Let $f:[0, \infty) \rightarrow[0, \infty)$ be defined as $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}}[\mathrm{y}] \mathrm{dy}$ where $[x]$ is the greatest integer less than or equal to $x$. Which of the following is true?

(C) $f$ is continuous at every point in $[0, \infty)$ and differentiable except at the integer points.

(A) $f$ is differentiable at every point in $[0, \infty)$

(B) $f$ is continuous everywhere except at the integer points in $[0, \infty)$.

(D) $f$ is both continuous and differentiable except at the integer points in $[0, \infty)$.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Continuity and Differentiability

Let

f : [0, \infty) \to [0, \infty)

be defined as

f (x) = \int_{0}^{x} [y] dy

where

[x]

is the greatest integer less than or equal to

x

. Which of the following is true?

(A) $f$ is differentiable at every point in $[0, \infty)$
(B) $f$ is continuous everywhere except at the integer points in $[0, \infty)$ .
(C) $f$ is continuous at every point in $[0, \infty)$ and differentiable except at the integer points.✓ Correct
(D) $f$ is both continuous and differentiable except at the integer points in $[0, \infty)$ .

Answer: (C)

For

x

[n, n + 1)

with

n

a non-negative integer,

f (x) = \int_{0}^{x} [y] d y = k = 0 \sum n - 1 k + n (x - n) = \frac{n ( n - 1 )}{2} + n (x - n)

The integrand is bounded, so the integral varies continuously with

x

; the pieces also meet, since the formula on

[n, n + 1)

tends to

\frac{n ( n - 1 )}{2} + n = \frac{n ( n + 1 )}{2}

x \to (n + 1)^{-}

, which is the value the next piece takes at

x = n + 1

. So

f

is continuous on

[0, \infty)

.
On

(n, n + 1)

the graph is a straight line of slope

n

. At an integer

x = n

with

n \geq 1

the left slope is

n - 1

and the right slope is

n

, so the two one-sided derivatives differ and

f

is not differentiable there.
Thus

f

is continuous at every point of

[0, \infty)

and differentiable except at the integer points.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option