Verified Solution • 2025–26 Session moderate

QJEEM23Ma4fb9: JEE Main 2023 Mathematics - Class 11 JEEM23Ma4fb9 Solution: Let and let the equation E be . Then the largest element in the …

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

Step-by-Step Verified Solution

D≥0⇒4−4∣λ−3∣≥0D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0

∣λ−3∣≤1|\lambda-3| \leq 1

−1≤λ−3≤1-1 \leq \lambda-3 \leq 1

2≤λ≤42 \leq \lambda \leq 4

∣x∣=2±4−4∣λ−3∣2|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}

=1±1−∣λ−3∣=1 \pm \sqrt{1-|\lambda-3|}

xlargest =1+1=2x_{\text {largest }}=1+1=2 , when λ=3\lambda=3

Largest element of S=2+3=5S=2+3=5

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring