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Question

x4(x1)(x2)=x2+3x+k+16x21x1k=

A
8
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B
1
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C
9
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D
7
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Solution

The correct option is D 7
The given equation
x4(x1)(x2)=x2+3x+k+16x21x1 .......... (1)
x4(x1)(x2)=(x2+3x+k)(x2)(x1)+16(x1)(x2)(x2)(x1)
x4=(x2+3x+k)(x2)(x1)+16(x1)(x2)
x4=(x2+3x+k)(x23x+2)+16x16x+2
x4=(x43x2+2x2+3x39x2+6x+kx23kx+2k)+16x16x+2
0=7x2+kx2+21x3kx14+2k
0=x2(k7)+3x(7k)+2(k7)
0=(x23x+2)(k7)
(x23x+2)(k7)=0
(x2)(x1)(k7)=0
x2=0 or x1=0 ork7=0
Now, x2 and x1 ........ (Since, the denominator in equation (1) becomes zero)
k7=0
Hence, k=7

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