NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

Q.1. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.1.\ \ Show\ that\ the\ function\ f:R\rightarrow {x\in R- 1< x< 1} \ defined\ by\ f( x) =\ \frac{x}{1+|x|} ,\ x\in R\ is\ one-one\ and\ onto\ function.\\ Solution:\ To\ show\ one-one:\ Case\ 1.\ Both\\ x_{1} ,x_{2} \ greater\ than\ or\ equal\ to\ 0\ Let\ us\ assume\ f( x_{1}) =f( x_{2}) \ ;\ \ x_{1} ,x_{2} \\ greater\ than\ or\ equal\ to\ 0.\ \Longrightarrow \frac{x_{1}}{1+x_{1}} =\frac{x_{2}}{1+x_{2}}\\\Longrightarrow x_{1} +x_{1} x_{2} =x_{2} +x_{1} x_{2}\ \Longrightarrow x_{1} =x_{2}\\ \Longrightarrow It\ is\ one-one.\ Case\ 2:\ Both\ x_{1} ,x_{2} < 0\ f( x_{1}) =f( x_{2})\\ \Longrightarrow \frac{x_{1}}{1-x_{1}} =\frac{x_{2}}{1-x_{2}}\ \Longrightarrow x_{1} -x_{1} x_{2} =x_{2} -x_{1} x_{2}\ \Longrightarrow x_{1} =x_{2}\\ \Longrightarrow It\ is\ one-one\ Case\ 3:\ x_{1} >0,\ x_{2} < 0\ or\ vice-versa\ Ifx_{1} >0,\ f( x_{1}) >0.\ If\ x_{2} < 0\ then\ f( x_{2}) < 0.\\ Thus,\ f( x_{1}) \neq f( x_{2})\\ So,\ this\ case\ is\ not\ possible.\\ Therefore,\ considering\ valid\ cases,\ f\ is\\ one-one.\ On-to\ or\ Surjective:\ If\ y\ is\ greater\ than\ or\ equal\ to\ 0:\ Let\ y=f( x)\\ \Longrightarrow y=\frac{x}{1+x}\ \Longrightarrow y+yx=x\ \Longrightarrow x=\frac{y}{1-y} .\\ Since\ 0< y< 1,\ 1-y >0,\\
For\ every\ y\in 0< y< 1,\ there\ exists\ x\ and\ hence\ it\ is\ surjective.\ Ify< 0,\ y=\frac{x}{1-x}\\ \Longrightarrow x=\frac{y}{1+y}\ As\ y >-1,\ so\ x\ is\ well\ defined.\\
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Therefore,\ f\ is\ surjective.
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Q.2. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.2.\ Show\ that\ the\ function\ f:R\rightarrow R\ given\ by\ f( x) =x^{3} \ is\ injective.\\
Solution:\\
Let\ f( x_{1}) =f( x_{2}) ,\ x_{1} ,\ x_{2} \ \in R\\
\Longrightarrow x_{1}^{3} =x_{2}^{3}\\
\Longrightarrow x_{1} =x_{2}\\
\Longrightarrow \ f\ is\ injective.\\
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Q.3. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.3.\ \ Given\ a\ non-empty\ set\ X,\ consider\ P( X) \ which\ is\ the\ set\ of\ all\ subsets\ of\ X.\\
Define\ the\ relationR\ in\ P( X) \ as\ follows:\\
For\ subsets\ A,\ B\ in\ P( X) ,\ ARB\ if\ and\ only\ if\ A\subset B.\ Is\ R\ an\ equivalence\ relation\ on\ P( X) ?\\
Justify\ your\ answer.\\
Solution:\\
Reflexivity:\ For\ any\ set\ A\in P( X) ,\\
A\subset A\ ( Every\ set\ is\ subset\ of\ itself) .\\
So,\ ARA\ is\ true.\\
\Longrightarrow it\ is\ reflexive.\\
Symmetric:\\
A\ relation\ is\ symmetric\ if\ ARB\Longrightarrow BRA.\ \\
For\ subsets\ A,B\in P( X) ,\ A\subset B\ does\ not\ necessarilyt\ imply\ B\subset A.\\
For\ example,\ if\ X={1,2,3} \ A={1,2} ,\ and\ B={1,2,3} \ then\ A\subset B\ ( ARB)\\
But\ B\ is\ not\ a\ subset\ of\ A,\ So\ B\ is\ not\ related\ to\ A.\\
\Longrightarrow it\ is\ not\ symmetric.\\
Transitive:\ If\ ARB\ ( A\subset B) \ and\ BRC\ ( B\subset C) \ then\ A\subset C\\
\Longrightarrow ARC\ is\ true.\\
Hence,\ it\ is\ transitive.\\
Since\ it\ is\ reflexive\ and\ transitive\ but\ not\ a\ symmetric\ relation.\\
Therefore,\ it\ is\ not\ an\ equivalence\ relation.
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Q.4. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.4.\ \ Find\ the\ number\ of\ all\ onto\ functions\ from\ the\ set\ {1,2,3,………,n} \ to\ itself.\\
Solution:\\
For\ every\ element\ in\ the\ given\ set,\ there\ must\ be\ preimage\ in\ the\ same\ set.\ \ \\
So,\ for\ the\ first\ element,\ say\ 1,\ it\ has\ n\ choices,\ the\ second\ element\ has\ n-1\ choices\\ and\ so\ on.\\
Therefore\ total\ number\ of\ onto\ functions\ is:\\
n( n-1)( n-2) ……….3.2.1\ =n!
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Q.5. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.5.\ \ Let\ A={-1,0,1,2} \ B={-4,-2,0,2} \ and\\ f,g:\ A\rightarrow B\ be\ functions\ defined\ by\\
f( x) =x^{2} -x,\ x\in A\ and\ g( x) =2|x-\frac{1}{2} |-1,\ x\in A.\ Are\ f\ and\ g\ equal?\ Justify\\ your\\
answer.\\
Solution:\\
We\ have\ f( x) =x^{2} -x,\ x\in A\\
So,\ f( -1) =( -1)^{2} -( -1) =1+1=2\ ,\ we\ get\ ( -1,2)\\
f( 0) =0^{2} -0=0,\ \ We\ get\ ( 0,0)\\
f( 1) =1^{2} -1=0,\ we\ get\ ( 1,0)\\
f( 2) =2^{2} -2=2,\ we\ get\ ( 2,2)\\
Now,\ g( x) =2|x-\frac{1}{2} |-1\\
\Longrightarrow g( -1) =2|-1-\frac{1}{2} |-1=2|-\frac{3}{2} |-1=2,\ we\ get\ ( -1,2)\\
g( 0) =0\ we\ get\ ( 0,0)\\
g( 1) =0\ we\ get\ ( 1,0)\\
g( 2) =2|2-\frac{1}{2} |-1=2\times \frac{3}{2} -1=2\ we\ get\ ( 2,2)\\
We\ see\ for\ all\ x\in A,\ f( x) =g( x) .\ Thus\ f\ and\ g\ are\ equal.
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Q.6. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

