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Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(a) Equivalence

(b) Reflexive and symmetric

(c) Transitive and symmetric

(d) Reflexive, transitive but not symmetric

Answer is (d)

Since n divides n, for all n in N, R is reflexive. R is not symmetric since for 3, 6 in N, 3 R 6 $\neq$ 6 R 3. R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n
divides m and m divides r, then n will devide r

Let P ={1,2,4,5}. The total number of distincts relations that can be defined on P is

(a) 16

(b) $2^{16}$

(c) 32

(d) 256

(b)

This is by formula

The function $f:[0,\infty)$ -> R given by $f(x)= \frac {x}{x+1}$ is

(a) one -one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) Neither one-one nor onto

Answer is (b)

$f(x)= \frac {x}{x+1}$

$\frac {df(x)}{dx}= \frac {1}{(x+1)^2}$

This is positive, So f(x) is a increasing function and it is one-one function

for onto

$y=\frac {x}{x+1}$

$x = \frac {y}{1-y}$

So y cannot be 1,So Range is subset of codomain

Hence not onto function

Let f : R -> R be defined by $f(x) = 2x - 1$. Then $f^{–1}(x)$ is given by

(a) $\frac {x-2}{2}$

(b) $\frac {x+1}{2}$

(c) $\frac {x-1}{2}$

(d) $\frac {2x-2}{3}$

Answer is (b)

Let y=2x- 1

$y+1=2x$

$x= \frac {y+1}{2}$

Therefore

$f^{–1}(x)= \frac {x+1}{2}$

The function f:R-> R given by $f(x)= \frac {x^2 -8}{x^2+2}$ is

(a) one -one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) Neither one-one nor onto

Answer is (d)

for one-one

$f(x_1)=f(x_2)$

$\frac {x_1^2 -8}{x_1^2+2}=\frac {x_2^2 -8}{x_2^2+2}$

$10x_1^2= 10x_2^2$

$x_1= \pm x_2$

Hence not a one-one function

For onto

$y=\frac {x^2 -8}{x^2+2}$

Solving this for x

$x= \sqrt {\frac {2y+8}{1-y}}$

So, y cannot assume 1

So not onto

let A={1,2,3} Which of the following relation defined on A is a reflexive relation?

(a) {(1, 1), (1, 2), (2, 1), (2, 2)}

(b) {(1, 2), (2, 1)}

(c) {(1, 1), (2, 2), (3, 3)}

(d) {(1, 2), (2, 3), (3, 1)}

Answer: (c)

{(1, 1), (2, 2), (3, 3)}

Which of the following is an onto function f:R -> R?

(a) $f(x) = x^2$

(b) $f(x) = 2x + 1$

(c) $f(x) = |x|$

(d) $f(x) = x^2 + 1$

Answer: (b)

f(x) = 2x + 1

For real numbers x and y, define xRy if and only if $x – y + \sqrt 2$ is an irrational number. Then the relation R is

(a) symmetric

(b) Reflexive

(c) Transitive

(d) None of these

Answer is (b)

$f(x) =x^2 +1$ and $g(x) = sin x$, then the value of $g \circ f$ is

(a) $sin^2x + 1$

(b) $sin (x^2 +1)$

(c) $x sin x + 1$

(d) None of these

$g \circ f= g(f(x)) = g(x^2 +1) =sin (x^2 +1)$

let f(x)=[x] and g(x) =|x|, the the value of $(g \circ f) (\frac {-5}{3}) - (f \circ g)(\frac {-5}{3}) $ is

(a) 1/2

(b) 1

(c) -1

(d) -2

$(g \circ f) (\frac {-5}{3}) - (f \circ g)(\frac {-5}{3})= g[f(\frac {-5}{3}] - f[g(\frac {-5}{3})]=g(-2) -f(\frac {5}{3}) =2 -1=1$

(i) Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = _____

(ii) if $f(x)=x^2 +2$ and $g(x)= 1 - \frac{1}{1-x}$, the $f \circ g$ is _____

(iii) The inverse of the function $f(x)= \frac {x}{x+2}$ is _______

(i) {(3,8) , (6,6) , ((9,4), (12,2)}

(ii) $\frac {3x^2 -4x+2}{(1-x)^2}$

(iii) $\frac {2x}{1-x}$

Let f : W -> W be defined as

$f(n)=\begin{cases} & \text{ n-1 if } n \ is \ odd \\ & \text{ n+1 if } n \ is \ even \end{cases}$

Then show that f is invertible.Also, find the inverse of f.

