Relations and Functions

30. Given, f (x)= $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$

We know that, for x $\in$ R.

So,  x²≥ 0 ⇒ $\frac{x^2}{x^2+1}\ge \frac{0}{x^2+1}\Rightarrow \frac{x^2}{x^2+1}\ge 0\Rightarrow f(x)\ge 0$

and x²+1>x²

⇒ $1>\frac{x^2}{x^2+1}$

⇒1 > f (x).

So, 0 ≤ f (x) < 1

∴ Range of f (x) = [0,1).

29. Given, f(x)=|x – 1|.

The given function is defined for all real number x.

Hence, domain of f(x)=R.

As f(x)=|x – 1|, x $\in$ R is a non-negative no.

Range of f(x)=[0, ?), if positive real numbers.

28. Given, f (x)=

The given fxⁿ is valid for all x such that x – 1 ≥ 0 ⇒x≥ 1

∴ Domain of f (x)= [1,∞)

As x ≥ 1

⇒ x – 1 ≥ 1 – 1

⇒ x – 1 ≥ 0

⇒ ≥ 0

⇒ f (x) ≥ 0

So, range of f (x)= [0,∞ )

27. Given, f (x)= $\frac{x^2+2x+1}{x^2-8x+12}$

The given function is valid if denominator is not zero.

So, if x² – 8x+12=0.

⇒ x² – 2x – 6x+12=0

⇒ x (x – 2) –6 (x – 2)=0

⇒ (x – 2) (x – 6)=0

⇒ x=2 and x=6.

So,  f (x) will be valid for all real number x except x=2,6.

∴ Domain of f (x)=R – {2,6}

26. Given, f(x)=x².

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

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 element 2 of the domain has more than one image i.e., 4 and 6.  

? g(x) is not a function.

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) = x, x is a real number.

As, f(x) = x, the range of f(x) is also real.

i.e., Range of f(x) = R.

22. Give, f (x) = 2x – 5.

(i) f (0)= (2 * 0) –5=0 – 5= –5

(ii) f (7)= (2 * 7) –5=14 – 5=9

(iii) f (–3)=2 * (–3) –5= –6 – 5= –11.