Please answer this question with explaination ✍️
⭐Explaination needed
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Answers
- A = { 1 , 2 , 3 }
- The relation R defined in A by aRb,
- If “|a² - b²| ≤ 5” .
- Set R .
.
- Domain of R .
- Range of R .
- And finally check that which Option is incorrect according to the question .
Given that,
- |a² - b²| ≤ 5
CASE - 1 :-
Let,
- a = 1
=> |1² - b²| ≤ 5
=> |b² - 1²| ≤ 5
☃️ If we put the value of b = 1 or 2, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 1 and 2 .
The possible sets are “ (1 , 1) and (1 , 2) ” .
CASE - 2 :-
Let,
- a = 2
=> |2² - b²| ≤ 5
=> |b² - 2²| ≤ 5
☃️ If we put the value of b = 1 or 2 or 3, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 1 , 2 and 3
The possible set are “ (2 , 1) , (2 , 2) , (2 , 3) ” .
CASE - 3 :-
Let,
- a = 3
=> |3² - b²| ≤ 5
=> |b² - 3²| ≤ 5
☃️ If we put the value of b = 2 or 3, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 2 and 3 .
The possible sets are “ (3 , 2) , (3 , 3) ” .
✯ Therefore, form the above three cases we can confirmed the set R, i.e.
We have know that,
We have know that,
☃️ Domain of a set = {x : (x , y) R}
We have know that,
☃️ Range of a set = {y : (x , y) R}
✍️ Hence, from the above four solution we have confirmed that the incorrect option is “D” .
Answer:
\mathcal{\pink{\underbrace{\blue{GIVEN:-}}}}
GIVEN:−
A = { 1 , 2 , 3 }
The relation R defined in A by aRb,
If “|a² - b²| ≤ 5” .
\mathcal{\pink{\underbrace{\blue{TO\:FIND:-}}}}
TOFIND:−
Set R .
\rm\bold{R^{-1}}R
−1
.
Domain of R .
Range of R .
And finally check that which Option is incorrect according to the question .
\mathcal{\pink{\underbrace{\blue{SOLUTION:-}}}}
SOLUTION:−
Given that,
|a² - b²| ≤ 5
CASE - 1 :-
Let,
a = 1
=> |1² - b²| ≤ 5
=> |b² - 1²| ≤ 5
☃️ If we put the value of b = 1 or 2, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 1 and 2 .
\rm\red{\therefore}∴ The possible sets are “ (1 , 1) and (1 , 2) ” .
CASE - 2 :-
Let,
a = 2
=> |2² - b²| ≤ 5
=> |b² - 2²| ≤ 5
☃️ If we put the value of b = 1 or 2 or 3, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 1 , 2 and 3
\rm\red{\therefore}∴ The possible set are “ (2 , 1) , (2 , 2) , (2 , 3) ” .
CASE - 3 :-
Let,
a = 3
=> |3² - b²| ≤ 5
=> |b² - 3²| ≤ 5
☃️ If we put the value of b = 2 or 3, then all outputs of the above equation are less than equal to 5 .
✍️ Hence, the values of b are 2 and 3 .
\rm\red{\therefore}∴ The possible sets are “ (3 , 2) , (3 , 3) ” .
✯ Therefore, form the above three cases we can confirmed the set R, i.e.
\rm{\pink{\overbrace{\underbrace{\purple{[1]\:R\:=\:\{(1\:,\:1)\:,\:(1\:,\:2)\:,\:(2\:,\:1)\:,\:(2\:,\:2)\:,\:(2\:,\:3)\:,\:(3\:,\:2)\:,\:(3\:,\:3)\}\:}}}}}
[1]R={(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}
We have know that,
\orange{\bigstar}★ \rm{\green{R^{-1}\:=\:\{(y\:,\:x):(x\:,\:y)\:\in\:R\}\:}}R
−1
={(y,x):(x,y)∈R}
\rm{\pink{\overbrace{\underbrace{\purple{[2]\:R^{-1}\:=\:\{(1\:,\:1)\:,\:(2\:,\:1)\:,\:(1\:,\:2)\:,\:(2\:,\:2)\:,\:(3\:,\:2)\:,\:(2\:,\:3)\:,\:(3\:,\:3)\}\:}}}}}
[2]R
−1
={(1,1),(2,1),(1,2),(2,2),(3,2),(2,3),(3,3)}
We have know that,
☃️ Domain of a set = {x : (x , y) \rm{\in}∈ R}
\rm{\pink{\overbrace{\underbrace{\purple{[3]\:Domain\:of\:R\:=\:{\:1\:,\:2\:,\:3\:}\:}}}}}
[3]DomainofR=1,2,3
We have know that,
☃️ Range of a set = {y : (x , y) \rm{\in}∈ R}
\rm{\pink{\overbrace{\underbrace{\purple{[4]\:Range\:of\:R\:=\:{\:1\:,\:2\:,\:3\:}\:}}}}}
[4]RangeofR=1,2,3
\rule{200}2
✍️ Hence, from the above four solution we have confirmed that the incorrect option is “D” .
\green\bigstar\:\rm{\gray{\boxed{\pink{Required\:Answer\:\longrightarrow\:(D)\:Range\:of\:R\:=\:\{5\}\:}}}}★
RequiredAnswer⟶(D)RangeofR={5}