The sum of all integral values of k ( k ≠ 0 ) for which the...

We have,

$\dfrac{2}{x - 1} - \dfrac{1}{x - 2} = \dfrac{2}{k}$

$⇒ \dfrac{2x - 4 - x + 1}{( x - 1 ) ( x - 2 )} = \dfrac{2}{k}$

$⇒ \dfrac{x - 3}{{x}^{2} - 3x + 2} = \dfrac{2}{k}$

$⇒ kx - 3k = 2{x}^{2} - 6x + 4$

$⇒ 2{x}^{2} - ( 6 + k ) x + 3k + 4 = 0$

For no real roots $D < 0$

$⇒ ( 6 + k{)}^{2} - 8 ( 3k + 4 ) < 0$

$⇒ {k}^{2} + 12k + 36 - 24k - 32 < 0$

$⇒ ( k - 6{)}^{2} - 32 < 0$

$⇒ \left|k - 6\right| < \sqrt{32}$

$⇒ 6 - \sqrt{32} < k < 6 + \sqrt{32}$

Integral value of $k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$

Sum$= \dfrac{11 \times 12}{2} = 66$