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Sets and Relations question

2023 · 31 Jan · Shift 2 · Q31
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  5. /2023 · 31 Jan · Shift 2 · Q31

Sets and Relations question

2023 · 31 Jan · Shift 2 · Q31

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Among the relations S={(a,b):a,b∈R−{0},2+ab>0}\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}\gt 0\right\}S={(a,b):a,b∈R−{0},2+ba​>0} and T={(a,b):a,b∈R,a2−b2∈Z}\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}T={(a,b):a,b∈R,a2−b2∈Z},
  1. A
    S\mathrm{S}S is transitive but T\mathrm{T}T is not
  2. B
    both S\mathrm{S}S and T\mathrm{T}T are symmetric
  3. C
    neither SSS nor TTT is transitive
  4. D
    TTT is symmetric but SSS is not
View written solutionFree

Correct answer: D

  1. Given relations

We have

S={(a,b):a,b∈R∖{0}, 2+ab>0}S=\{(a,b): a,b\in \mathbb{R}\setminus\{0\},\ 2+\frac{a}{b}>0\}S={(a,b):a,b∈R∖{0}, 2+ba​>0}

and

T={(a,b):a,b∈R, a2−b2∈Z}.T=\{(a,b): a,b\in \mathbb{R},\ a^2-b^2\in \mathbb{Z}\}.T={(a,b):a,b∈R, a2−b2∈Z}.

We must check symmetry and transitivity to identify the correct option.

  1. Analyse relation SSS

The condition for (a,b)∈S(a,b)\in S(a,b)∈S is

2+ab>0  ⟺  ab>−2.2+\frac{a}{b}>0 \quad \iff \quad \frac{a}{b}>-2.2+ba​>0⟺ba​>−2.

(i) Is SSS symmetric?

For symmetry, if (a,b)∈S(a,b)\in S(a,b)∈S, then we must also have (b,a)∈S(b,a)\in S(b,a)∈S.

Take

a=1,b=−1.a=1,\quad b=-1.a=1,b=−1.

Then

2+ab=2+1−1=1>0,2+\frac{a}{b}=2+\frac{1}{-1}=1>0,2+ba​=2+−11​=1>0,

so (1,−1)∈S(1,-1)\in S(1,−1)∈S.

Now check (−1,1)(-1,1)(−1,1):

2+ba=2+−11=1>0.2+\frac{b}{a}=2+\frac{-1}{1}=1>0.2+ab​=2+1−1​=1>0.

This example does not disprove symmetry, so try another one.

Take

a=1,b=−13.a=1,\quad b=-\frac{1}{3}.a=1,b=−31​.

Then

2+ab=2+1−1/3=2−3=−1<0,2+\frac{a}{b}=2+\frac{1}{-1/3}=2-3=-1<0,2+ba​=2+−1/31​=2−3=−1<0,

so not in SSS.

We need an example where (a,b)∈S(a,b)\in S(a,b)∈S but (b,a)∉S(b,a)\notin S(b,a)∈/S.

Take

a=−1,b=1.a=-1,\quad b=1.a=−1,b=1.

Then

2+ab=2−1=1>0,2+\frac{a}{b}=2-1=1>0,2+ba​=2−1=1>0,

so (−1,1)∈S(-1,1)\in S(−1,1)∈S.

But

2+ba=2+1−1=1>0,2+\frac{b}{a}=2+\frac{1}{-1}=1>0,2+ab​=2+−11​=1>0,

still in SSS.

Try a more extreme ratio. Let

a=−32,b=1.a=-\frac{3}{2},\quad b=1.a=−23​,b=1.

Then

2+ab=2−32=12>0,2+\frac{a}{b}=2-\frac{3}{2}=\frac12>0,2+ba​=2−23​=21​>0,

so (−32,1)∈S\left(-\frac32,1\right)\in S(−23​,1)∈S.

Now reverse:

2+ba=2+1−3/2=2−23=43>0.2+\frac{b}{a}=2+\frac{1}{-3/2}=2-\frac23=\frac43>0.2+ab​=2+−3/21​=2−32​=34​>0.

Still in SSS.

Let us reason algebraically instead. If (a,b)∈S(a,b)\in S(a,b)∈S, then

ab>−2.\frac{a}{b}>-2.ba​>−2.

For (b,a)∈S(b,a)\in S(b,a)∈S, we need

ba>−2.\frac{b}{a}>-2.ab​>−2.

These are not equivalent in general.

Counterexample: let

ab=−14.\frac{a}{b}=-\frac14.ba​=−41​.

Then

2+ab=2−14=74>0,2+\frac{a}{b}=2-\frac14=\frac74>0,2+ba​=2−41​=47​>0,

so (a,b)∈S(a,b)\in S(a,b)∈S.

But then

ba=−4,\frac{b}{a}=-4,ab​=−4,

so

2+ba=2−4=−2<0.2+\frac{b}{a}=2-4=-2<0.2+ab​=2−4=−2<0.

Hence (b,a)∉S(b,a)\notin S(b,a)∈/S.

A concrete choice is

a=−1,b=4.a=-1,\quad b=4.a=−1,b=4.

Then

2+−14=74>02+\frac{-1}{4}=\frac74>02+4−1​=47​>0

and

2+4−1=−2<0.2+\frac{4}{-1}=-2<0.2+−14​=−2<0.

So SSS is not symmetric.

(ii) Is SSS transitive?

For transitivity, if (a,b)∈S(a,b)\in S(a,b)∈S and (b,c)∈S(b,c)\in S(b,c)∈S, then we must have (a,c)∈S(a,c)\in S(a,c)∈S.

