In each of the following cases, state whether the function is on

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 Multiple Choice QuestionsShort Answer Type

371.

Check the injectivity and surjectivity of the following functions:
f : N → N given by f(x) : x2 


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372.

Check the injectivity and surjectivity of the following functions.

f : Z → Z given by f(x) = x2

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373.

Show that the Signum Function f : R → R, given by

                    straight f left parenthesis straight x right parenthesis space open curly brackets table row cell 1 comma space if space straight x space 0 end cell row cell 0 comma space if space straight x space 0 end cell row cell negative 1 comma space if space straight x space 0 end cell end table close curly brackets

is neither one-one nor onto

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374.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

f : R → R defined by f(x) = 3 – 4x

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375.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

f : R → R defined by f(x) = 1 + x2


f : R → R defined by

f(x) = 1 + x2

Let  straight x subscript 1 comma space straight x subscript 2 space element of space straight R space such that straight f left parenthesis straight x subscript 1 right parenthesis equals straight f left parenthesis straight x subscript 2 right parenthesis


rightwards double arrow space space space 1 plus straight x subscript 1 superscript 2 space equals space 1 plus straight x subscript 2 superscript 2
rightwards double arrow space space space straight x subscript 1 superscript 2 space equals space straight x subscript 2 superscript 2
rightwards double arrow space space space straight x subscript 1 equals plus-or-minus space straight x subscript 2

therefore space space space straight f left parenthesis straight x subscript 1 right parenthesis space equals space straight f left parenthesis straight x subscript 2 right parenthesis does not imply that x1 = x2 for instance
f(1) = (f - 1) = 2

therefore  f is not one-one

Consider an element of  - 2 in co domain R.
It is seen that f(x) = 1 + x2 is positive for all x element of straight R
Thus there does not exists any x in domain R such that  f(x) = - 2
therefore    f is not onto
Hence, f is neither one-one nor onto.
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376. Let A = {1. 2. 3}. Then show that the number of relations containing (1,2) and (2. 3) which are reflexive and transitive but not symmetric is four.
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377. Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y) : {x, y} ⊂ {1,4, 7 }} or ⊂ {2, 5, 8} or {x, y,}⊂ {3, 6, 9}}. Show that R1 = R2.
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378.

Consider the binary operation A on the set {1, 2, 3, 4, 5} defined by a ∧ b = min. {a, b}. Write the operation table of the operation ∧.

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379. Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as

straight a space asterisk times space straight b space equals space open curly brackets table attributes columnalign left end attributes row cell straight a plus straight b space space space space space space space space space space space space space space space space if space straight a plus straight b less than 6 end cell row cell straight a plus straight b minus 6 space space space space space space space space space space space space if space straight a plus straight b greater or equal than 6 end cell end table close

Show that zero is the identity for this operation and each element a of the set is invertible with 6 – a being the inverse of a.
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380.

Let A = Q x Q. Let be a binary operation on A defined by (a, b) * (c, d) = (ac, ad + b).

Then

(i) Find the identify element of (A, *)
(ii) Find the invertible elements of (A, *)

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