The set of all a 
R for which the equation x |x-l| + |x + 2|+a = 0 has exactly one real root, is
)
, 
)
,-3)
Text Solution
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Given
x|x-1|+ |x + 2| +a = 0
Now taking, Case I: x < -2, we get
a = x 2 + 2
And y = x 2 + 2 is decreasing 
Now taking Case II: -2 < x < 1 we get,
-x 2 + x + x + 2 + a = 0
a = x 2 - 2x - 2
And y = x 2 - 2x - 2
So, 
Hence, y is decreasing 
[-2,1)
Now taking Case III: x 
1 we get,
x 2 -x + x + 2 + a = 0
a = -(x 2 + 2)
And y = - (x 2 + 2) is decreasing 
Hence, from all the cases we can say that nature of function is continuously decreasing so, it will cut x-axis only one time,
Exactly one real root 
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