Question

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

Answer 1

Correct Answer: 5

To solve the equation $E: |x|^2 - 2|x| + |\lambda-3| = 0$, we first redefine it in terms of $y = |x|$. This transforms the equation to $y^2 - 2y + |\lambda-3| = 0$, a quadratic. Solving this involves the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-2$, and $c=|\lambda-3|$. The discriminant $\Delta = b^2 - 4ac = 4 - 4|\lambda-3|$. For real solutions, $\Delta \ge 0$, so $|\lambda-3| \le 1$. This gives $2 \le \lambda \le 4$. Exploring integer solutions $y=1$ and $y=2$, as they are potential integers satisfying $y = |x|$:

• If $y=1$, the equation is $1 - 2 + |\lambda-3| = 0 \Rightarrow |\lambda-3| = 1$, hence $\lambda = 4$ or $\lambda = 2$.
• If $y=2$, the equation becomes $4 - 4 + |\lambda-3| = 0 \Rightarrow |\lambda-3| = 0$, thus $\lambda = 3$.

Integer solutions for $x$ are then:

$\lambda$	$x$	$x+\lambda$
4	-1, 1	3, 5
2	-1, 1	1, 3
3	-2, 2	1, 5

The largest element is $x + \lambda = 5$, which indeed is within the range (5,5). Therefore, the largest element in the set $S$ is 5, confirming it fits the expected range.