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JEE Main 2013 Paper

Section: Mathematics

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Question : 51 of 90

Marks: +1, -0

The number of values of

k

, for which the system of equations:

(k+1)x+8y=4k

k x+(k+3) y=3k-1

has no solution, is

1
2
3
infinite

Solution:

From the given system, we have

\; \frac{k+1}{k} = \; \frac{8}{k+3} \neq \; \frac{4k}{3k-1}

( as System has no solution)

\Rightarrow k^{2}+4k+3=8k

\Rightarrow k=1,3

If

k=1

then

\; \frac{8}{1+3} \neq \; \frac{4.1}{2}

which is false

And if

k=3

Then

\; \frac{8}{6} \neq \; \frac{4.3}{9-1}

which is true, therefore

k=3

Hence for only one value of

k

.

System has no solution.

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