Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\) be defined as \(f(x)=\begin{cases}x+a, & x<0 \\ |x-1|, & x \geq 0\end{cases}\) and \(g(x)=\begin{cases}x+1, & x<0 \\ (x-1)^{2}+b, & x \geq 0\end{cases}\), where \(a, b\) are non-negative real numbers. If \((g \circ f)(x)\) is continuous for all \(x \in \mathbb{R}\), then \(a+b\) is equal to ______

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Continuity and Differentiability

Let

f : R \to R

and

g : R \to R

be defined as

f (x) = {x + a, ∣ x - 1∣, x < 0 x \geq 0

and

g (x) = {x + 1, (x - 1)^{2} + b, x < 0 x \geq 0

, where

a, b

are non-negative real numbers.
If

(g \circ f) (x)

is continuous for all

x \in R

, then

a + b

is equal to ______

Answer: 1

For

x \geq 0

the inner value

f (x) = ∣ x - 1∣

is never negative, so

(g \circ f) (x) = (∣ x - 1∣ - 1)^{2} + b

there, which is continuous, and

(g \circ f) (0) = b

.
For

x < 0

the inner value is

f (x) = x + a

, negative for

x < - a

and non-negative for

- a \leq x < 0

, since

a \geq 0

.
If

a > 0

, continuity at

x = - a

equates the left limit

x + a + 1 \to 1

with the right limit

(x + a - 1)^{2} + b \to 1 + b

, forcing

b = 0

; continuity at

x = 0

then equates

(a - 1)^{2} + b

with

b

, forcing

a = 1

.
If

a = 0

, then

(g \circ f) (x) = x + 1

on the whole of

x < 0

, and continuity at

x = 0

equates

1

with

b

, forcing

b = 1

.
In both admissible cases

a + b = 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer