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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 ..... .

Similar Questions

Consider the following two statements

$I$. Any pair of consistent liner equations in two variables must have a unique solution.

$II$. There do not exist two consecutive integers, the sum of whose squares is $365$.Then,

If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided

Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes

If the sum of all the roots of the equation $e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0$ is $\log _{ e } P$, then $p$ is equal to

Let $S$ be the set of all $\alpha \in R$ such that the equation, $cos\,2 x + \alpha \,sin\, x = 2\alpha -7$ has a solution. Then $S$ is equal to