We need

2+ab>0,2+bc>0,2+\frac{a}{b}>0, \qquad 2+\frac{b}{c}>0,2+ba​>0,2+cb​>0,

but this does not necessarily imply

2+ac>0.2+\frac{a}{c}>0.2+ca​>0.

Take

a=1,b=1,c=−1.a=1,\quad b=1,\quad c=-1.a=1,b=1,c=−1.

Then

2+ab=2+1=3>0,2+\frac{a}{b}=2+1=3>0,2+ba​=2+1=3>0,

so (1,1)∈S(1,1)\in S(1,1)∈S. Also,

2+bc=2+1−1=1>0,2+\frac{b}{c}=2+\frac{1}{-1}=1>0,2+cb​=2+−11​=1>0,

so (1,−1)∈S(1,-1)\in S(1,−1)∈S.

But

2+ac=2+1−1=1>0,2+\frac{a}{c}=2+\frac{1}{-1}=1>0,2+ca​=2+−11​=1>0,

so this still works.

Try another choice. We want a/c<−2a/c<-2a/c<−2 while keeping a/b>−2a/b>-2a/b>−2 and b/c>−2b/c>-2b/c>−2.

Take

a=1,b=−1,c=1.a=1,\quad b=-1,\quad c=1.a=1,b=−1,c=1.

Then

2+ab=2−1=1>0,2+\frac{a}{b}=2-1=1>0,2+ba​=2−1=1>0,

so (1,−1)∈S(1,-1)\in S(1,−1)∈S. Also,

2+bc=2−1=1>0,2+\frac{b}{c}=2-1=1>0,2+cb​=2−1=1>0,

so (−1,1)∈S(-1,1)\in S(−1,1)∈S. But

2+ac=2+1=3>0,2+\frac{a}{c}=2+1=3>0,2+ca​=2+1=3>0,

so again transitivity holds in this example.

Try a ratio-based construction:

  • Choose ab=−12>−2\dfrac{a}{b}=-\dfrac12 > -2ba​=−21​>−2,
  • Choose bc=10>−2\dfrac{b}{c}=10 > -2cb​=10>−2.

Then

ac=ab⋅bc=(−12)(10)=−5,\frac{a}{c}=\frac{a}{b}\cdot \frac{b}{c}=\left(-\frac12\right)(10)=-5,ca​=ba​⋅cb​=(−21​)(10)=−5,

so

2+ac=2−5=−3<0.2+\frac{a}{c}=2-5=-3<0.2+ca​=2−5=−3<0.

Thus transitivity fails.

A concrete choice is

a=−1,b=2,c=15.a=-1,\quad b=2,\quad c=\frac15.a=−1,b=2,c=51​.

Then

ab=−12  ⟹  2+ab=32>0,\frac{a}{b}=-\frac12 \implies 2+\frac{a}{b}=\frac32>0,ba​=−21​⟹2+ba​=23​>0,

so (−1,2)∈S(-1,2)\in S(−1,2)∈S.

Also,

bc=21/5=10  ⟹  2+bc=12>0,\frac{b}{c}=\frac{2}{1/5}=10 \implies 2+\frac{b}{c}=12>0,cb​=1/52​=10⟹2+cb​=12>0,

so (2,15)∈S\left(2,\frac15\right)\in S(2,51​)∈S.

But

ac=−11/5=−5  ⟹  2+ac=−3<0,\frac{a}{c}=\frac{-1}{1/5}=-5 \implies 2+\frac{a}{c}=-3<0,ca​=1/5−1​=−5⟹2+ca​=−3<0,

so (−1,15)∉S\left(-1,\frac15\right)\notin S(−1,51​)∈/S.

Hence SSS is not transitive.

So for SSS:

  • not symmetric,
  • not transitive.
  1. Analyse relation TTT

The relation is

T={(a,b):a2−b2∈Z}.T=\{(a,b): a^2-b^2\in \mathbb{Z}\}.T={(a,b):a2−b2∈Z}.

(i) Is TTT symmetric?

Suppose (a,b)∈T(a,b)\in T(a,b)∈T. Then

a2−b2∈Z.a^2-b^2\in \mathbb{Z}.a2−b2∈Z.

Now

b2−a2=−(a2−b2).b^2-a^2=-(a^2-b^2).b2−a2=−(a2−b2).

Since the negative of an integer is also an integer,

b2−a2∈Z.b^2-a^2\in \mathbb{Z}.b2−a2∈Z.

Therefore,

(b,a)∈T.(b,a)\in T.(b,a)∈T.

So TTT is symmetric.

(ii) Is TTT transitive?

Suppose (a,b)∈T(a,b)\in T(a,b)∈T and (b,c)∈T(b,c)\in T(b,c)∈T. Then

a2−b2∈Z,b2−c2∈Z.a^2-b^2\in \mathbb{Z}, \qquad b^2-c^2\in \mathbb{Z}.a2−b2∈Z,b2−c2∈Z.

Adding these,

(a2−b2)+(b2−c2)=a2−c2∈Z.(a^2-b^2)+(b^2-c^2)=a^2-c^2\in \mathbb{Z}.(a2−b2)+(b2−c2)=a2−c2∈Z.

Hence

(a,c)∈T.(a,c)\in T.(a,c)∈T.

So TTT is transitive.

Thus for TTT:

  • symmetric,
  • transitive.
  1. Check the options

  • A: SSS is transitive but TTT is not.
    False.

  • B: both SSS and TTT are symmetric.
    False, because SSS is not symmetric.

  • C: neither SSS nor TTT is transitive.
    False, because TTT is transitive.

  • D: TTT is symmetric but SSS is not.
    True.

  1. Final answer

The correct option is

D\boxed{D}D​
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