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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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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$