JEE Main 2021 — Mathematics Calculus

$$g[f(x)]=[\begin{matrix}f(x)+1 & f(x)<0 \\ (f(x)-1{)}^{2}+b & f(x)\geq 0\end{matrix}$$g[f(x)]=[f(x)+1(f(x)−1)2+b​f(x)<0f(x)≥0​

$$g[f(x)]=[\begin{matrix}x+a+1 & x+a<0;x<0 \\ |x-1|+1 & |x-1|<0;x\geq 0 \\ (x+a-1{)}^{2}+b & x+a\geq 0;x<0 \\ (|x-1|-1{)}^{2}+b & |x-1|\geq 0;x\geq 0\end{matrix}$$g[f(x)]=[x+a+1∣x−1∣+1(x+a−1)2+b(∣x−1∣−1)2+b​x+a<0;x<0∣x−1∣<0;x≥0x+a≥0;x<0∣x−1∣≥0;x≥0​

$$g[f(x)]=[\begin{matrix}x+a+1 & x\in (-\infty ,-a);x\in (-\infty ,0) \\ |x-1|+1 & x\in \phi \\ (x+a-1{)}^{2}+b & x\in [-a,\infty );x\in (-\infty ,0) \\ (|x-1|-1{)}^{2}+b & x\in R;x\in [0,\infty )\end{matrix}$$g[f(x)]=[x+a+1∣x−1∣+1(x+a−1)2+b(∣x−1∣−1)2+b​x∈(−∞,−a);x∈(−∞,0)x∈ϕx∈[−a,∞);x∈(−∞,0)x∈R;x∈[0,∞)​

$$g[f(x)]=[\begin{matrix}x+a+1 & x\in (-\infty ,-a) \\ (x+a-1{)}^{2}+b & x\in [-a,0) \\ (|x-1|-1{)}^{2}+b & x\in [0,\infty )\end{matrix}$$g[f(x)]=[x+a+1(x+a−1)2+b(∣x−1∣−1)2+b​x∈(−∞,−a)x∈[−a,0)x∈[0,∞)​

$g(f(x))$g(f(x)) is continuous

at x=-a& at $x=0$x=0

1=b+1&(a-1{)}^{2}+b=b

b=0&a=1

$$\Rightarrow a+b=1$$⇒a+b=1