A) \[\frac{2\sin x}{\sqrt{\sin 2x}}\]
B) \[\frac{2\cos x}{\sqrt{\cos 2x}}\]
C) \[\frac{2\cos x}{\sqrt{\sin 2x}}\]
D) \[\frac{2\sin x}{\sqrt{\cos 2x}}\]
Correct Answer: B
Solution :
Let the roots of the equation\[{{x}^{2}}-x-k=0\]are \[\alpha \]and\[{{\alpha }^{2}}\]. Then, \[\alpha +{{\alpha }^{2}}=1\] ..(i) and \[\alpha .{{\alpha }^{2}}={{\alpha }^{3}}=-k\] \[\Rightarrow \] \[\alpha ={{(-k)}^{1/3}}\] On putting this value of a in Eq. (i), we get \[{{(-k)}^{1/3}}+{{(-k)}^{2/3}}=1\] ?(ii) On cubing both sides, we get \[(-k)+{{(-k)}^{2}}+3k[{{(-k)}^{1/3}}+{{(-k)}^{2/3}}]=1\] \[\Rightarrow \] \[-k+{{k}^{2}}-3k({{k}^{2/3}}-{{k}^{1/3}})=1\] \[\Rightarrow \] \[{{k}^{2}}-k-3k(1)=1\] [Using Eq.(ii)] \[\Rightarrow \] \[{{k}^{2}}-4k-1=0\] \[\Rightarrow \] \[lk=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5}\]
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