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Q.

Find the value of k for which the given simultaneous equations have infinitely many solutions: kx-y+3-k=0;4x-ky+k=0.


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a

k=3

b

k=4

c

k=2

d

k=1 

answer is C.

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Detailed Solution

Given that, kx-y+3-k=0;4x-ky+k=0.
Compare the equations kx-y+3-k=0and 4x-ky+k=0 with the general equations a1x+b1y+c1=0 and a2x+b2y+c2=0 respectively.
Here, a1a2=k4,b1b2=-1-k=1k and c1c2=3-kk.  For a pair of linear equations to have infinitely many solutions: a1a2=b1b2=c1c2
So, we have k4=1k=3-kk
We first solve k4=1k,
Which gives k2=4,
k=±2.
Also, 1k=3-kk
k=3k-k2
k2-2k=0,
k(k-2)=0 ,which means k=0 or k=2.
Therefore, the value of k, that satisfies both the conditions, is k=2.
Hence, for k=2, the pair of linear equations has infinitely many solutions.
Therefore, option (3) is correct.
 
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