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

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Accepted Answer

"${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$

$x \in R - \{ 1,2\}$

$\Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$

$\Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$

for x $\ne$ 3, $k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$

$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$

& $x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$

$$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$$

for no real roots

$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\}$

Integral k$\in${1, 2 ..... 11}

Sum of k = 66"

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

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

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