JEE Main 202228 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Continuity and Differentiability

Let f: ℝ → ℝ be defined as f(x) = [ cases [ e ^ x ], & x where a,...

Let

f: \mathbb{R} \to \mathbb{R}

be defined as

f\left(x\right) = \left[

\begin{cases}\left[{e}^{x}\right], & x < 0 \\ a{e}^{x} + \left[x - 1\right], & 0 \le x < 1 \\ b + \left[\sin \left(\pix\right)\right], & 1 \le x < 2 \\ \left[{e}^{- x}\right] - c, & x \ge 2\end{cases}

\right)<br />where

a, b, c \in \mathbb{R} and \left[t\right]

denotes greatest integer less than or equal to

t$. Then, which of the following statements is true?

Options

AThere exists $a, b, c \in \mathbb{R}$ such that $f$ is continuous of $\mathbb{R}$ .
BIf $f$ is discontinuous at exactly one point, then $a + b + c = 1$
CIf $f$ is discontinuous at exactly one point, then $a + b + c \ne 1$ .
D $f$ is discontinuous at atleast two points, for any values of $a, b$ and $c$ .

Correct answer

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

Step-by-step solution

Given $f\left(x\right) = \left[$ \begin{cases}\left[{e}^{x}\right], & x < 0 \\ a{e}^{x} + \left[x - 1\right], & 0 \le x < 1 \\ b + \left[\sin \left(\pix\right)\right], & 1 \le x < 2 \\ \left[{e}^{- x}\right] - c, & x \ge 2\end{cases} $\right)</p><p>i.e.$ f $\left(x\right) =$ \left\{ $\begin{cases}0, & x < 0 \\ a{e}^{x} - 1, & 0 \le x < 1 \\ b, & x = 1 \\ b - 1, & 1 < x < 2 \\ - c, & x \ge 2\end{cases}$ \right)

Here, we know $f\left(x\right)$ is discontinuous at $x = 1$ as $b \ne b - 1$

For $f$ \left(x\right) $to be continuous at$ x = 0; a = 1 $</p><p>For$ f $\left(x\right)$ to be continuous at $x = 2; b + c = 1$

Hence, $a + b + c = 2$

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