Let $$\lambda \in \mathbb{R}$$ 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 _____

We need to find the largest element in the set $$S = \{x + \lambda : x \text{ is an integer solution of } E\}$$, where equation $$E$$ is $$|x|^2 - 2|x| + |\lambda - 3| = 0$$.

Let $$t = |x| \geq 0$$. The equation becomes:

This is a quadratic in $$t$$. Using the quadratic formula:

For $$t$$ to be real: $$1 - |\lambda - 3| \geq 0$$, i.e., $$|\lambda - 3| \leq 1$$, which gives $$2 \leq \lambda \leq 4$$.

For $$t \geq 0$$: We need $$1 \pm \sqrt{1 - |\lambda - 3|} \geq 0$$. Since $$\sqrt{1 - |\lambda-3|} \leq 1$$, the value $$1 - \sqrt{1-|\lambda-3|} \geq 0$$ always. So both roots are non-negative.

Since $$t = |x|$$, for each valid $$t > 0$$, we get $$x = \pm t$$. For $$t = 0$$, we get $$x = 0$$. We need $$t$$ to be a non-negative integer (since $$x$$ must be an integer and $$t = |x|$$).

For $$t$$ to be an integer, we need $$\sqrt{1 - |\lambda-3|}$$ to yield integer values of $$t = 1 \pm \sqrt{1 - |\lambda-3|}$$.

Case 1: $$|\lambda - 3| = 0$$, i.e., $$\lambda = 3$$. Then $$t = 1 \pm 1$$, giving $$t = 0$$ or $$t = 2$$. Integer values of $$x$$: $$0, \pm 2$$. Elements of $$S$$: $$\{0+3, 2+3, -2+3\} = \{3, 5, 1\}$$. Largest = 5.

Case 2: $$|\lambda - 3| = 1$$, i.e., $$\lambda = 2$$ or $$\lambda = 4$$. Then $$t = 1 \pm 0 = 1$$. Integer value of $$x$$: $$\pm 1$$.

If $$\lambda = 2$$: $$S = \{1+2, -1+2\} = \{3, 1\}$$. Largest = 3.

If $$\lambda = 4$$: $$S = \{1+4, -1+4\} = \{5, 3\}$$. Largest = 5.

Now check other values where $$t$$ is integer. We need $$1 - |\lambda - 3| = k^2$$ for some value that makes $$1 \pm k$$ a non-negative integer.

If $$k = 0$$: $$|\lambda-3| = 1$$, covered above.

If $$k = 1$$: $$|\lambda-3| = 0$$, covered above.

For non-integer $$k$$, $$t$$ would not be integer.

From Case 1 ($$\lambda = 3$$): largest element in $$S$$ is 5.

From Case 2 ($$\lambda = 4$$): largest element in $$S$$ is 5.

From Case 2 ($$\lambda = 2$$): largest element in $$S$$ is 3.

The largest element in $$S$$ is 5.