Maths

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

Sol:

Given that, g = { (1, 1), (2, 3), (3, 5), (4, 7)}.

Here, each element of domain has unique image. So, g is a function.

Now given that, g (x ) = $\alpha$ x + $\beta$

For (1,1) g (1) = $\alpha$ + $\beta$

$\alpha$ + $\beta$ = 1… (i)

For (2,3)   g (2) = 2 $\alpha$ + $\beta$

2 $\alpha$ + $\beta$ = 3 … (ii)

From Equations. (i) and (ii),

2 (1 - $\beta$ ) + $\beta$ = 3

2 - 2 $\beta$ + $\beta$ = 3

2 - $\beta$ = 3

$\beta$ = - 1

If $\beta$ = - 1, then $\alpha$ = 2

$\therefore$ $\alpha$ = 2,  $\beta$ = - 1

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

Sol:

$\begin{array}{l}Given that\hfill \end{array}$$\hfill$$\begin{array}{l}f(x)=x^2- 3x+ 2\hfill \end{array}$ $\begin{array}{ll}f\left\{f(x)\right\} =f\left(x^2-3x+2\right)\hfill & \hfill \end{array}$

$\begin{array}{ll}= {\left(x^2-3x+2\right)}^2- 3\left(x^2-3x+2\right) + 2\hfill & \hfill \end{array}$

$\begin{array}{ll}=x^4+ 9x^2+ 4 - 6x^3- 12x+ 4x^2- 3x^2+ 9x- 6 + 2\hfill & \hfill \end{array}$

$\begin{array}{l}=x^4+ 10x^2- 6x^3- 3x\hfill \end{array}$

$\begin{array}{l}Hence f\left\{f(x)\right\} =x^4- 6x^3+ 10x^2- 3x\hfill \end{array}$

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

Sol:

$\begin{array}{l}Given that, A = \left\{a,b,c,d\right\}\hfill \end{array}$

$\begin{array}{l}and f = \left\{\left(a,b\right),\left(b,d\right),\left(c,a\right),\left(d,c\right)\right\}\hfill \end{array}$

$\begin{array}{l}{f}^{-1}= \left\{\left(b,a\right),\left(d,b\right),\left(a,c\right),\left(c,d\right)\right\}\hfill \end{array}$

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

Sol:

$\begin{array}{l}\begin{array}{l}f\left(x\right) = 2x - 3\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Let f\left(x\right)= y = 2x – 3\hfill \end{array}\\ \Rightarrow y+3=2x\Rightarrow x=\frac{y\text{?}+3}{2}\\ \therefore f^{-1}(y)=\frac{y+3}{2}0r{f}^{-1}(x)=\frac{x+3}{2}\end{array}$

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

Sol:

$\begin{array}{l}f(x)=2x+1andg(x)=x^2-2\\ \therefore gof=g\left[f(x)\right]\\ ={\left[2x+1\right]}^2-2\\ =4x^2+4x+1-2\\ =4x^2+4x-1\\ Hence,gof=4x^2+4x-1.\end{array}$

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

Sol:

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

Sol: R = { (a, a), (b, c), (a, b)}.

To make R as reflexive we must add (b, b) and (c, c) to R. Also, to make R as transitive we must add

(a, c) to R.

So, minimum number of ordered pair is to be added are (b, b), (c, c), (a, c).

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

Sol:

$\begin{array}{l}(i):Giventhat\\ a\ast b=1+ab\forall a,b\in R\\ andb\ast a=1+ba\forall a,b\in R\\ a\ast b=b\ast a=1+ab\\ So,\ast iscommutative.\\ Nowa\ast (b\ast c)=(a\ast b)\ast c\forall a,b,c\in R\\ L.H.S.a\ast (b\ast c)=a\ast (1+bc)=1+a(1+bc)=1+a+abc\\ R.H.S.(a\ast b)\ast c=(1+ab)\ast c=1+(1+ab).c=1+c+abc\\ So,\ast isnotassociative.\\ Hence,\ast iscommutativebutnotassociative.\end{array}$

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

Sol:

$\begin{array}{l}(i)a\ast b=a-b\in Qa,b\in Q,\\ So,\ast isbinaryoperation.\\ a\ast b=a-bandb\ast a=b-a\forall a,b\in Q\\ a-b\ne b-a\\ So,\ast isnotcommutative.\\ (ii)a\ast b={{\displaystyle a}}^2+{{\displaystyle b}}^2\in Q,So,\ast isbinaryoperation.\\ a\ast b=b\ast a\\ \Rightarrow a^2+b^2=b^2+a^2\forall a,b\in Q\\ Whichistrue.So,\ast iscommutative.\\ (iii)a\ast b=a+ab\in Q,So,\ast isbinaryoperation.\\ a\ast b=a+abandb\ast a=b+ba\\ a+ab\ne b+ba\Rightarrow a\ast b\ne b\ast a\forall a,b\in Q.\\ So,\ast isnotcommutative.\\ (iv)a\ast b={{\displaystyle (a-b)}}^2\in Q,So,\ast isbinaryoperation.\\ a\ast b={{\displaystyle (a-b)}}^{2}andb\ast a={{\displaystyle (b-a)}}^2\\ a\ast b=b\ast a\Rightarrow {{\displaystyle (a-b)}}^2={{\displaystyle (b-a)}}^2\forall a,b\in Q.\\ So,\ast iscommutative.\end{array}$

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

Sol:

$\begin{array}{l}(i)fog\Rightarrow f\left[\left(g(x\right)\right]=[g]{(x)}^2+3\left[\left(g(x\right)\right]+1\\ ={(2x-3)}^2+3(2x-3)+1\\ =4x^2+9-12x+6x-9+1=4x^2-6x+1\\ (ii)gof\Rightarrow g\left[f(x)\right]=2\left[{{\displaystyle x}}^2+3x+1\right]-3\\ =2{{\displaystyle x}}^2+6x+2-3=2{{\displaystyle x}}^2+6x+1\\ (iii)fof\Rightarrow f\left[\left(f(x\right)\right]={{\displaystyle \left[f(x)\right]}}^2+3\left[f(x)\right]+1\\ =({{\displaystyle {{\displaystyle x}}^2+3x+1)}}^2+3({{\displaystyle x}}^2+3x+1)+1\\ ={{\displaystyle x}}^4+9{{\displaystyle x}}^2+1+6{{\displaystyle x}}^3+6x+2{{\displaystyle x}}^2+3{{\displaystyle x}}^2+9x+3+1\\ ={{\displaystyle x}}^4+6{{\displaystyle x}}^3+14{{\displaystyle x}}^2+15x+5\\ (iv)gog\Rightarrow g\left[g(x)\right]=2\left[g(x)\right]-3=2(2x-3)-3=4x-6-3=4x-9\end{array}$