NCERT Solutions · Class 12 Mathematics Relations and Functions

First check that \(\displaystyle f\) really lands inside the stated codomain: since \(\displaystyle |x|<1+|x|\) for every real \(\displaystyle x\), \[|f(x)|=\frac{|x|}{1+|x|}<1,\] so \(\displaystyle f(x)\in\{x\in\mathbf{R}:-1<x<1\}\).Remove the modulus by splitting at \(\displaystyle 0\): \[f(x)=\frac{x}{1+x}\ \ (x\ge 0),\qquad f(x)=\frac{x}{1-x}\ \ (x<0).\] Since the denominator \(\displaystyle 1+|x|\) is always positive, \(\displaystyle f(x)\) carries the sign of \(\displaystyle x\): \(\displaystyle f(x)>0\) for \(\displaystyle x>0\), \(\displaystyle f(x)<0\) for \(\displaystyle x<0\), and \(\displaystyle f(x)=0\) only for \(\displaystyle x=0\).One-one. By definition, we must show \(\displaystyle f(x_1)=f(x_2)\Rightarrow x_1=x_2\). Suppose \(\displaystyle f(x_1)=f(x_2)\). By the sign remark above, \(\displaystyle x_1\) and \(\displaystyle x_2\) cannot have opposite signs, and one of them is \(\displaystyle 0\) only if both are. So \(\displaystyle x_1,x_2\) lie on the same branch — this is the step to state, because otherwise the case work below is not justified.If \(\displaystyle x_1,x_2\ge 0\): \[\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}\ \Rightarrow\ x_1(1+x_2)=x_2(1+x_1)\ \Rightarrow\ x_1+x_1x_2=x_2+x_1x_2\ \Rightarrow\ x_1=x_2.\] If \(\displaystyle x_1,x_2<0\): \[\frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}\ \Rightarrow\ x_1(1-x_2)=x_2(1-x_1)\ \Rightarrow\ x_1-x_1x_2=x_2-x_1x_2\ \Rightarrow\ x_1=x_2.\] Hence \(\displaystyle f\) is one-one.Onto. Take any \(\displaystyle y\) with \(\displaystyle -1<y<1\) and produce a pre-image.If \(\displaystyle y=0\): \(\displaystyle f(0)=0\).If \(\displaystyle 0<y<1\): put \(\displaystyle x=\dfrac{y}{1-y}\). Here \(\displaystyle 1-y>0\) and \(\displaystyle y>0\), so \(\displaystyle x>0\) and \(\displaystyle |x|=x\); then \[f(x)=\frac{y/(1-y)}{1+\dfrac{y}{1-y}}=\frac{y}{(1-y)+y}=y.\]If \(\displaystyle -1<y<0\): put \(\displaystyle x=\dfrac{y}{1+y}\). Here \(\displaystyle 1+y>0\) and \(\displaystyle y<0\), so \(\displaystyle x<0\) and \(\displaystyle |x|=-x\); then \[f(x)=\frac{x}{1-x}=\frac{y/(1+y)}{1-\dfrac{y}{1+y}}=\frac{y}{(1+y)-y}=y.\]So every element of \(\displaystyle \{x\in\mathbf{R}:-1<x<1\}\) is a value of \(\displaystyle f\), i.e. \(\displaystyle f\) is onto.Being both one-one and onto, \(\displaystyle f\) is a bijection from \(\displaystyle \mathbf{R}\) onto \(\displaystyle \{x\in\mathbf{R}:-1<x<1\}\).

By definition, \(\displaystyle f\) is injective (one-one) if \(\displaystyle f(x_1)=f(x_2)\Rightarrow x_1=x_2\).Let \(\displaystyle x_1,x_2\in\mathbf{R}\) with \(\displaystyle f(x_1)=f(x_2)\), i.e. \(\displaystyle x_1^{3}=x_2^{3}\). Then \[x_1^{3}-x_2^{3}=0\ \Rightarrow\ (x_1-x_2)\left(x_1^{2}+x_1x_2+x_2^{2}\right)=0.\]The step a student usually skips is dealing with the quadratic factor, so examine it by completing the square: \[x_1^{2}+x_1x_2+x_2^{2}=\left(x_1+\frac{x_2}{2}\right)^{2}+\frac{3}{4}x_2^{2}.\] This is a sum of two squares, hence \(\displaystyle \ge 0\), and it vanishes only when \(\displaystyle x_2=0\) and \(\displaystyle x_1+\dfrac{x_2}{2}=0\), i.e. only when \(\displaystyle x_1=x_2=0\).Two cases follow from the factorisation: In either case \(\displaystyle x_1=x_2\). Hence \(\displaystyle f(x)=x^{3}\) is injective on \(\displaystyle \mathbf{R}\).

