| DIRECTION: Each of these questions contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Choose the correct answer (Only one option is correct) from the following - |
| Statement-1: Integral part of\[{{\left( \sqrt{3}+1 \right)}^{2n+1}}\] is even where \[n\in I.\] |
| Statement-2: Integral part of any integral power of the expression of the form of \[p+\sqrt{q}\]is even. |
A) Statement-1 is false, Statement-2 is true.
B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D) Statement-1 is true, Statement-2 is false.
Correct Answer: D
Solution :
\[{{(\sqrt{3}+1)}^{2n+1}}\] \[{{=}^{2n+1}}{{C}_{0}}{{(\sqrt{3})}^{2n+1}}{{+}^{2n+1}}{{C}_{1}}{{(\sqrt{3})}^{2n}}\] \[{{+}^{2n+1}}{{C}_{2}}{{(\sqrt{3})}^{2n-1}}+......{{+}^{2n+1}}{{C}_{2n+1}}...\] (1) \[{{(\sqrt{3}-1)}^{2n+1}}{{=}^{2n+1}}{{C}_{0}}{{(\sqrt{3})}^{2n+1}}{{-}^{2n+1}}{{C}_{1}}{{(\sqrt{3})}^{2n}}\] \[{{+}^{2n+1}}{{C}_{2}}{{(\sqrt{3})}^{2n-1}}+{{.....}^{2n+1}}{{C}_{2n+1}}\] ..(2) Subtracting (2) from (1) \[{{(\sqrt{3}+1)}^{2n+1}}-{{(\sqrt{3}-1)}^{2n+1}}\] \[=2{{(}^{2n+1}}{{C}_{1}}{{(\sqrt{3})}^{2n}}+{{C}_{3}}{{(\sqrt{3})}^{2n-1}}+...)\] \[I+f-{{f}_{1}}=\] even integer Now, \[0<f<1\Rightarrow -1<f-{{f}_{1}}<1\Rightarrow f-{{f}_{1}}=0\] \[0<{{f}_{1}}<1.\]so 1 =even integer
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