If x-4-x2-5x-2k-2-x-2-1-x-k - then k-

The correct option is B 3 writing x 4x^2 5x 2k in the partial fractionswe get 2x 2 1x+k #160; (given RHS)So, 2 is one of the solution ...

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If x-4/x2-5...

Question

If $\frac{x - 4}{x^{2} - 5 x - 2 k} = \frac{2}{x - 2} - \frac{1}{x + k}$ , then $k =$

- 3

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$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P (X = x)$	$0$	$K$	$2 K$	$2 K$	$3 K$	$K^{2}$	$2 K^{2}$	$7 K^{2} + K$

P (0 < X < 5) =

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} - 1 & x^{2} + 2 x + 1 & 2 x^{2} + 3 x + 1 2 x^{2} + x - 1 & 2 x^{2} + 5 x - 3 & 4 x^{2} + 4 x - 3 6 x^{2} - x - 2 & 6 x^{2} - 7 x + 2 & 12 x^{2} - 5 x - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

X

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$k^{2}$	$2 k^{2}$	$7 k^{2} + k$

k

lim x \to 1 \frac{x^{4} - 1}{x - 1} = lim x \to k \frac{x^{3} - k^{3}}{x^{2} - k^{2}}

k

\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}

\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}

2 k

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