JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, |z+5|≤4, which is equation of circle.
|z+5|≤4‌
⇒‌(x+5)2+y2≤16. . . (i)
‌ and ‌‌z(1+i)+z(1−i)≥−10
⇒‌(z+z)+i(z−z)≥−10
⇒‌x−y+5≥0 . . . (ii)
From Eqs. (i) and (ii), region bounded by inequalities areNow, |z+1|2=|z−(−1)|2
Maximum value of |z+1|2 will be equal to (AC)2.
Now, (x+5)2+y2=16
and x−y+5=0
Given, y=±2√2
and x=±2√2−5
∴ Coordinates are
A(−2√2−5,−2√2)
B(2√2−5,2√2)
and ‌‌C(−1,,
AC2‌‌=(2√2+4)2+(2√2)2
‌‌=32+16√2
Given, that maximum value of
|z+1|2‌ is ‌α+β√2
⇒‌α+β√2‌=32+16√2
⇒‌α‌=32,β=16
∴‌α+β‌=32+16=48