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Let $\text{k}$ be a constant. The equations $\text{kx + y} = 3$ and $\text{4x + ky} = 4$ have a unique solution if and only if 

  1. $|\text{k}| \neq 2$
  2. $|\text{k}| = 2$
  3. $\text{k} \neq 2$
  4. $\text{k} = 2$
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Given that, the equations.

  • $ kx + y = 3 \; \longrightarrow (1) $
  • $ 4x + ky = 4 \; \longrightarrow (2) $

For a system of equations.

  • $ a_{1}x + b_{1}y + c_{1} = 0 $
  • $ a_{2}x + b_{2}y + c_{2} = 0 $

to have unique solution, the condition to be satisfied is 

$$ \boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}} $$

We can re-write the equation $(1)$ and $(2)$ as follows:

  • $ kx + y – 3 = 0$
  • $ 4x + ky – 4 =0 $

For unique solution $: \frac{k}{4} \neq \frac{1}{k} \neq \frac{-3}{-4} $

Taking first two terms,

$ \frac{k}{4} \neq \frac{1}{\text{k}} $

$ \Rightarrow k^{2} \neq 4 $

$ \Rightarrow k \neq \sqrt{4} $

$ \Rightarrow k \neq \pm 2 $

$ \Rightarrow \boxed{|k| \neq 2} $

Correct Answer$: \text{A} $ 

$\textsf{PS:}$

$1.$ For a system of equation $ a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0 $ to have no solution, the condition to be satisfied is, 

$$ \boxed{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}} $$

$2.$ For a system of equation $ a_{1}x + b_{1}y + c_{1} =0 ; \; a_{2}x + b_{2}y + c_{2} = 0 $ to have infinitely many solutions, we must have 

$$ \boxed{\frac{a_{1}}{a_{2}} =  \frac{b_{1}}{b_{2}} =  \frac{c_{1}}{c_{2}}} $$

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