Relations and Functions

26. Given, f(x)=x².

$\frac{f(1.1)-f(1)}{1.1-1}=\frac{{(1.1)}^2-1^2}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

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

25. Given, f(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 3\hfill \\ 3x,\hfill & 3\le x\le 10\hfill \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f  has one and only one image.

 ? f(x) is a function.

Given, g(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 2\hfill \\ 3x,\hfill & 2\le x\le 10\hfill \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the e

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

A = N × N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) 

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (

24. (i) f(x)=2 – 3x, x $\in$ R, x>0.

Given, x>0

3x>3 × 0

3x>0

(–1) × 3x<(–1) × 0.

–3x<0

2 – 3x<0+2

2 – 3x<2

i.e., f(x) < 2

Hene, range of f(x) = (– ?, 2)

(ii) Given, f(x) = x²+2, x is a real number.

Since, x is a real number,

x² ≥ 0 (x²=0 for x=0)

x²+2 ≥ 0+2

x²+2 ≥ 2

f(x) ≥ 2

?Range of f(x) = [2, ?) 

(iii) Given, f(x