### Quadratic Equation Question 2

#### Question 2 - 24 January - Shift 1

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: \text{x is an integer solution of E}$ is

## Show Answer

#### Answer: (5)

#### Solution:

#### Formula: Roots of equations

$ |x|^2-2|x|+|\lambda-3|=0 $

For real roots $\Delta \geq 0$

$ \begin{gathered} \Rightarrow|\lambda-3| \leq 1 \\ \Rightarrow 2 \leq \lambda \leq 4 \end{gathered} $

When $\lambda=4 \Rightarrow|x|^2-2|x|+1=0$

$ \Rightarrow|x|=1 \Rightarrow x=-1,1 $

Largest element $=x+\lambda=5$