Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $$\mathrm{f}(\mathrm{x})=\left[\begin{array}{ll}{\left[\mathrm{e}^{\mathrm{x}}\right],} & \mathrm{x}<0 \\ a \mathrm{e}^{\mathrm{x}}+[\mathrm{x}-1], & 0 \leq \mathrm{x}<1 \\ \mathrm{~b}+[\sin (\pi \mathrm{x})], & 1 \leq \mathrm{x}<2 \\ {\left[\mathrm{e}^{-\mathrm{x}}\right]-\mathrm{c},} & \mathrm{x} \geq 2\end{array}\right.$$ where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{R}$ and $[\mathrm{t}]$ denotes greatest integer less than or equal to $t$. Then, which of the following statements is true?

(C) If f is discontinuous at exactly one point, then $a+b+c \neq 1$.

(A) There exists $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{R}$ such that f is continuous of $\mathbb{R}$.

(B) If f is discontinuous at exactly one point, then $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$.

(D) f is discontinuous at atleast two points, for any values of $\mathrm{a}, \mathrm{b}$ and c.

Continuity and Differentiability — JEE Main · 2022 · 28 Jun 2022 · Shift 1 Question with Solution

Let

f : R \to R

be defined as

f (x) = [e^{x}], a e^{x} + [x - 1], b + [sin (π x)], [e^{- x}] - c, x < 0 0 \leq x < 1 1 \leq x < 2 x \geq 2

where

a, b, c \in R

and

[t]

denotes greatest integer less than or equal to

t

. Then, which of the following statements is true?

(A)There exists $a, b, c \in R$ such that f is continuous of $R$ .
(B)If f is discontinuous at exactly one point, then $a + b + c = 1$ .
(C)If f is discontinuous at exactly one point, then $a + b + c \neq = 1$ .✓ Correct
(D)f is discontinuous at atleast two points, for any values of $a, b$ and c.

Answer: (C)

Examine the function at each junction. For

x < 0

e^{x}

lies strictly between

0

and

1

, so

[e^{x}] = 0

and the left-hand limit at

0

0

; the middle branch at

0^{+}

a e^{x} + [x - 1] = a - 1

because

[x - 1] = - 1

[0, 1)

, so continuity at

0

needs

a = 1

. At

x = 1

the middle branch tends to

a e - 1

from the left, the value there is

b + [sin π] = b

, and the next branch tends to

b - 1

from the right since

sin π x

is just below zero for

x

slightly above

1

; these three cannot all agree, so the function is discontinuous at

x = 1

for every choice of constants. At

x = 2

the left limit is

b - 1

while the branch for

x \geq 2

gives

[e^{- x}] - c = - c

, so continuity there needs

b + c = 1

. Because

x = 1

always breaks continuity, a single point of discontinuity requires

x = 0

and

x = 2

both continuous, forcing

a = 1

and

b + c = 1

, whence

a + b + c = 2

. This is not

1

, so if the function is discontinuous at exactly one point then

a + b + c \neq = 1

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option