The pair of linear equations $2 x + k y = k, 4 x + 2 y = k + 1$ has infinitely many solutions if

k = 1

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k \neq 1

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k = 2

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k = 4

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Solution

The correct option is B $k = 1$
The equation are
$2 x + k y - k = 0$
$4 x + 2 y - (k + 1) = 0$
Here, $a_{1} = 2, b_{1} = k, c_{1} = - k$
and $a_{2} = 4, b_{2} = 2, c_{2} = - (k + 1)$
For the system to have infinite solutions,
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{2}{4} = \frac{k}{2} = \frac{- k}{- (k + 1)}$
Taking,
$\frac{2}{4} = \frac{k}{2}$
$\Rightarrow 4 k = 4$
$\Rightarrow k = 1$
Taking,
$\frac{k}{2} = \frac{- k}{- (k + 1)}$
$\Rightarrow k + 1 = 2 \Rightarrow k = 1$
So, $k = 1$ is the answer

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