NCERT Exemplar · Class 12 Mathematics Relations and Functions

(i) \(\displaystyle f(x)=\dfrac{x}{2}\) \[f(x_1)=f(x_2)\Rightarrow \frac{x_1}{2}=\frac{x_2}{2}\Rightarrow x_1=x_2 \quad\text{(one-one)} \] \[f(x)=1\Rightarrow x=2\notin A \quad\text{(not onto)} \] (ii) \(\displaystyle g(x)=|x|\) \[g(-1)=g(1)=1 \quad\text{(not one-one)} \] \[|x|=-1 \ \text{is impossible} \quad\text{(not onto)} \] (iii) \(\displaystyle h(x)=x|x|\) \[h(x)=\begin{cases} x^2, & x\ge 0\\ -x^2, & x<0\end{cases} \] Equal images need equal signs: \[x_1,x_2\ge 0:\ x_1^2=x_2^2\Rightarrow x_1=x_2 \] \[x_1,x_2<0:\ -x_1^2=-x_2^2\Rightarrow x_1=x_2 \] For \(\displaystyle y\in[-1,1]\): \[y\ge 0:\ x=\sqrt{y}\in[0,1],\quad h(x)=y \] \[y<0:\ x=-\sqrt{-y}\in[-1,0),\quad h(x)=-x^2=-(-y)=y \] (iv) \(\displaystyle k(x)=x^2\) \[k(-1)=k(1)=1 \quad\text{(not one-one)} \] \[x^2=-1 \ \text{has no real solution} \quad\text{(not onto)} \] Answer: (i) one-one, not onto; (ii) neither; (iii) bijective; (iv) neither.

(i) \(\displaystyle x>y\) \[x>x \ \text{is false} \Rightarrow \text{not reflexive} \] \[(2,1)\in R,\ \ (1,2)\notin R \Rightarrow \text{not symmetric} \] \[x>y,\ y>z \Rightarrow x>z \Rightarrow \text{transitive} \] (ii) \(\displaystyle x+y=10\) \[1+1=2\ne 10 \Rightarrow \text{not reflexive} \] \[x+y=10 \Rightarrow y+x=10 \Rightarrow \text{symmetric} \] \[(1,9),(9,1)\in R,\ \ (1,1)\notin R \Rightarrow \text{not transitive} \] (iii) \(\displaystyle xy\) is a perfect square \[x\cdot x=x^2 \Rightarrow \text{reflexive} \] \[xy=yx \Rightarrow \text{symmetric} \] Let \(\displaystyle xy=m^2\) and \(\displaystyle yz=n^2\): \[xz=\frac{m^2n^2}{y^2}=\left(\frac{mn}{y}\right)^2 \] A rational whose square is a natural number is itself an integer, so \(\displaystyle xz\) is a perfect square: transitive. (iv) \(\displaystyle x+4y=10\) \[y=1\Rightarrow x=6;\quad y=2\Rightarrow x=2;\quad y\ge 3\Rightarrow x\le -2 \] \[R=\{(6,1),\,(2,2)\} \] \[(1,1)\notin R \Rightarrow \text{not reflexive} \] \[(6,1)\in R,\ \ (1,6)\notin R \Rightarrow \text{not symmetric} \] No pair \(\displaystyle (1,z)\) lies in \(\displaystyle R\), so the only chain is \(\displaystyle (2,2),(2,2)\) and \(\displaystyle (2,2)\in R\): transitive. Answer: (i) transitive only; (ii) symmetric only; (iii) reflexive, symmetric and transitive; (iv) transitive only.

Reflexive: for \(\displaystyle (a,b)\in A\times A\), \[a+b=b+a \Rightarrow (a,b)\,R\,(a,b) \] Symmetric: let \(\displaystyle (a,b)\,R\,(c,d)\). \[a+d=b+c \Rightarrow c+b=d+a \Rightarrow (c,d)\,R\,(a,b) \] Transitive: let \(\displaystyle (a,b)\,R\,(c,d)\) and \(\displaystyle (c,d)\,R\,(e,f)\). \[a+d=b+c,\qquad c+f=d+e \] \[a+d+c+f=b+c+d+e \Rightarrow a+f=b+e \] \[\Rightarrow (a,b)\,R\,(e,f) \] So \(\displaystyle R\) is an equivalence relation. Class of \(\displaystyle (2,5)\): \(\displaystyle (c,d)\) with \(\displaystyle (2,5)\,R\,(c,d)\), \[2+d=5+c \Rightarrow d=c+3 \] With \(\displaystyle c,d\in\{1,\ldots,9\}\): \(\displaystyle c=1,\ldots,6\). \[[(2,5)]=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\} \] Answer: \(\displaystyle R\) is an equivalence relation and \(\displaystyle [(2,5)]=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}\).

