Verified Solution • 2025–26 Session moderate QJEEM24Ma1a63: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1a63 Solution: If the set has elements and ∑ n = 1 m ( 1 − i n ! ) = x + i y , … If the set R={(a,b):a+5b=42,a,b∈N}R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}R={(a,b):a+5b=42,a,b∈N} has mmm elements and ∑n=1m(1−in!)=x+iy, where i=−1i=\sqrt{-1}i=−1 , then the value of m+x+ym+x+ym+x+y is Multiple Choice Options A 12 B 4 C 8 D 5 Step-by-Step Verified Solution R={(a,b):a+5b=42}R=\{(a, b): a+5 b=42\}R={(a,b):a+5b=42} Then R={(2,8),(7,7),(12,6),(17,5),(22,4),(27,3),(32,2),(37,1)}R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}R={(2,8),(7,7),(12,6),(17,5),(22,4),(27,3),(32,2),(37,1)} amp; and ∑n=1m=8(1−in!)=x+iyamp;∴∑n=18(1−in!)=8−(i+i2+i6+1+1+1+1+1)amp;=5−iamp;∴x=5,y=−1amp;x+y+m=5−1+8=12\begin{aligned}& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\& =5-i \\& \therefore x=5, y=-1 \\& x+y+m=5-1+8=12\end{aligned}amp; and n=1∑m=8(1−in!)=x+iyamp;∴n=1∑8(1−in!)=8−(i+i2+i6+1+1+1+1+1)amp;=5−iamp;∴x=5,y=−1amp;x+y+m=5−1+8=12 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