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$x + y = 1$, ${x^2} + {y^2} = 2$, ${x^{11}} + {y^{11}} = ?$

We have, $2xy = {(x + y)^2} - ({x^2} + {y^2}) = 1 - 2 =  - 1$

Now, ${(x - y)^2} = {(x + y)^2} - 4xy = 1 + 2 = 3$

$x - y =  \pm \sqrt 3 $

Solving for x & y,

One of the two variables would be $\frac{{1 + \sqrt 3 }}{2}$ and the other would be $\frac{{1 - \sqrt 3 }}{2}$.

Now, ${x^{11}} + {y^{11}} = \frac{1}{{{2^{11}}}}[{(1 + \sqrt 3 )^{11}} + {(1 - \sqrt 3 )^{11}}]$

$ = \frac{2}{{{2^{11}}}}({}^{11}{C_0} + {}^{11}{C_2} \times 3 + {}^{11}{C_4} \times 9 + {}^{11}{C_6} \times 27 + {}^{11}{C_8} \times 81 + {}^{11}{C_{10}} \times 243)$

$ = \frac{1}{{{2^{10}}}}(1 + 165+2970+12474+13365+2673)$

$ = \frac{{31648}}{{{2^{10}}}} = \frac{{989}}{{32}}$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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