JEE Main 16 March 2021 Shift 2 Solved Paper

Given, f(x)={

x+a	x<0
|x−1|	x≥0.

g(x)={

x+1	‌	x<0
(x−1)2+b	‌	x≥0.

g[f(x)]={

f(x)+1	f(x)<0
[f(x)−1]2+b,	f(x)≥0.

Case I x+a<0 and x<0⇒x<−a
Case II |x−1|<0 and x≥0⇒ Not possible
f(x)≥0
Case I x+a≥0 and x<0⇒x∈[−a,0)
Case II |x−1|≥0 and x≥0⇒x≥0
This is continuous function.
Since, g[f(x)] is continuous for all x∈R
So, g(f(x)) will be continuous at x=−a and x=0
Now, at x=−a
LHL=RHL= value of function
⇒‌‌(a−1)2+b=b
⇒(a−1)2=0
⇒‌‌a=1
Hence, a+b=1