Question

PYQ jee mains

Chapter:- Function​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:R_1 = \{(a,b) \in \: {R}^{2} : {a}^{2} + {b}^{2} \in \: Q \}$

We know,

Relation R is said to be transitive iff (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R.

Now,

Let we assume a, b, c ∈ R

Let

$\rm :\longmapsto\:a = 1 + \sqrt{2} \in \: R$

$\rm :\longmapsto\:b = 1 - \sqrt{2} \in \: R$

$\rm :\longmapsto\:c = {8}^{ \frac{1}{4} }$

Now,

Consider,

$\rm :\longmapsto\: {a}^{2} + {b}^{2}$

$\rm \: = \: {(1 + \sqrt{2} )}^{2} + {(1 - \sqrt{2} )}^{2}$

$\rm \: = \:1 + 2 + 2 \sqrt{2} + 1 + 2 - 2 \sqrt{2}$

$\rm \: = \:6$

$\rm :\implies\: {a}^{2} + {b}^{2} \in \: Q$

$\rm :\implies\:(a,b) \: \in \: R$

Consider

$\rm :\longmapsto\: {b}^{2} + {c}^{2}$

$\rm \: = \: {(1 - \sqrt{2} )}^{2} + {8}^{ \frac{1}{2} }$

$\rm \: = \:1 + 2 - 2 \sqrt{2} + \sqrt{8}$

$\rm \: = \:3 - 2 \sqrt{2} + \sqrt{2 \times 2 \times 2}$

$\rm \: = \:3 - 2 \sqrt{2} + 2\sqrt{2 }$

$\rm \: = \:3$

$\rm :\implies\: {b}^{2} + {c}^{2} \in \: Q$

$\rm :\implies\:(b,c) \: \in \: R$

Now,

Consider,

$\rm :\longmapsto\: {a}^{2} + {c}^{2}$

$\rm \: = \: {(1 + \sqrt{2} )}^{2} + {8}^{ \frac{1}{2} }$

$\rm \: = \:1 + 2 + 2 \sqrt{2} + \sqrt{8}$

$\rm \: = \:3 + 2 \sqrt{2} + \sqrt{2 \times 2 \times 2}$

$\rm \: = \:3 + 2 \sqrt{2} + 2\sqrt{2 }$

$\rm \: = \:3 + 2 \sqrt{2}$

$\rm :\implies\: {a}^{2} + {c}^{2} \cancel\in \: Q$

$\rm :\implies\:(a,c) \: \cancel\in \: R$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: R_1 \: is \: not \: transitive}}$

Consider,

$\rm :\longmapsto\:R_2 = \{(a,b) \in \: {R}^{2} : {a}^{2} + {b}^{2} \cancel \in \: Q \}$

Let assume that

$\rm :\longmapsto\:a = 1 + \sqrt{2} \in \: R$

$\rm :\longmapsto\:b = \sqrt{2} \in \: R$

$\rm :\longmapsto\:c = 1 - \sqrt{2} \in \: R$

Consider,

$\rm :\longmapsto\: {a}^{2} + {b}^{2}$

$\rm \: = \: {(1 + \sqrt{2}) }^{2} + {( \sqrt{2} )}^{2}$

$\rm \: = \:1 + 2 + 2 \sqrt{2} + 2$

$\rm \: = \:5 + 2 \sqrt{2}$

$\rm :\implies\: {a}^{2} + {b}^{2} \cancel\in \: Q$

$\rm :\implies\:(a,b) \: \in \: R$

Consider,

$\rm :\longmapsto\: {b}^{2} + {c}^{2}$

$\rm \: = \: {(\sqrt{2}) }^{2} + {( 1 - \sqrt{2} )}^{2}$

$\rm \: = \:2 + 1 + 2 - 2 \sqrt{2}$

$\rm \: = \:5 - 2\sqrt{2}$

$\rm :\implies\: {b}^{2} + {c}^{2} \cancel\in \: Q$

$\rm :\implies\:(b,c) \: \in \: R$

Consider,

$\rm :\longmapsto\: {a}^{2} + {c}^{2}$

$\rm \: = \: {(1 + \sqrt{2} )}^{2} + {(1 - \sqrt{2} )}^{2}$

$\rm \: = \:1 + 2 + 2 \sqrt{2} + 1 + 2 - 2 \sqrt{2}$

$\rm \: = \:6$

$\rm :\implies\: {a}^{2} + {c}^{2} \in \: Q$

$\rm :\implies\:(a,c) \: \cancel\in \: R$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: R_2 \: is \: not \: transitive}}$

Therefore,

$\: \: \: \: \: \: \underbrace{ \boxed{ \bf\implies \:\bf \: Option \: (a) \: is \: correct \quad}}$