Here,

f : W -> W

is such that, if n is odd,

$f \circ f (n)= f (f (n)) =f (n -1)= n-1 +1= n $

and if n is even,

$f \circ f (n)= f (f (n)) =f (n +1)= n+ 1 =1= n $

Hence

$f \circ f =I$
This implies that f is invertible and $f^{-1} = f$

Show that f : [– 1, 1] -> R , given by $f(x) =\frac {x}{x+2}$ is one-one.

Find the inverse of the function f : [– 1, 1] -> Range of f

For one-one

$f(x_1) = f(x_2)$

$\frac {x_1}{x_1+2}=\frac {x_2}{x_2+2}$

$x_1=x_2$

Hence it is one-one function

For onto

let $y= \frac {x}{x+2}$

or

$x=\frac {2y}{1-y}$

From this we see that,y cannot take the value 1

So it is not onto on R

But we can make it onto function, by defining the function as

f : [– 1, 1] -> Range of f , given by $f(x) =\frac {x}{x+2}$

Also

$f(x) = f(\frac {2y}{1-y})= y$

Now Since this is both one-one and onto, we can have the inverse and inverse will be

$f^{-1} (x) = \frac {2x}{1-x}$

Show that the relation R in the set N X N defined by (a,b)R(c,d) if $a^2+d^2 =b^2+ c^2$ for all a,b,c,d in N is an equivalence relation

Let (a,b) in N x N

Now as

$a^2 + b^2 = b^2 + a^2$

Thre (a,b)R(a,b)

Hence, R is reflexive.

Let (a, b), (c, d) in N x N be such that (a, b)R(c, d)

$a^2+d^2 =b^2+ c^2$

$c^2 + b^2=a^2+d^2$

So ,(c, d)R(a, b)

Hence, R is symmetric.

Let (a, b), (c, d), (e, f ) N N be such that (a, b)R(c, d), (c, d)R(e, f ).

$a^2+d^2 =b^2+ c^2$

$c^2 + f^2=d^2 +e^2$

Adding both

$a^2 + f^2= e^2 + b^2$

(a,b) R (e,f)

Here , R is transitive also

Functions f , g : R -> R are defined, respectively, by $f(x) = x^2 + 3x + 1$,g(x) = 2x – 3, find

(i) $f \circ g$

(ii) $g \circ f$

(iii)$f \circ f$

(iv) $g \circ g$

(i) $f \circ g = f(g(x)) = f(2x-3) = (2x-3)^2 + 3(2x-3) +1=4x^2 -6x +1$

(ii)$g \circ f = g(f(x))= g(x^2 + 3x + 1) = 2(x^2 + 3x +1) -3 = 2x^2 + 6x-1$

(iii) $f \circ f = f(f(x)) = f(x^2 + 3x +1) = (x^2 + 3x+1)^2 + 3(x^2 + 3x+1) + 1=x^4 + 6x^3 + 14x^2 + 15x + 5$

(iv) $g \circ g= g(g(x) = g(2x-3) = 2(2x-3) -3 = 4x-9$

Show that the relation R in the set A X A defined by (a,b)R(c,d) if $a+d =b+c$ for all a,b,c,d in A is an equivalence relation. Here A={1,2,3...10}

Let (a,b) in A x A

Now as

$a + b = b + a$

Thre (a,b)R(a,b)

Hence, R is reflexive.

Let (a, b), (c, d) in N x N be such that (a, b)R(c, d)

$a+d =b+ c$

$c + b=a+d$

So ,(c, d)R(a, b)

Hence, R is symmetric.

Let (a, b), (c, d), (e, f ) N N be such that (a, b)R(c, d), (c, d)R(e, f ).

$a+d =b + c$

$c+ f=d +e$

Adding both

$a + f= e + b$

(a,b) R (e,f)

Here , R is transitive also

If $f(x) =\frac {4x+3}{6x-4}$ and $x \ne \frac {-2}{3}$ , show that fof(x) = x for all x $x \ne \frac {-2}{3}$ . Also, find the inverse of f

$f(x) =\frac {4x+3}{6x-4}$

Now

$f \circ f (x) = f(f(x)) = f(\frac {4x+3}{6x-4}) = \frac { 4\frac {4x+3}{6x-4} + 3}{6\frac {4x+3}{6x-4} - 4} = \frac { 16x + 12 +18x -12}{24x + 18 -24x+ 16} = x$

For inverse

$y=\frac {4x+3}{6x-4}$

Solving

$x= \frac {4y+3}{6y-4}$

So inverse is $\frac {4x+3}{6x-4}$

Let A= R -{-1} and * be the binary operation on A defined by

a*b = a+b+ab

(i) Show that * is commutative and Associative

(ii) Find the identity element for * on A

(iii) Prove that every element of A is invertible

(i)Commutative:

a∗b=a+b+ab

b*a =b+a +ab=a +b +ab

So , ∗ is commutative.

