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Q.

$A = {x : - 1 \leq x \leq 1} . f : A \to A d e f i n e d b y f (x) = x | x | . T h e n f i s$

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An Intiative by Sri Chaitanya

a

A bijiection

b

An injection but not surjection

c

A surjection but not an injection

d

neither an injection nor a surjection

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = x | x | = {\begin{cases} - x^{2} - 1 \leq x \\ x^{2} 0 \leq x \leq 1 \end{cases} \\ d i f f e r e n t e \lim e n t s h a v e d i f f e r e n t Im a g e s a n d \\ R a n g e = C o d o m a i n \\ ∴ f i s o n e - o n e a n d o n t o \end{array}$

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A={x: −1≤x≤1}. f: A→A defined by f(x)=x|x|. Then f is