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$x^4y^5+x^5y^4=810$

$x^3y^6+x^6y^3=945$

$\{x,y\}=?$

We have, $x^4y^4(x+y)=810$ and
$x^3y^3(x^3+y^3)=945$ or $x^3y^3 (x+y)(x^2-xy+y^2)=945$

On division, $\frac {x^2-xy+y^2}{xy}=\frac {945}{810}=\frac {7}{6}$

$\Rightarrow \frac {x}{y}-1+\frac {y}{x}=\frac {7}{6}$

Let, $\frac {y}{x}=v$

So, $\frac {1}{v}-1+v=\frac {7}{6}$

$\Rightarrow v+\frac {1}{v}=\frac {7}{6}+1=\frac {13}{6}$

Thus, $6v^2+6=13v$

$\Rightarrow 6v^2-4v-9v+6=0$

$\Rightarrow (3v-2)(2v-3)=0$

$\Rightarrow v=\frac {3}{2},\frac {2}{3}$

Since, $v=\frac {y}{x}=\frac {3}{2}$, y=3k & x=2k

So, $(2k)^4.(3k)^4.(2k+3k)=810$

$\Rightarrow 8k^9 =1$

$\Rightarrow k=\frac {1}{2^{1/3}}$

Now, $y=3k=3.2^{-1/3}$ and $x=2k=2.2^{-1/3}=2^{2/3}$

When $v=\frac {2}{3}$, it can be shown that the values of x and y would interchange, that is $x=3.2^{-1/3}$ and $y=2^{2/3}$.

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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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