Let $\mathrm{f}, \mathrm{g}: \mathbf{R} \rightarrow \mathbf{R}$ be functions defined by $f(x)=\left\{\begin{array}{ll}{[x]} & , \quad x<0 \\ |1-x|, & x \geq 0\end{array}\right.$ and $g(x)= \begin{cases}e^{x}-x & x<0 \\ (x-1)^{2}-1 & , x \geq 0\end{cases}$ where $[x]$ denote the greatest integer less than or equal to x . Then, the function fog is discontinuous at exactly :

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Continuity and Differentiability

Let

f, g : R \to R

be functions defined by

f (x) = {[x] ∣1 - x ∣,, x < 0 x \geq 0

and

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

where

[x]

denote the greatest integer less than or equal to x . Then, the function fog is discontinuous at exactly :

Answer: (B)

To find discontinuities of

f \circ g

, we check:

g (x) = 0

when

x = 0

x = 2

.

At

x = 0

g (0^{-}) = e^{0} - 0 = 1

g (0) = 0

f (g (0^{-})) = f (1) = ∣1 - 1∣ = 0

f (g (0)) = f (0) = ∣1 - 0∣ = 1

f \circ g

is discontinuous at

x = 0

.

At

x = 2

g (2^{-}) = (2 - 1)^{2} - 1 = 0

g (2^{+}) = (2 + ϵ - 1)^{2} - 1 > 0

f (g (2^{-})) = f (0) = 1

f (g (2^{+})) = ∣1 - (1 + ϵ)^{2} + 1∣ \to 1

ϵ \to 0^{+}

Actually,

g (2) = 0

, and checking limits shows discontinuity.

Therefore,

f \circ g

is discontinuous at exactly two points:

x = 0

and

x = 2

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option