A) \[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}-1}\]
B) \[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}-1}\]
C) \[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}+1}\]
D) \[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\]
Correct Answer: D
Solution :
\[\mu +i\lambda =\frac{{{(p+i)}^{2}}}{2p-i}=\frac{({{p}^{2}}-1+2pi)(2p+i)}{(2p-i)(2p+i)}\] \[=\frac{2p({{p}^{2}}-2)+i(5{{p}^{2}}-1)}{4{{p}^{2}}+1}\] \ \[{{\mu }^{2}}+{{\lambda }^{2}}=\frac{4{{p}^{2}}{{({{p}^{2}}-2)}^{2}}+{{(5{{p}^{2}}-1)}^{2}}}{{{(4{{p}^{2}}+1)}^{2}}}\] \[=\frac{4{{p}^{6}}+6{{p}^{2}}+9{{p}^{4}}+1}{{{(4{{p}^{2}}+1)}^{2}}}\] \[=\,\,\frac{{{p}^{4}}(4{{p}^{2}}+1)+2{{p}^{2}}(4{{p}^{2}}+1)+(4{{p}^{2}}+1)}{{{(4{{p}^{2}}+1)}^{2}}}\] \[=\frac{{{p}^{4}}+2{{p}^{2}}+1}{4{{p}^{2}}+1}=\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\].You need to login to perform this action.
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