1.

The equation `(|x^2-4x|+3)/(x^2+|x-5|)=1` hasA. no real solutionB. exactly one real solutionsC. two real solutionsD. three real solutions

Answer» Correct Answer - D
Let us consider the following cases:
CASE I When `x le0` or ,`4lex le5`
In this case, we have
`|x^(2)-4x|=x^(2)-4x` and `|x-5=-(-5)`.
So, the given equation reduces to
`(x^(2)-4x+3)/(x^(2)-x+5)=1`
`implies x^(2)-4x+3=x^(2)-x+5 implies-3x=2 implies x=-(2)/(3)`
CASE II When `0 lt x lt 4`
In this case, we have
`|x^(2)-4x|=-(x^(2)-4x)` and `|x-5|=-(x-5)`.
So, the given equation reduces to
`-x^(2)+4x+3=x^(2)-x+5`
`implies 2x^(2)-5x+2=0 implies(2x-1)(x-2)=0 implies x=(1)/(2),2`
CASE III When ` x gt 5`
In this case, we have
`|x^(2)-4=x^(2)-4x` and `|x-5|=x-5`
So, the equation reduces to
`x^(2)-4x+3=x^(2)+x-5 implies-5x=-8 impliesx=(8)/(5)`
But, it does not belong to `(5,oo)`.
Hence, the given equation has three real solutions.


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