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 = {$x+\lambda:x$ is an integer solution of E} is ______
Answer (integer)
5
Solution
$D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0$
<br/><br/>
$|\lambda-3| \leq 1$
<br/><br/>
$-1 \leq \lambda-3 \leq 1$
<br/><br/>
$2 \leq \lambda \leq 4$
<br/><br/>
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
<br/><br/>
$=1 \pm \sqrt{1-|\lambda-3|}$
<br/><br/>
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
<br/><br/>
Largest element of $S=2+3=5$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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