JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 14)
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
5
Explanation
$D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0$
$|\lambda-3| \leq 1$
$-1 \leq \lambda-3 \leq 1$
$2 \leq \lambda \leq 4$
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
$=1 \pm \sqrt{1-|\lambda-3|}$
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
Largest element of $S=2+3=5$
$|\lambda-3| \leq 1$
$-1 \leq \lambda-3 \leq 1$
$2 \leq \lambda \leq 4$
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
$=1 \pm \sqrt{1-|\lambda-3|}$
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
Largest element of $S=2+3=5$
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