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Question

For the given equations cos(4y3x2)cos(4y+3x+2)=2+2 ln(k4255) and cos(4y3x2)+cos(4y+3x+2)=2k+8 to have real solutions (x,y), the possible value of k is

A
4
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B
-4
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C
-5
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D
none of these
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Solution

The correct option is B -4
2sin(4y)sin(3x+2)=2+2 ln(k4255)

1+ln(k4255)=sin(4y)sin(3x+2)1

k42551

k42564k4

and 2cos(4y)cos(3x+2)=2k+8

k+41k3

So k=4

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