrow a-b=-2$⇒a−b=−2 or $a-b=\frac{1}{2}$a−b=21​ (not possible)

Now for $a-b=-2$a−b=−2 possible pair will be,

$$\Rightarrow (a,b)\in {(1,3),(2,4),\ldots .(8,10)}$$⇒(a,b)∈(1,3),(2,4),….(8,10)

$\Rightarrow 8$⇒8 elements belonging to $R$R

Total number of elements will be$=18$=18