Associative:

a∗(b *c)= a *(b + c +bc)

=a + b + c +bc + a(b + c +bc) = a+b+c +ab + ab+ac + bc +abc

(a∗b) *c)=(a +b + ab) * c

=a +b + ab + c + (a +b + ab) c= a+b+c +ab + ab+ac + bc +abc

Hence ∗ is associative.

(ii) Identity element:

Let (e) be the identity element

then by definition

a* e = a = e*a

a+ e + ae=a=e+a + ae
So e+ ae=0
e(1+a)=0
Hence e=0

So (0,0) is the identity element

(iii)Inverse:

Let i is inverse of element a ∈ A

Now by definition.

a* i= 0 =i* a

a+i + ai=0 = i+ a+ia

So, $i = \frac {a}{1+a}$

So we have inverse for all elements

Prove that the function f:[0, $\infty$) -> R given by $f(x) = 9x^2+ 6x – 5$ is not invertible. Modify the codomain of the function f to make it invertible, and hence find $f^{–1}$

$f(x) = 9x^2+ 6x – 5$

for one-one

$f(x_1) = f(x_2)$

$9x_1^2 + 6x_1 -5= 9x_2^2 + 6x_2 -5$

$9(x_1^2 - x_2^2) + 6(x_1 - x_2) =0$

$9(x_1 -x_2)(x_1 + x_2) + 6(x_1 - x_2) =0$

$((x_1 - x_2)[9(x_1 + x_2) +6]=0$

either $x_1=x_2$

or $9(x_1 + x_2) + 6=0$

Since $x \geq 0$, this cannot be true.

so $x_1=x_2$

This is one-one function

For onto

$f(x) =y= 9x^2+ 6x – 5$

$ 9x^2+ 6x – 5 -y=0$

Solving the quadratic equation

$x=\frac {-1 \pm \sqrt {6+y}}{3}$

Since $x \geq 0$

$x=\frac {-1 + \sqrt {6+y}}{3}$

So y cannot have value lessor than -6. So it is not onto function

Now $6 + y \geq $ or $y \geq -6$,

Also $\sqrt {6+y} \geq 1$ as $x \geq 0$

so, $6+y \geq 1$ or $y \geq -5$

So, Range of the function will be (-5,$\infty$) .

So we can define the function as

function f:[0, $\infty$) -> (-5,$\infty$) given by $f(x) = 9x^2+ 6x – 5$

Now for any value of y in (-5,$\infty$) , there exists an value $\frac {-1 + \sqrt {6+y}}{3}$ in [0, $\infty$) such that

$f(x) = 9(\frac {-1 + \sqrt {6+y}}{3})^2 + 6(\frac {-1 + \sqrt {6+y}}{3}) -5= y$

So this is both one-one and onto .hence invertible

Inverse is given by

$f^{–1} (x) =\frac {-1 + \sqrt {6+x}}{3}$

Check whether the relation R in the set R of real numbers, defined by R = {(a, b) : 1 + ab > 0}, is reflexive, symmetric or transitive

Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides (a – b)} is an equivalence relation

Prove that the relation R in the set A = {1, 2, 3, 4, 5, 6, 7} given by R = {(a, b) : |a – b| is even} is an equivalence relation

Show by examples that the relation R in , defined by

R = {(a, b) : $a \le b^3$} is neither reflexive nor transitive.

Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y) R (u, v) if and only if xv = yu. Show that R is an equivalence relation

Clearly, (x, y) R (x, y), ∀ (x, y) A, since xy = yx. This shows that R is reflexive.

Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y). This shows that R is symmetric.

Similarly, (x, y) R (u, v) , xv = yu or v/u=y/x,

(u, v) R (a, b), ub = va or va=ub or v/u=b/a

So y/x=b/a or ay=bx or xb=ya

so (x,y) R (a, b)

This shows that R is transitive

Therefore R is an equivalence relation

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