A relation is an equivalence relation only if it is reflexive, symmetric and transitive; test all three for \(\displaystyle \mathrm{A}\,\mathrm{R}\,\mathrm{B}\iff \mathrm{A}\subset \mathrm{B}\) on \(\displaystyle \mathrm{P(X)}\).Reflexive. For every subset \(\displaystyle \mathrm{A}\) of \(\displaystyle \mathrm{X}\), each element of \(\displaystyle \mathrm{A}\) belongs to \(\displaystyle \mathrm{A}\), so \(\displaystyle \mathrm{A}\subset \mathrm{A}\), i.e. \(\displaystyle \mathrm{A}\,\mathrm{R}\,\mathrm{A}\). So R is reflexive.Symmetric. Here is where it fails, and the hypothesis "\(\displaystyle \mathrm{X}\) non-empty" is exactly what lets us build the counterexample. Since \(\displaystyle \mathrm{X}\neq\varphi\), choose some \(\displaystyle a\in \mathrm{X}\) and take \[\mathrm{A}=\varphi,\qquad \mathrm{B}=\{a\},\] both of which are members of \(\displaystyle \mathrm{P(X)}\). Then \(\displaystyle \varphi\subset\{a\}\), so \(\displaystyle \mathrm{A}\,\mathrm{R}\,\mathrm{B}\); but \(\displaystyle a\in \mathrm{B}\) and \(\displaystyle a\notin \mathrm{A}\), so \(\displaystyle \mathrm{B}\not\subset \mathrm{A}\), i.e. \(\displaystyle \mathrm{B}\,\mathrm{R}\,\mathrm{A}\) is false. Hence R is not symmetric.Transitive. Let \(\displaystyle \mathrm{A}\subset \mathrm{B}\) and \(\displaystyle \mathrm{B}\subset \mathrm{C}\). If \(\displaystyle x\in \mathrm{A}\), then \(\displaystyle x\in \mathrm{B}\) (first inclusion), hence \(\displaystyle x\in \mathrm{C}\) (second inclusion). So \(\displaystyle \mathrm{A}\subset \mathrm{C}\), i.e. \(\displaystyle \mathrm{A}\,\mathrm{R}\,\mathrm{C}\). So R is transitive.R is reflexive and transitive but not symmetric, therefore R is not an equivalence relation on \(\displaystyle \mathrm{P(X)}\).

Let \(\displaystyle \mathrm{S}=\{1,2,3,\ldots,n\}\) and let \(\displaystyle f:\mathrm{S}\to\mathrm{S}\) be onto.Step $\displaystyle 1$: on a finite set, onto forces one-one. The image of \(\displaystyle f\) is the list of \(\displaystyle n\) values \[f(\mathrm{S})=\{f(1),f(2),\ldots,f(n)\},\] so \(\displaystyle f(\mathrm{S})\) has at most \(\displaystyle n\) distinct elements, with exactly \(\displaystyle n\) only if the values \(\displaystyle f(1),\ldots,f(n)\) are all different. Since \(\displaystyle f\) is onto, \(\displaystyle f(\mathrm{S})=\mathrm{S}\), which has \(\displaystyle n\) elements; so the \(\displaystyle n\) values must indeed be all different, i.e. \(\displaystyle f\) is one-one. (This is special to a finite set mapped to itself — it is false for infinite sets.)Step $\displaystyle 2$: count the bijections. So the onto functions from \(\displaystyle \mathrm{S}\) to itself are precisely the one-one onto functions, i.e. the arrangements of \(\displaystyle 1,2,\ldots,n\) in the \(\displaystyle n\) places \(\displaystyle f(1),f(2),\ldots,f(n)\). Choose the values one at a time, none repeating: \[f(1):\ n\ \text{choices},\quad f(2):\ n-1,\quad f(3):\ n-2,\ \ldots,\quad f(n):\ 1.\] By the multiplication principle the total is \[n(n-1)(n-2)\cdots 2\cdot 1=n!.\]The number of onto functions from \(\displaystyle \{1,2,\ldots,n\}\) to itself is \(\displaystyle n!\).

