If the set R=\ (a, b) ; a+5 b=42, a, b in mathbb N \ has m elements and sum n=1 m (1+i n! )=x+i y , where I=√(-1) , then the value of m+x+y is :

Mathematics · Complex Numbers

JEE Main 2024 — 8 April, Shift 1 — Question 16

If the set $R=\{(a, b) ; a+5 b=42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^{m}\left(1+i^{n!}\right)=x+i y$ ,

where $I=\sqrt{-1}$ , then the value of $m+x+y$ is :

$\mathrm{a}+5 \mathrm{~b}=42, \mathrm{a}, \mathrm{b} \in \mathrm{N}$

$\mathrm{a}=42-5 \mathrm{~b}, \mathrm{~b}=1, \mathrm{a}=37$

$\mathrm{b}=2, \mathrm{a}=32$

$\mathrm{b}=3, \mathrm{a}=27$

$\vdots$

$\mathrm{b}=8, \mathrm{a}=2$

R has " 8 " elements $\Rightarrow \mathrm{m}=8$

$\sum_{n=1}^{8}\left(1-i^{n!}\right)=x+i y$

for $\mathrm{n} \geq 4, \mathrm{i}^{\mathrm{n}!}=1$

$\Rightarrow(1-\mathrm{i})+\left(1-\mathrm{i}^{2!}\right)+\left(1-\mathrm{i}^{3!}\right)$

$=1-\mathrm{I}+2+1+1$

$=5-\mathrm{I}=\mathrm{x}+\mathrm{iy}$

$m+x+y=8+5-1=12$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.