A) \[f\]
B) \[{{f}^{-1}}\]
C) \[f(x)=\left\{ \begin{matrix} 2x & x<0 \\ 2x+1 & x\ge 0 \\ \end{matrix}, \right.\]
D) None of the above
Correct Answer: A
Solution :
Let a = first AM between 1 and 5 \[{{x}^{2}}+{{y}^{2}}-8x-6y=0\] \[2{{x}^{2}}+2{{y}^{2}}-x-7y=0\] b = second GM between 1 and 5 \[{{x}^{2}}+{{y}^{2}}-6x-10y=0\] \[{{\left( px+\frac{1}{x} \right)}^{n}}\] and c = third HM between 1 and 5 \[n\in N\] \[\frac{5}{2}\] Now, \[a+b+c=2+\sqrt{5}+\frac{5}{2}\] \[ab+bc+ca=2\sqrt{5}+\frac{5\sqrt{5}}{2}+5\] And \[abc=5\sqrt{5}\] The cubic equation having a, b, c as its roots, is \[{{x}^{3}}-(a+b+c){{x}^{2}}+(ab+bc+ca)x-abc=0\] \[\Rightarrow \]\[{{x}^{3}}-\left( \frac{9}{2}+\sqrt{5} \right){{x}^{2}}+\left( \frac{9\sqrt{5}}{2}+5 \right)x-5\sqrt{5}=0\]
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