JEE MAIN - Mathematics (2013 (Offline) - No. 18)
The number of values of $$k$$, for which the system of equations : $$$\matrix{
{\left( {k + 1} \right)x + 8y = 4k} \cr
{kx + \left( {k + 3} \right)y = 3k - 1} \cr
} $$$
has no solution, is
has no solution, is
infinite
1
2
3
Explanation
From the given system, we have
$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$
( as System has no solution)
$$ \Rightarrow {k^2} + 4k + 3 = 8k$$
$$ \Rightarrow k = 1,3$$
If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false
And if $$k = 3$$
Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$
Hence for only one value of $$k.$$ System has no solution.
$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$
( as System has no solution)
$$ \Rightarrow {k^2} + 4k + 3 = 8k$$
$$ \Rightarrow k = 1,3$$
If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false
And if $$k = 3$$
Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$
Hence for only one value of $$k.$$ System has no solution.
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