A) \[2x-5\]
B) \[|x|\]
C) \[{{x}^{2}}\]
D) \[{{x}^{2}}+1\]
Correct Answer: A
Solution :
\[|x|\] is not one-one; \[\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{4x}{(\sqrt{{{x}^{2}}+8x+3}+\sqrt{{{x}^{2}}+4x+3}}\] is not one-one; \[{{x}^{2}}+1\] is not one-one. But \[2x-5\] is one-one because \[f(x)=f(y)\Rightarrow 2x-5=2y-5\Rightarrow x=y\] Now \[f(x)=2x-5\] is onto. \\[f(x)=2x-5\] is bijective.
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