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Question

If the system of equations kx+3y(k3)=0,12x+kyk= has infinitely many solutions, then k=

A
6
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B
6
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C
0
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D
None of these
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Solution

The correct option is A 6
For infinitely many solution
a1a2=b1b2=c1c2
K12=3K=(K3)K
By condition (i) and (ii)
K2=36K=±6 ...(i)
By, condition (i) and (ii)
3K=K23K
K26K=0
K(K6)=0
K=0 or K=6 ...(ii)
So, by equation (i) and (ii)
K=6

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