Q. 134.9( 7 Votes )

Prove that

Answer :

f is invertible if it is one-one as well as onto.


f(x) = 9x2 + 6x – 5


f’(x) = 18x + 6 > 0 for all x > 0


So, f(x) is one-one.


Now, let y R, then for any x,


f(x) = y if y = 9x2 + 6x – 5


y = (3x)2 + 2(3x) (1) + (1)2 – 5 – 12


y = (3x+1)2 – 6






As x[0,) which means x is a positive real number.


So, x cannot be equal to


Now, fory = - 6


does not belongs to[0,)


Hence, f(x) is not onto


f(x) is not invertible


Now procedure to make it invertible.


Since, x ≥ 0, therefore





Redefining, f: [0, ) [ –5, ) makes f onto


Now, checking if f: [0, ) [ –5, ) is one – one as well as onto.


f(x) = 9x2 + 6x – 5


f’(x) = 18x + 6


f’(x) > 0 for all x [0, )


So, f(x) is one-one in [0, ) … (1)


As, f(x) is an increasing function.


As, x [0, )


So, minimum value of f(x) is at x = 0,


f(0) = 9(0) + 6(0) – 5


f (0) = -5


Also, as x , f(x) .


Also, f(x) is continuous for all x, so f(x) attains all values in


[–5, ).


range of f(x) = co-domain of f(x).


So, f is onto. … (2)


From (1) and (2)-


f(x) is bijective.


Hence f is invertible and f-1: [-5, ) [0, )



OR


Reflexive:


R is reflexive, as 1 + a.a = 1 + a2 > 0


Because square of a number is always positive and 1 added to a positive number is a positive number.


(a, a) R aR


Symmetric:


If (a, b) R then, 1 + ab> 0


1 + ba> 0 (b, a) R


Hence, R is symmetric.


Transitive:


Let a=-8, b=-1,


Since, 1 + ab = 1 + (–8) (–1) = 9> 0


(a, b) R


also,


(b, c) R


but,


Hence, R is not transitive.


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