If f: R → R is a function defined by f(x) = [x - 1] ((2x - 1)/(2))π, where [·] denotes the greatest integer function, then f is:

Correct answer: D. continuous for every real x Doubtful points are x = n, n ∈ I L . H . L = _ x → n ^ - [x - 1] ((2x - 1)/(2))π = (n - 2) ((2n - 1)/(2))π = 0 R . H . L = _ x → n ^ + [x - 1] ((2x - 1)/(2))π = (n - 1) ((2n - 1)/(2))π = 0 f(n) = 0 Hence continuous.

JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Continuity and Differentiability

If f: R → R is a function defined by f(x) = [x - 1] ((2x -...

f: R \to R

is a function defined by

f\left(x\right) =

\left[x - 1\right]\cos \left(\dfrac{2x - 1}{2}\right)\pi,

where \left[\cdot\right]

denotes the greatest integer function, then

f

is:

Options

Adiscontinuous only at $x = 1$
Bdiscontinuous at all integral values of $x$ except at $x = 1$
Ccontinuous only at $x = 1$
Dcontinuous for every real $x$

Correct answer

D. continuous for every real $x$

Step-by-step solution

Doubtful points are $x = n, n \in I$

$L . H . L = \lim_{x \to {n}^{-}}\left[x - 1\right]\cos \left(\dfrac{2x - 1}{2}\right)\pi = \left(n - 2\right)\cos \left(\dfrac{2n - 1}{2}\right)\pi = 0$

$R . H . L = \lim_{x \to {n}^{+}}\left[x - 1\right]\cos \left(\dfrac{2x - 1}{2}\right)\pi = \left(n - 1\right)\cos \left(\dfrac{2n - 1}{2}\right)\pi = 0$

$f\left(n\right) = 0$

Hence continuous.

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