If the equation |x+1|+|x−3|=k has exactly two solution then k lies in
A
(4,∞)
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B
[4,∞)
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C
R
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D
(−∞,4)
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Solution
The correct option is A(4,∞) y=|x+1|+|x−3|
Here the turning points are −1,3 y=⎧⎨⎩−(x+1)−(x−3)x<−1(x+1)−(x−3)−1≤x<3(x+1)+(x−3)x≥3⇒y=⎧⎨⎩2−2xx<−14−1≤x<32x−2x≥3
Graph will be
So when k>4, then the given expression has exactly two solutions.
Hence, k∈(4,∞)