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  • Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective: (i) f(x) = x/2 (ii) g(x) = |x|

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:

(i) f(x) = x/2 (ii) g(x) = \vert x \vert \\

(iii) h(x) = x \vert x \vert  (iv) k(x) = x\textsuperscript{2}\\
 

Answers (1)

Here, A = [-1, 1]

(i) f: [-1, 1] \rightarrow [-1, 1], f (x) = x/2\\

If, f(x\textsubscript{1}) = f(x\textsubscript{2})\\

x\textsubscript{1}/2 = x\textsubscript{2}/2\\

So, f(x) is one-one.
Also, x \in [-1, 1]\\

x/2 = f(x) = [-1/2, 1/2]\\

Hence, the range is a subset of co-domain A

So, f(x) is not onto.

Therefore, f (x) is not bijective.

(ii)

 \\g(x) = \vert x \vert \\ \text{Let} g(x\textsubscript{1}) = g (x\textsubscript{2})\\ \vert x\textsubscript{1} \vert = \vert x\textsubscript{2} \vert \\ x\textsubscript{1}= \pm x\textsubscript{2}\\

So, g(x) is not one-one.
Also, g(x) = \vert x \vert \geq 0, \text{for all real x}\\

Hence, the range is [0, 1], which is subset of co-domain ‘A’

So, f(x) is not onto.

Therefore, f(x) is not bijective.

(iii)

\\\\h(x) = x \vert x \vert \\ \text{Let } h(x\textsubscript{1}) = h(x\textsubscript{2})\\ x\textsubscript{1} \vert x\textsubscript{1} \vert = x\textsubscript{2} \vert x\textsubscript{2} \vert \\ \text{If }x\textsubscript{1}, x\textsubscript{2}> 0\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1}\textsuperscript{2}- x\textsubscript{2}\textsuperscript{2}= 0\\ (x\textsubscript{1} - x\textsubscript{2})(x\textsubscript{1} + x\textsubscript{2}) = 0\\ x\textsubscript{1 }= x\textsubscript{2 }(as x\textsubscript{1}+ x\textsubscript{2} \neq 0)\\ \text{Again }, x\textsubscript{1}, x\textsubscript{2 }< 0, and x\textsubscript{1}= x\textsubscript{2}\\ \text{Therefore }, x\textsubscript{1 }and x\textsubscript{2 }\text{of opposite sign}, x\textsubscript{1 } \neq x\textsubscript{2 }.\\

Hence, f(x) is one-one.

\\\text{For x} \in [0, 1], f(x) = x\textsubscript{2} \in [0, 1]\\ \text{For x} < 0, f(x) = -x\textsubscript{2} \in [-1, 0)\\

Hence, the range is [-1, 1].

So, h(x) is onto.

Therefore, h(x) is bijective.

(iv)

\\k(x) = x^{2}\\ \text{Let }k (x\textsubscript{1}) = k (x\textsubscript{2})\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1 }= \pm x\textsubscript{2}\\

Therefore, k(x) is not one-one.

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