Q.7. NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

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Q.7.\ Let\ A={1,2,3} .\ \ Then\ number\ of\ equivalence\ relations\ containing( 1,2) \ is\\
( A) \ 1\ ( B) \ 2\ ( C) \ 3\ ( D) \ 4\\
Solution:\ To\ be\ reflexive\ we\ must\ have\ ( 1,1) ,\ ( 2,2) ,\ ( 3,3)\\
To\ be\ symmetric\ \ we\ must\ have\ ( 2,1) \ as\ we\ already\ have\ ( 1,2) .\\
To\ be\ transitive\ we\ have\ ( a,b) \in R\ and\ ( b,c) \in R\ then\ ( a,c) \in R\\
So\ for\ an\ equivalence\ relation\ the\ first\ relation\ we\ can\ consider:\\
R_{1} ={( 1,1) ,\ ( 2,2) ,\ ( 3,3) ,\ ( 1,2) ,\ ( 2,1)}\\
[ for\ transitivity\ we\ have\ ( 1,2) ,\ ( 2,1) ,\ ( 1,1)]\\
R_{2} ={( 1,1) ,\ ( 2,2) ,\ ( 3,3) ,\ ( 1,2) ,\ ( 2,1) ,\ ( 2,3) ,\ ( 3,2)( 1,3)}\\
Only\ these\ 2\ relations\ are\ possible.\\
Ans.\ ( B)
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NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise solution

The NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise Solutions have been prepared keeping in mind the Relations and Functions Chapter 1 topic of Class 12 Maths. This exercise combines application-based questions and logical questions. Throughout this chapter, practice solutions are provided to develop problem-solving skills and a deeper understanding. We have tried to provide step-by-step solutions to make it easy for students to understand, and we hope to meet your expectations.

Class 12 maths NCERT ch1 miscellaneous ex. Solution

This well-explained NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise Solution helps students deepen their understanding of relations and functions. Prepared with correct methodology and step-by-step explanations, it covers all important steps and patterns necessary for CBSE board exams. Regular practice with this solution will also help students build confidence.

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