Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{A}=\{\mathrm{x} \in \mathbb{R}:-1 \leq \mathrm{x} \leq 1\} \& \mathrm{f}: \mathrm{A} \rightarrow \mathrm{A}$ be a mapping defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}| .$ Then $\mathrm{f}$ is
Options:
Solution:
1831 Upvotes
Verified Answer
The correct answer is:
bijective
Hint:
$f(x)=x|x|=\left\{\begin{array}{c}-x^{2} x \in[-1,0) \\ x^{2} x \in[0,1]\end{array}\right.$
$f(x)=x|x|=\left\{\begin{array}{c}-x^{2} x \in[-1,0) \\ x^{2} x \in[0,1]\end{array}\right.$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.