Relations and Functions

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}(i)Giventhat-1\le x\le 1\\ Let{{\displaystyle x}}_1,{{\displaystyle x}}_2\in f(x)\\ f({{\displaystyle x}}_1)=\frac{1}{{{\displaystyle x}}_1}andf({{\displaystyle x}}_2)=\frac{1}{{{\displaystyle x}}_2}\\ f({{\displaystyle x}}_1)=f({{\displaystyle x}}_2)\Rightarrow \frac{1}{{{\displaystyle x}}_1}=\frac{1}{{{\displaystyle x}}_2}\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,f(x)isone-onefunction.\\ Letf(x)=y=\frac{x}{2}\Rightarrow x=2y\\ Fory=1,x=2\notin \left[-1,1\right]\\ So,f(x)isnotonto.Hence,f(x)isnotbijectivefunction.\\ (ii)Here,g(x)=\left|x\right|\\ g({{\displaystyle x}}_1)=g({{\displaystyle x}}_2)\Rightarrow \left|{{\displaystyle x}}_1\right|=\left|{{\displaystyle x}}_2\right|\Rightarrow {{\displaystyle x}}_1=\pm {{\displaystyle x}}_2\\ So,g(x)isnotone-onefunction.\\ Letg(x)=y=\left|x\right|\Rightarrow x=\pm y\notin A\forall y\in A\\ So,g(x)isnotontofunction.\\ Hence,g(x)isnotbijectivefunction.\\ (iii)Here,h(x)=x\left|x\right|\\ h({{\displaystyle x}}_1)=h({{\displaystyle x}}_2)\\ \Rightarrow {{\displaystyle x}}_1\left|{{\displaystyle x}}_1\right|={{\displaystyle x}}_2\left|{{\displaystyle x}}_2\right|\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,h(x)isone-onefunction.\\ Now,leth(x)=y=x\left|x\right|={{\displaystyle x}}^{2}or-{{\displaystyle x}}^2\\ \Rightarrow \text{\&thi}\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,A\in R-\{3\},B=R-\{1\}\\ Giventhatf:A\to Bdefinedbyf(x)=\frac{x-2}{x-3}\forall x\in A.\\ Let{{\displaystyle x}}_1,{{\displaystyle x}}_2\in f(x)\\ \therefore f({{\displaystyle x}}_1)=f({{\displaystyle x}}_2)\\ \Rightarrow \frac{{{\displaystyle x}}_1-2}{{{\displaystyle x}}_1-3}=\frac{{{\displaystyle x}}_2-2}{{{\displaystyle x}}_2-3}\\ \Rightarrow ({{\displaystyle x}}_1-2)({{\displaystyle x}}_1-3)=({{\displaystyle x}}_2-2)({{\displaystyle x}}_2-3)\\ \Rightarrow -{{\displaystyle x}}_1=-{{\displaystyle x}}_2\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,itisinjectivefunction.\\ Now,Lety=\frac{x-2}{x-3}\\ \Rightarrow xy-3y=x-2\Rightarrow xy-x=3y-2\\ \Rightarrow x(y-1)=3y-2\Rightarrow x=\frac{3y-2}{y-1}\\ f(x)=\frac{x-2}{x-3}=\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}\Rightarrow \frac{3y-2-2y+2}{3y-2-3y+3}\Rightarrow y\\ \Rightarrow f(x)=y\in B.\\ So,f(x)issurjectivefunction.\\ Hence,f(x)isbijectivefunction.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:$\begin{array}{l}\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,A=\{2,3,4\}andB=\{2,5,6,7\}\\ \left(i\right)Letf:A\to BbethemappingfromAtoB\\ f=\{\left(x,y\right):y=x+3\}\\ \therefore f=\{\left(2,5\right),\left(3,6\right),\left(4,7\right)\}whichisaninjectivemapping.\\ \left(ii\right)Letg:A\to BbethemappingfromA\to Bsuchthat\\ \therefore g=\{\left(2,5\right),\left(3,5\right),\left(4,2\right)\}whichisnotaninjectivemapping.\\ \left(iii\right)Leth:B\to AbethemappingfromBtoA\\ h=\{\left(y,x\right):x=y-2\}\\ \therefore h=\{\left(5,3\right),\left(6,4\right),\left(7,3\right)\}whichisthemappingfromBtoA.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given\text{?}\text{?}thatx\in N,y\in Nand2x+y=41\\ \therefore DomainofR=\{1,2,3,4,5,\dots ,20\}\\ andRange=\{39,37,35,33,31,\dots ,1\}\\ Here,\left(3,3\right)\notin R\\ as2\times 3+3\ne 41\\ So,Risnotreflexive.\\ Risnotsymmetricas\left(2,37\right)\in Rbut\left(37,2\right)\notin R\\ Risnottransitiveas\left(11,19\right)\in Rand\left(19,3\right)\in R\\ but\left(11,3\right)\notin R.\\ Hence,Risneitherreflexive,norsymmetricandnortransitive.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$Giventhat:$
$\begin{array}{l} A=\{1,2,3\}\\ and R=\{(1,1),(2,2),\left(3,3\right),(1,2),\left(2,3\right),(1,3)\}\\ Here,1R1,2R2and3R3,So,Risreflexive.\\ 1R2but2\cancel{R}1or2R3but3\cancel{R}2,\\ So,Risnotsymmetric.\\ 1R1and1R2\Rightarrow 1R3,So,Ristransitive.\\ Hence,thecorrectansweris(a).\end{array}$

35. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 × 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b $\in$ z

So, ab=(–1) × (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1,

34. Given,

A={1,2,3,4}

B={1,5,9,11,15,16}

f={(1,5),(2,9),(3,1),(4,5),(2,11)}.

(i) As every element of f is an element of A × B

We can clearly say that f $\subset$ A × B.

?f is a relation from A to B.

(ii) As the element 2 of the domain has two image i.e., 9 and 11. f is not a function.

33. Given, R= { (a, b): a, b $\in$ N and a = b²}

(i) Let a = 2 $\in$ N

Then b = 2² = 4 $\in$ N

but a ≠ b.

Hence the given statement is not true.

(ii) For a=b² the inverse b=a² may not hold true

Example (4,2) $\in$ R, a=4, b=2 and a=b²

but (2,4) $\notin$ R.

Hence, the given statement is not true.

(iii) I

32. Given, f(x) = (ax + b)

= {(1,1),(2,3),(0, – 1),(–1, –3)} .

As (1,1) $\in$ f.

Then, f(1)=1   [? f(x) = y for (x, y)]

a × 1+b=1

a+b=1…… (1)

and (0, – 1) $\in$ f .

Then, f(0)= –1

a× 0+b= –1

b= –1…….(2)

Putting value of (2) in (1) we gets

a – 1=1

a=1+1

a=2

So, (a, b)=(2, –1)