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 quite right,but specifically $|

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$0$

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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 quite right,but specifically $|x+y| - |x-y| = 2x$ if $x \ge |y|$ and $2y$ if $|y| > x$. Here,let $y = |2x-1|$. The expression is $f(x) = |x+y| - |x-y|$. If $x \ge |2x-1|$,then $f(x) = 2|2x-1|$. If $x Solving $x = |2x-1|$: $x = 2x-1 \implies x=1$ or $x = -(2x-1) \implies 3x=1 \implies x=1/3$. For $x \in [1/3, 1]$,$x \ge |2x-1|$ is false,so $f(x) = 2x$. For $x For $x > 1$,$f(x) = 2|2x-1| = 4x-2$. Analyzing the function $f(x)$: $1$) If $x $2$) If $1/3 \le x \le 1$,$f(x) = 2x$. $3$) If $x > 1$,$f(x) = 4x-2$. Wait,let's re-evaluate: If $x If $1/3 \le x \le 1$,$|2x-1| = 2x-1$. Then $f(x) = |x+2x-1| - |x-(2x-1)| = |3x-1| - |1-x| = (3x-1) - (1-x) = 4x-2$. If $x > 1$,$|2x-1| = 2x-1$. Then $f(x) = |x+2x-1| - |x-(2x-1)| = |3x-1| - |1-x| = (3x-1) - (x-1) = 2x$. The function $f(x)$ is $2x$ for $x  1$. At $x=1/3$,$f(x) = 2/3$. At $x=1$,$f(x) = 2$. The graph increases from $-\infty$ to $2/3$,then increases faster to $2$,then increases to $\infty$. Since the function is strictly increasing,it can have at most one solution for any $K$. Thus,there are no values of $K$ for which it has three solutions. The answer 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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