Ncert Solutions Maths class 12th

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

Sol:

$Given that,$
$\begin{array}{l}f = \left\{\left(1,2\right),\left(3,5\right),\left(4,1\right)\right\}\hfill \end{array}$

$and$
$\begin{array}{l}g = \left\{\left(2,3\right),\left(5,1\right),\left(1,3\right)\right\}\hfill \end{array}$

$\begin{array}{l}Now, fog \left(2\right) = f \left\{g\left(2\right)\right\} = f \left(3\right) = 5\hfill \end{array}$

$\begin{array}{l}fog \left(5\right) = f \left\{g\left(5\right)\right\} = f \left(1\right) = 2\hfill \end{array}$

$\begin{array}{l}fog \left(1\right) = f \left\{g\left(1\right)\right\} = f \left(3\right) = 5\hfill \end{array}$

$\begin{array}{l}fog = \left\{\left(2,5\right),\left(5,2\right),\left(1,5\right)\right\}\hfill \end{array}$

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

Sol:

$\begin{array}{l}\left(i\right)Given set of ordered pair is \left\{\left(x,y\right):xisaperson,yisthemotherofx\right\}.\hfill \end{array}$$\begin{array}{l}It represents a function.\hfill \end{array}$

$\begin{array}{l}Here, the image of distinct elements of x under f are not distinct, so it is not an injective but it is a surjective.\hfill \end{array}$

$\begin{array}{l}\left(ii\right)Set of ordered pairs = \left\{\left(a,b\right):aisaperson,bisanancestorofa\right\}\hfill \end{array}$

$\begin{array}{ll}Here, each element of domain does not have a unique image.\hfill & \hfill \end{array}$

$\begin{array}{l}So, it does not represent function.\hfill \end{array}$

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

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:

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