\(\displaystyle f\) is invertible if some \(\displaystyle g:B\to A\) satisfies \[g\circ f=I_A,\qquad f\circ g=I_B \] Invertible \(\displaystyle \Rightarrow\) one-one and onto. One-one: \[f(x_1)=f(x_2)\Rightarrow g(f(x_1))=g(f(x_2))\Rightarrow x_1=x_2 \] Onto: for \(\displaystyle y\in B\), take \(\displaystyle x=g(y)\in A\). \[f(x)=f(g(y))=y \] One-one and onto \(\displaystyle \Rightarrow\) invertible. For each \(\displaystyle y\in B\), onto gives some \(\displaystyle x\in A\) with \(\displaystyle f(x)=y\), and one-one makes it unique. So \(\displaystyle g\) is a well-defined function \(\displaystyle B\to A\): \[g(y)=x \iff f(x)=y \] \[f(g(y))=f(x)=y \Rightarrow f\circ g=I_B \] For \(\displaystyle x\in A\), put \(\displaystyle y=f(x)\). Its only preimage is \(\displaystyle x\): \[g(f(x))=g(y)=x \Rightarrow g\circ f=I_A \] Answer: \(\displaystyle g\) is an inverse of \(\displaystyle f\), so \(\displaystyle f\) is invertible if and only if it is one-one and onto.

(i) \[(f\circ g)(x) = f(2x-3) = (2x-3)^2 + 3(2x-3) + 1 \] \[= 4x^2 - 12x + 9 + 6x - 9 + 1 = 4x^2 - 6x + 1 \](ii) \[(g\circ f)(x) = g(x^2+3x+1) = 2(x^2+3x+1) - 3 = 2x^2 + 6x - 1 \](iii) \[(f\circ f)(x) = (x^2+3x+1)^2 + 3(x^2+3x+1) + 1 \] \[= (x^4 + 6x^3 + 11x^2 + 6x + 1) + (3x^2 + 9x + 3) + 1 \] \[= x^4 + 6x^3 + 14x^2 + 15x + 5 \](iv) \[(g\circ g)(x) = g(2x-3) = 2(2x-3) - 3 = 4x - 9 \]Answer: (i) \(\displaystyle 4x^2-6x+1\); (ii) \(\displaystyle 2x^2+6x-1\); (iii) \(\displaystyle x^4+6x^3+14x^2+15x+5\); (iv) \(\displaystyle 4x-9\).

(i) \[a*b = a-b, \qquad b*a = b-a \] \[1*0 = 1 \ne -1 = 0*1 \quad \text{(not commutative)} \](ii) \[a*b = a^2+b^2 = b^2+a^2 = b*a \quad \text{(commutative)} \](iii) \[1*0 = 1+0 = 1, \qquad 0*1 = 0+0 = 0 \quad \text{(not commutative)} \](iv) \[a*b = (a-b)^2 = (b-a)^2 = b*a \quad \text{(commutative)} \]Answer: (ii) and (iv) are commutative.

(i) commutative but not associative.\[a*b = 1+ab = 1+ba = b*a \quad \text{(commutative)} \] \[(a*b)*c = 1 + (1+ab)c = 1 + c + abc \] \[a*(b*c) = 1 + a(1+bc) = 1 + a + abc \] \[(1*0)*0 = 1*0 = 1, \qquad 1*(0*0) = 1*1 = 2 \quad \text{(not associative)} \]

(C) equivalence\[a \cong a \quad \text{(reflexive)} \] \[a \cong b \Rightarrow b \cong a \quad \text{(symmetric)} \] \[a \cong b,\ b \cong c \Rightarrow a \cong c \quad \text{(transitive)} \]Congruence has all three properties.

(B) transitive but not symmetric\[a \text{ brother of } b,\ b \text{ sister of } a \Rightarrow (a,b)\in R,\ (b,a)\notin R \quad \text{(not symmetric)} \] \[a \text{ brother of } b,\ b \text{ brother of } c\ (a \ne c) \Rightarrow a \text{ brother of } c \quad \text{(transitive)} \]

(D) $\displaystyle 5$Each equivalence relation on \(\displaystyle A\) is a partition of \(\displaystyle A\) into classes.\[\{1\},\{2\},\{3\} \] \[\{1,2\},\{3\} \qquad \{1,3\},\{2\} \qquad \{2,3\},\{1\} \] \[\{1,2,3\} \] \[1 + 3 + 1 = 5 \]