The number of positive integral values of K f | vedclass

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(A) Let $f(x) = |x + |2x - 1|| - |x - |2x - 1||$. We know that $|a+b| - |a-b| = 2 	ext{sgn}(b) \min(|a|, |b|)$ is not directly applicable,but we can

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(A) Let $f(x) = |x + |2x - 1|| - |x - |2x - 1||$. We know that $|a+b| - |a-b| = 2 	ext{sgn}(b) \min(|a|, |b|)$ is not directly applicable,but we can analyze the function $f(x)$ by cases:Case $1$: $x \ge 1$. Then $|2x-1| = 2x-1$. So $f(x) = |x + 2x - 1| - |x - (2x - 1)| = |3x - 1| - |-x + 1| = (3x - 1) - (x - 1) = 2x$.Case $2$: $1/2 \le x Case $3$: $0 \le x Case $4$: $x Thus,$f(x) = 2x$ for $x The graph of $f(x)$ increases from $-\infty$ to $1$ at $x=1/2$,then increases from $0$ to $2$ as $x$ goes from $1/2$ to $1$,then continues to increase for $x > 1$.For $K$ to have exactly three solutions,$K$ must be such that the horizontal line $y=K$ intersects the graph at three points. Looking at the behavior,there is no such $K$ that yields exactly three solutions. Thus,the number of such positive integral values is $0$.

The number of positive integral values of $K$ for which the equation $K = |x + |2x - 1|| - |x - |2x - 1||$ has exactly three real solutions is:

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