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 $S=$
${x+\lambda: x$ 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 $\lambda=4 \Rightarrow|\mathrm{x}|^2-2|\mathrm{x}|+1=0$
$ \Rightarrow|x|=1 \Rightarrow x=-1,1 $
Largest element $=x+\lambda=5$
