Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective:

(i) f (x) = \( \dfrac{x}{2}\)

(ii) g (x) = |x|

(iii) h (x) = x2

Asked by Sakshi | 1 year ago |  68

1 Answer

Solution :-

(i) Given f: A → A, given by f (x) =\( \dfrac{ x}{2}\)

Now we have to show that the given function is one-one and on-to

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

\( \dfrac{ x}{2}= \dfrac{ y}{2}\)

x = y

So, f is one-one.

Surjection test:

Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)

f(x) = y

\( \dfrac{ x}{2}=y\)

x = 2y, which may not be in A.

For example, if y = 1, then

x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

(ii) Given g: A → A, given by g (x) = |x|

Now we have to show that the given function is one-one and on-to

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

g(x) = g(y)

|x| = |y|

x = ± y

So, f is not one-one.

Surjection test:

For y = -1, there is no value of x in A.

So, g is not onto.

So, g is not bijective.

(iii) Given h: A → A, given by h (x) = x2

Now we have to show that the given function is one-one and on-to

Injection test:

Let x and y be any two elements in the domain (A), such that h(x) = h(y).

h(x) = h(y)

x2 = y2

x = ±y

So, f is not one-one.

Surjection test:

For y = – 1, there is no value of x in A.

So, h is not onto.

So, h is not bijective.

Answered by Aaryan | 1 year ago

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