A) One-one into
B) One-one onto
C) Many one into
D) Many one onto
Correct Answer: C
Solution :
Let \[L{f}'\,(2)\ne R{f}'\,(2)\] Þ \[\frac{{{x}^{2}}-4}{{{x}^{2}}+4}=\frac{{{y}^{2}}-4}{{{y}^{2}}+4}\] Þ \[\frac{{{x}^{2}}-4}{{{x}^{2}}+4}-1=\frac{{{y}^{2}}-4}{{{y}^{2}}+4}-1\,\,\,\Rightarrow \,\,{{x}^{2}}+4={{y}^{2}}+4\] Þ \[x=\pm y\], \\[f(x)\] is many-one. Now for each \[y\in (-1,\,1),\] there does not exist \[x\in X\] such that\[f(x)=y\]. Hence f is into.
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