JEE Main 2022 — Mathematics Algebra

Given, a relation $R$R on $N$N given by $R=$R={$(x,y):3x+\alpha y$(x,y):3x+αy is a multiple of $7$7},

Now for $R$R to be reflexive $\Rightarrow xRx$⇒xRx

$$\Rightarrow 3x+\alpha x=7x$$⇒3x+αx=7x

$$\Rightarrow (3+\alpha )x=7K$$⇒(3+α)x=7K

$$\Rightarrow 3+\alpha =7\lambda$$⇒3+α=7λ

$$\Rightarrow \alpha =7\lambda -3=7N+4,{\text{where }K,\lambda ,N\in I}$$⇒α=7λ−3=7N+4,where K,λ,N∈I

So, when $\alpha$α divided by $7$7, remainder is $4$4.

Now $R$R to be symmetric $xRy\Rightarrow yRx$xRy⇒yRx

$$3x+\alpha y=7{N}_{1},3y+\alpha x=7{N}_{2}$$3x+αy=7N1​,3y+αx=7N2​

$$\Rightarrow (3+\alpha )(x+y)=7({N}_{1}+{N}_{2})=7{N}_{3}$$⇒(3+α)(x+y)=7(N1​+N2​)=7N3​

Which holds when $3+\alpha$3+α is multiple of $7$7

So, $\alpha =7N+4$α=7N+4 (as did earlier)

Now, for $R$R to be transitive

xRy&yRz\Rightarrow xRz.

$$\Rightarrow 3x+\alpha y=7{N}_{1}....(1)$$⇒3x+αy=7N1​....(1)

$$\Rightarrow 3y+\alpha z=7{N}_{2}......(2)$$⇒3y+αz=7N2​......(2)

And $3x+\alpha z=7{N}_{3}.....(3)$3x+αz=7N3​.....(3)

Now subtracting equation $(3)-(2)$(3)−(2) we get,

$$3x+7{N}_{2}-3y=7{N}_{3}$$3x+7N2​−3y=7N3​

Now putting the value of $3x$3x from equation $(1)$(1) we get, $7{N}_{1}-\alpha y+7{N}_{2}-3y=7{N}_{3}$7N1​−αy+7N2​−3y=7N3​

$$\Rightarrow 7({N}_{1}+{N}_{2})-(3+\alpha )y=7{N}_{3}$$⇒7(N1​+N2​)−(3+α)y=7N3​

$$\Rightarrow (3+\alpha )y=7N$$⇒(3+α)y=7N

Which is true again when $3+\alpha$3+α divisible by $7$7, i.e. when $\alpha$α divided by $7$7, remainder is $4$4 .