Question

Previous year Question of jee mains
Chapter:- Function​

Answer 1

EXPLANATION.

If R = {(x, y) : x, y ∈ Z, x² + 3y² ≤ 8 } is a relation on the set of integers Z.

As we know that,

Domain of R⁻¹ = Range of R.

⇒ x² + 3y² ≤ 8.

Put the value of y = 0 in the equation, we get.

⇒ x² + 3(0)² ≤ 8.

⇒ x² ≤ 8.

Possible value of x = 0, ± 1, ± 2.

⇒ (0)² ≤ 8. [satisfied].

⇒ (1)² ≤ 8. [satisfied].

⇒ (-1)² ≤ 8. [satisfied].

⇒ (2)² ≤ 8. [satisfied].

⇒ (-2)² ≤ 8. [satisfied].

Put the value of y = ± 1 in the equation, we get.

⇒ x² + 3y² ≤ 8.

⇒ x² + 3(± 1)² ≤ 8.

⇒ x² + 3 ≤ 8.

⇒ x² ≤ 5.

Possible value of x = 0, ± 1, ± 2.

⇒ (0)² ≤ 5. [satisfied].

⇒ (1)² ≤ 5. [satisfied].

⇒ (-1)² ≤ 5. [satisfied].

⇒ (2)² ≤ 5. [satisfied].

⇒ (-2)² ≤ 5. [satisfied].

If we put y = ± 2 in the equation, we get.

⇒ x² + 3y² ≤ 8.

⇒ x² + 3(±2)² ≤ 8.

⇒ x² + 12 ≤ 8.

⇒ x² ≤ 8 - 12.

⇒ x² ≤ - 4.

It is not possible.

∴ Range of R = {-1,0,1}.

Domain of R⁻¹ = {-1,0,1}.

Option [D] is correct answer.

Answer 2

Step-by-step explanation:

R = { (x,y) : x, y € Z , x²+3y² ≤ 8 }

let x = 1 and y = 1 ;

→ 1² + 3(1)²

= 1+3

= 4 ≤ 8

So, (1,1) € R

again,

take x = 2 and y = 1

→ 2²+3(1)²

= 4+3

= 7 ≤ 8

So, (2,1) € R

Similarly, (0,0) , (0,1) , (1,0) , (2,0) , (-1,0), (-2,0) ,(0,-1) , (-1,-1) ,(-2,-1) satisfies the equation x²+3y²≤8.

∵R = {(0,0),(0,1),(1,0),(1,1),(2,0),(2,1)(-1,0),(-2,0),(0,-1),(-1,-1),(-2,-1)}

→ Range if R = {0,1,-1}

we know,

$Domain \: of \: {R}^{ - 1} = Range \: of \: R$

$→ Domain \: of \: {R}^{ - 1} = {0,1,-1 \: }$