Ncert Solutions Maths class 12th

Since every set is a subset of itself, ARA for all A ∈ P (X).

∴R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒

Define $f:{\rm N}\to {\rm N}$ by

$f\left(x\right)=x+1$

And,  $g:{\rm N}\to {\rm N}$ by,

$g\left(x\right)\left\{\begin{array}{cc}\hfill x-1\hfill & \hfill if,x>1\hfill \\ \hfill 1\hfill & \hfill f,x=1\hfill \end{array}\right\}$

We first show that g is not onto.

For this, consider element 1 in co-domain N. it is clear that this element is not an image of any of the elements in domain.

$\therefore f$ is not onto.

Now,  $gof:{\rm N}\to {\rm N}$ is defined by,

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(x+1\right)=\left(x+1\right)-1\left[x,in{\rm N}=>\left(x+1\right)>1\right]\\ =x\end{array}$

Then, it is clear that for $y\in {\rm N}$ , there exists

Define $f:N\to Z$ as $f\left(x\right)$ and $g:Z\to Z$ as $g\left(x\right)=\left|x\right|$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l}g\left(-1\right)=\left|-1\right|=1\\ g\left(1\right)=\left|1\right|=1\\ \therefore g\left(-1\right)=g\left(1\right),but,-1\ne 1\end{array}$

$\therefore g$ is not injective.

Now, $gof:N\to Z$ is defined as

$gof\left(x\right)=y\left(f\left(x\right)\right)=y\left(x\right)=\left|x\right|$

Let $x,y\in N$ such that $gof\left(x\right)-gof\left(y\right)$

$\Rightarrow \left|x\right|=\left|y\right|$

Since $x,y\in N$ , both are positive.

$\therefore \left|x\right|=\left|y\right|\Rightarrow x=y$

Hence, gof is injective

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³  … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

It is given that $f:R\to \left\{x\in R:-1<x<1\right\}$ is defined as $f\left(x\right)=\frac{x}{1+\left|x\right|},x\in R.$

Suppose $f\left(x\right)=f\left(y\right)$ , where $x,y\in R.$

$\Rightarrow \frac{x}{1+x}=\frac{y}{1-y}\Rightarrow 2xy=x-y$

Since x is positive and y is negative:

$x>y\Rightarrow x-y>0$

But, 2xy is negative.

Then, $2xy\ne x-y$ .

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can als

It is given that $f:R\to R$ is defined as $f\left(x\right)=x^2-3x+2.$

$\begin{array}{l}f\left(f\left(x\right)\right)=f\left(x^2-3x+2\right)\\ ={\left(x^2+3x+2\right)}^2-3\left(x^2-3x+2\right)+2\\ =x^4+9x^2+4-6x^2-12x+4x^2-3x^2+9x-6+2\\ =x^4-6x^2+10x^2-3x\end{array}$

It is given that:

$f:W\to W$ is defined as $f\left(n\right)=\{\begin{array}{l}n-1,if.n,odd\\ n+1,if.n,even\end{array}$

One-one:

Let, $f\left(n\right)=f\left(m\right).$

It can be observed that if n is odd and m is even, then we will have n-1=m+1.

$\Rightarrow n-m=2$

However, the possibility of n being even and m being odd can also be ignored under a similar argument.

$\therefore$ Both n and m must be either odd or even.

Now, if both n

It is given that $f:R\to R$ is defined as $f\left(x\right)=10x+7.$

One-one:

$\begin{array}{l}Let,f\left(x\right)=f\left(y\right),where,x,y\in R\\ \Rightarrow 10x+7=10y+7\\ \Rightarrow x=y\end{array}$

$\therefore f$ is a one-one function.

Onto:

$\begin{array}{l}For,y\in R,let,y=10x+7\\ \Rightarrow x=\frac{y-7}{10}\in R\end{array}$

Therefore, for any $y\in R$ ,there exists $x=\frac{y-7}{10}\in R$

Such that

$f\left(x\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$

$\therefore f$ is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define $g:R\to R$ as $g\left(y\right)=\frac{y-7}{10}$

Now, we have

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(10x+7\right)=\frac{\left(10x+7\right)-7}{10}=10\frac{x}{10}=10\\ And\\ fog\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y\end{array}$

Hence, the required function

On N, the operation * is defined as a * b = a³ + b³.

For, a, b, ∈ N, we have:

a * b =  a³ + b³ = b³ + a³  = b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

  (1*2)*3 = (1³+2³)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

A

(i) Define an operation * on N as:

a * b = a + b  a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

The