The number of integers a in the interval [1,2 | vedclass

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Question

(d)

Given,

$x+y=a \Rightarrow a \in[1,2014]$

and $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

$	herefore \quad \frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

Vedclass · Accepted Answer

$2014$

Vedclass · Answer

(d)

Given,

$x+y=a \Rightarrow a \in[1,2014]$

and $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

$	herefore \quad \frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

$\Rightarrow x^2 y-x^2+x y^2-y^2=4(x-1)(y-1)$

$\Rightarrow x^2 y+x y^2=4(x y-(x+y)+1)+x^2+y^2$

$\Rightarrow \quad x y(x+y)=4(x y-a+1) +(x+y)^2-2 x y$

$\Rightarrow \quad x y(a)=4 x y-4 a+4+a^2-2 x y$

$\Rightarrow \quad x y a-2 x y=a^2-4 a+4$

$\Rightarrow x y \quad(a-2)=(a-2)^2$

$\Rightarrow \quad(a-2)^2-x y(a-2)=0$

$\Rightarrow \quad(a-2)(a-2-x y)=0$

$a=2$ or $x y=a-2$

$x(a-x)=a-2 \quad[\because y=a-x]$

$x^2-a x+a-2=0$

$\Rightarrow(a 	imes b)(c+1)+b(c+1)+a(c+1)+(c+1)=30$

$\Rightarrow(c+1)(a 	imes b+b+a+1)=30$

$\Rightarrow(c+1)(b(a+1)+1(a+1))=30$

$\Rightarrow \quad(a+1))(b+1)(c+1)=30$

$1 \leq a \leq 9,0 \leq b, c \leq 9$

$\because$ Total number of solution $=18$

The number of integers $a$ in the interval $[1,2014]$ for which the system of equations $x+y=a$, $\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ has finitely many solutions is

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