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
(b)
This is by formula
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
Answer is (b)
Let y=2x- 1
$y+1=2x$
$x= \frac {y+1}{2}$
Therefore
$f^{–1}(x)= \frac {x+1}{2}$
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
Answer: (c)
{(1, 1), (2, 2), (3, 3)}
Answer: (b)
f(x) = 2x + 1
Answer is (b)
$g \circ f= g(f(x)) = g(x^2 +1) =sin (x^2 +1)$
$(g \circ f) (\frac {-5}{3}) - (f \circ g)(\frac {-5}{3})= g[f(\frac {-5}{3}] - f[g(\frac {-5}{3})]=g(-2) -f(\frac {5}{3}) =2 -1=1$
(i) {(3,8) , (6,6) , ((9,4), (12,2)}
(ii) $\frac {3x^2 -4x+2}{(1-x)^2}$
(iii) $\frac {2x}{1-x}$
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 $f^{-1} = f$
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
$f^{-1} (x) = \frac {2x}{1-x}$
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
(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) $g \circ g= g(g(x) = g(2x-3) = 2(2x-3) -3 = 4x-9$
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
$f(x) =\frac {4x+3}{6x-4}$
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 $\frac {4x+3}{6x-4}$
(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
$f(x) = 9x^2+ 6x – 5$
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
$f^{–1} (x) =\frac {-1 + \sqrt {6+x}}{3}$
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