The sum of all integral values of mathrm{k}(m | vedclass

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Question

$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$

$x \in R-\{1,2\}$

$\Rightarrow k(2 x-4-x+1)=2\left(x^{2}-3 x+2\right)$

$\Rightarrow k(x-3)=2\left(x

Vedclass · Accepted Answer

$66$

Vedclass · Answer

$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$

$x \in R-\{1,2\}$

$\Rightarrow k(2 x-4-x+1)=2\left(x^{2}-3 x+2ight)$

$\Rightarrow k(x-3)=2\left(x^{2}-3 x+2ight)$

$	ext { for } x 
eq 3, \quad \mathrm{k}=2\left(\mathrm{x}-3+\frac{2}{\mathrm{x}-3}+3ight)$

$\mathrm{x}-3+\frac{2}{\mathrm{x}-3} \geq 2 \sqrt{2}, \forall \mathrm{x}>3$

$\, \mathrm{x}-3+\frac{2}{\mathrm{x}-3} \leq-2 \sqrt{2}, \forall \mathrm{x}<-3$

$\left(\mathrm{x}-3+\frac{2}{\mathrm{x}-3}+3ight) \in(-\infty, 6-4 \sqrt{2}] \cup[6+4 \sqrt{2}, \infty)$

for no real roots

$\mathrm{k} \in(6-4 \sqrt{2}, 6+4 \sqrt{2})-\{0\}$

Integral $\mathrm{k} \in\{1,2 \ldots \ldots 11\}$

Sum of $\mathrm{k}=66$

The sum of all integral values of $\mathrm{k}(\mathrm{k} \neq 0$ ) for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots, is ..... .

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