| Direction: For the following questions. Choose the correct answer to the codes [a], [b], [c] and [d] defined as follows. |
| Statement I If \[\sin \,x+{{\sin }^{2}}x=1,\] then value of the expression \[{{\cos }^{12}}\,x+3\,{{\cos }^{10}}x+3\,{{\cos }^{8}}x+{{\cos }^{6}}x-1\] |
| Statement II \[{{\cos }^{2}}\,x=\frac{\sqrt{5}-1}{2}\]. |
A) Statement I is true. Statement II is also true and Statement II is the correct explanation of Statement I.
B) Statement I is true/Statement II is also true and Statement II is not the correct explanation of Statement I.
C) Statement I is true. Statement II is false.
D) Statement I is false, Statement II is true.
Correct Answer: D
Solution :
I. Given, \[{{\sin }^{2}}x+\sin \,x=1\] \[\Rightarrow \] \[\sin \,x=1-{{\sin }^{2}}x={{\cos }^{2}}x\] Now, the given expression is \[{{\cos }^{12}}x+3\,{{\cos }^{10}}x+3\,{{\cos }^{8}}\,x+{{\cos }^{6}}x-1\] \[={{\cos }^{6}}x\,[{{\cos }^{6}}x+3\,{{\cos }^{4}}x+3\,{{\cos }^{2}}x+1]-1\] \[={{\cos }^{6}}\,x{{({{\cos }^{2}}x+1)}^{3}}-1\] \[=\,{{\sin }^{3}}x\,{{(\sin \,x+1)}^{3}}-1\] [from Eq. (i)] \[={{({{\sin }^{2}}x+\sin x)}^{3}}-1\] \[={{(1)}^{3}}-1=1-1=0\] II. \[{{\sin }^{2}}x+\sin \,x-1=0\] Now, \[\sin \,x=\frac{-1\pm \,\sqrt{1+4}}{2}\] \[=-\frac{-1\pm \,\sqrt{5}}{2}=\frac{-1+\sqrt{5}}{2}\] \[\therefore \] \[{{\cos }^{2}}x=1-{{\sin }^{2}}x=1-\,{{\left( \frac{\sqrt{5}-1}{2} \right)}^{2}}\] \[=1-\,\left( \frac{5+1-2\sqrt{5}}{4} \right)\] \[=\frac{-2+2\sqrt{5}}{4}\] \[=\frac{\sqrt{5}-1}{2}\]
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