Relations and Functions

Total number of possible relation = $2^{n^2}=2^4=16$  

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability =  $\frac{5}{16}$  

Kindly consider then following figure

$l={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dx+}{\displaystyle \underset{-1/2}{\overset{1}{\int }}\left|2x\right|dx}$

let  $l_1={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dxput2x=t\Rightarrow dx=\frac{dt}{2}}$

$\therefore l=l_1+l_2=0+\frac{5}{8}=\frac{5}{8}\Rightarrow 8l=5$

36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

f(11)=11 [? prime factor of 11 = 11]

f(12)=3 [? prime factor of 12 = 2, 3]

f(13)=13 [? prime factor of 13 = 13]

?Range of f=set of all image of f(x) = {3,5,11,13}.

35. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 * 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b $\in$ z

So, ab=(–1) * (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1, –2).

?The element 1 has two image i.e., 2 and –2.

Hence, f is not a function.

34. Given,

A={1,2,3,4}

B={1,5,9,11,15,16}

f={(1,5),(2,9),(3,1),(4,5),(2,11)}.

(i) As every element of f is an element of A * B

We can clearly say that f $\subset$ A * B.

?f is a relation from A to B.

(ii) As the element 2 of the domain has two image i.e., 9 and 11. f is not a function.

33. Given, R= { (a, b): a, b $\in$ N and a = b²}

(i) Let a = 2 $\in$ N

Then b = 2² = 4 $\in$ N

but a ≠ b.

Hence the given statement is not true.

(ii) For a=b² the inverse b=a² may not hold true

Example (4,2) $\in$ R, a=4, b=2 and a=b²

but (2,4) $\notin$ R.

Hence, the given statement is not true.

(iii) If (a, b) $\in$ R

a=b²…… (1)

and (b, c) $\in$ R

b=c²……. (2)

so for (1) and (2),

a= (c²)²=c⁴.

is, a ≠c²,

Hence, (a, c) $\notin$ R.

? The given statement is false.

32. Given, f(x) = (ax + b)

= {(1,1),(2,3),(0, – 1),(–1, –3)} .

As (1,1) $\in$ f.

Then, f(1)=1   [? f(x) = y for (x, y)]

a * 1+b=1

a+b=1…… (1)

and (0, – 1) $\in$ f .

Then, f(0)= –1

a* 0+b= –1

b= –1…….(2)

Putting value of (2) in (1) we gets

a – 1=1

a=1+1

a=2

So, (a, b)=(2, –1)

31. Given, f(x) = x+1. and g(x) = 2x – 3.

So, (f +g)(x) = f(x)+g(x) = (x+1)+(2x – 3) = x+1+2x – 3 = 3x – 2

(f – g)(x) = f(x) –g(x) = (x+1)–(2x–3) = x+1 – 2x+3 = 4 – x

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3} such that x\ne \frac{3}{2}$