Let lambda in R and let the equation E be |x| | vedclass

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(C) The given equation is $|x|^2 - 2|x| + |\lambda - 3| = 0$.This can be rewritten as $(|x| - 1)^2 - 1 + |\lambda - 3| = 0$,or $(|x| - 1)^2 = 1 - |\la

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$5$

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(C) The given equation is $|x|^2 - 2|x| + |\lambda - 3| = 0$.This can be rewritten as $(|x| - 1)^2 - 1 + |\lambda - 3| = 0$,or $(|x| - 1)^2 = 1 - |\lambda - 3|$.For $x$ to be a real solution,we must have $1 - |\lambda - 3| \ge 0$,which implies $|\lambda - 3| \le 1$,so $2 \le \lambda \le 4$.Since $x$ must be an integer,$(|x| - 1)^2$ must be a perfect square,i.e.,$0$ or $1$.Case $1$: $(|x| - 1)^2 = 0 \implies |x| = 1 \implies x = \pm 1$. Then $1 - |\lambda - 3| = 0 \implies |\lambda - 3| = 1 \implies \lambda = 4$ or $\lambda = 2$.If $\lambda = 4, x = 1, -1$,then $x + \lambda$ can be $1+4=5$ or $-1+4=3$.If $\lambda = 2, x = 1, -1$,then $x + \lambda$ can be $1+2=3$ or $-1+2=1$.Case $2$: $(|x| - 1)^2 = 1 \implies |x| - 1 = \pm 1 \implies |x| = 2$ or $0 \implies x = \pm 2, 0$. Then $1 - |\lambda - 3| = 1 \implies |\lambda - 3| = 0 \implies \lambda = 3$.If $\lambda = 3, x = 2, -2, 0$,then $x + \lambda$ can be $2+3=5, -2+3=1, 0+3=3$.The set $S = \{5, 3, 1\}$. The largest element is $5$.

Let $\lambda \in R$ and let the equation $E$ be $|x|^2 - 2|x| + |\lambda - 3| = 0$. Then the largest element in the set $S = \{x + \lambda : x \text{ is an integer solution of } E\}$ is $..........$

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