Two functions \(\displaystyle f,g:\mathrm{A}\to\mathrm{B}\) are equal precisely when \(\displaystyle f(a)=g(a)\) for every \(\displaystyle a\in\mathrm{A}\). Here \(\displaystyle \mathrm{A}=\{-1,0,1,2\}\) has only four elements, so check all four. Take care with the modulus in \(\displaystyle g(x)=2\left|x-\frac12\right|-1\): the sign inside the bars changes at \(\displaystyle x=\frac12\), so evaluate \(\displaystyle \left|x-\frac12\right|\) numerically at each point rather than dropping the bars.\(\displaystyle x=-1\): \[f(-1)=(-1)^{2}-(-1)=1+1=2,\qquad g(-1)=2\left|-1-\tfrac12\right|-1=2\cdot\tfrac32-1=2.\]\(\displaystyle x=0\): \[f(0)=0^{2}-0=0,\qquad g(0)=2\left|0-\tfrac12\right|-1=2\cdot\tfrac12-1=0.\]\(\displaystyle x=1\): \[f(1)=1^{2}-1=0,\qquad g(1)=2\left|1-\tfrac12\right|-1=2\cdot\tfrac12-1=0.\]\(\displaystyle x=2\): \[f(2)=2^{2}-2=2,\qquad g(2)=2\left|2-\tfrac12\right|-1=2\cdot\tfrac32-1=2.\]All four values agree (and each lies in \(\displaystyle \mathrm{B}=\{-4,-2,0,2\}\), so both are indeed functions \(\displaystyle \mathrm{A}\to\mathrm{B}\)). Since \(\displaystyle f(a)=g(a)\ \forall a\in\mathrm{A}\), \(\displaystyle f\) and \(\displaystyle g\) are equal functions.Remark (why it is a coincidence of the domain, not an identity): solving \(\displaystyle x^{2}-x=2\left|x-\frac12\right|-1\) gives, for \(\displaystyle x\ge\frac12\), \(\displaystyle x^{2}-x=2x-2\Rightarrow x^{2}-3x+2=0\Rightarrow x=1,2\); and for \(\displaystyle x<\frac12\), \(\displaystyle x^{2}-x=-2x\Rightarrow x^{2}+x=0\Rightarrow x=0,-1\). The two formulas agree at exactly \(\displaystyle -1,0,1,2\) — which is exactly the set \(\displaystyle \mathrm{A}\). On any larger domain they would not be equal.

Write \(\displaystyle \mathrm{A}=\{1,2,3\}\); a relation on A is a subset of \(\displaystyle \mathrm{A}\times\mathrm{A}\), which has \(\displaystyle 9\) ordered pairs.Step $\displaystyle 1$: pairs forced into R. R must be reflexive, so it contains \(\displaystyle (1,1),(2,2),(3,3)\). It is given to contain \(\displaystyle (1,2)\) and \(\displaystyle (1,3)\), and symmetry then forces \(\displaystyle (2,1)\) and \(\displaystyle (3,1)\). So every admissible R contains \[\mathrm{R}_0=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}.\]Step $\displaystyle 2$: what is left to choose. Of the nine pairs, only \(\displaystyle (2,3)\) and \(\displaystyle (3,2)\) are still free. Symmetry means they must be included together or omitted together — so there are just two candidates.Step $\displaystyle 3$: apply the "not transitive" condition. Exactly one relation qualifies.Answer: (A) 1.

Use the fact that an equivalence relation on a set corresponds to a partition of that set into equivalence classes, and count the partitions instead of the pairs.Let R be an equivalence relation on \(\displaystyle \mathrm{A}=\{1,2,3\}\) containing \(\displaystyle (1,2)\). Then \(\displaystyle 1\) and \(\displaystyle 2\) lie in the same equivalence class. The only question left is where \(\displaystyle 3\) sits: either in a class of its own, or in the same class as \(\displaystyle 1\) and \(\displaystyle 2\). That gives two partitions, hence two relations.\(\displaystyle \{1,2\},\{3\}\): \[\mathrm{R}_1=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}.\]\(\displaystyle \{1,2,3\}\): \[\mathrm{R}_2=\mathrm{A}\times\mathrm{A}.\]Both are reflexive, symmetric and transitive, and both contain \(\displaystyle (1,2)\).The step worth stating is why nothing else can occur: if R contained \(\displaystyle (1,2)\) together with, say, \(\displaystyle (2,3)\), transitivity would force \(\displaystyle (1,3)\) and symmetry the reverses, giving \(\displaystyle \mathrm{A}\times\mathrm{A}=\mathrm{R}_2\) again — no third possibility arises. Similarly \(\displaystyle (1,3)\in\mathrm{R}\) forces \(\displaystyle \mathrm{R}=\mathrm{R}_2\).Answer: (B) 2.