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Question

Number of solutions of the equation |x1|+|x2|+|x3|=7 is

A
1
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B
2
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C
3
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D
0
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Solution

The correct option is B 2
|x1|+|x2|+|x3|=7
critical points
x1=0x=1
x2=0x=2
x3=0x=3

On the basis of above points, we have Four regions
x<1, x[1,2], x(2,3], x>3

case-1, x<1
For x<1,
|x1|=(x1), |x2|=(x2), |x3|=(x3)
Now equation will be :
(x1)(x2)(x3)=7
3x+6=7
x=13 Also 13<1
So,x=13 (1)

case-2, x[1,2]
For x[1,2],
|x1|=(x1), |x2|=(x2), |x3|=(x3)
Now equation will be :
(x1)(x2)(x3)=7
x+4=7
x=3 but 3[1,2]
So, x=ϕ (2)

case-3, x(2,3]
For x(2,3],
|x1|=(x1), |x2|=(x2), |x3|=(x3)
Now equation will be :
(x1)+(x2)(x3)=7
x=7
x=7 but 7(2,3]
So, x=ϕ (3)

case-4, x>3
For x>3,
|x1|=(x1), |x2|=(x2), |x3|=(x3)
Now equation will be :
(x1)+(x2)+(x3)=7
3x6=7
x=133 Also 133>3
So, x=133 (4)

By (1)(2)(3)(4)
x={13,133}

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