JEE Main 2020 - 5 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = x \cdot [\frac{x}{2}]$ , for $- 10 < x < 10$ , where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

Q: Wait, why is $x=0$ continuous when all other $2k$ points are discontinuous?

Great question! Continuity requires the LHL, RHL, and function value to be equal. At $x=0$, $f(0)=0$. The LHL approaches $0$ from the negative side, and the RHL approaches $0$ from the positive side. Because the multiplier $x$ is zero, it 'crushes' the jump of the GIF, making the function continuous at the origin. It is the only point where this happens!

Q: Why does the LHL at $x=2k$ give $[k - \frac{h}{2}] = k-1$?

Think of the number line. If $k$ is an integer, $k - \text{tiny}$ is a number like $4.999$. The greatest integer of $4.999$ is $4$, which is $k-1$. This is the fundamental behavior of the GIF: it always rounds down to the nearest integer below the value.

Q: Can I just graph this to find the answer?

Visualizing is a powerful tool, but in JEE Advanced, you must back it up with algebra. The graph helps you identify the candidates, but the limit calculation is the proof. Always use the limit definition to be absolutely certain, especially when the function involves a product of a variable and a GIF.

Enter Numerical Value:

Progressive Hints

Hint 1

The Greatest Integer Function, or

[t]

, is like a staircase that jumps at every integer. When you multiply this staircase by

x

, you are essentially scaling the steps. Visualize where the 'steps' occur by setting the input of the bracket,

\frac{x}{2}

, to an integer

k

Hint 2

To check for continuity at a point

x = a

, you must verify if the Left Hand Limit (LHL), Right Hand Limit (RHL), and the function value

f (a)

are all equal. For the GIF, the LHL and RHL often differ because the value inside the bracket approaches an integer from below or above.

Hint 3

Set

x = 2 k

for an integer

k

. Calculate the LHL by approaching

2 k

from the left (i.e.,

x = 2 k - h

) and the RHL by approaching from the right (i.e.,

x = 2 k + h

). The condition for continuity is

f (2 k) = LHL = RHL

. Solve for

k

to see which points are actually continuous.

Visualized Solution

$A na l y z in g t h e F u n c t i o n f (x)$

Given: $f (x) = x \cdot [\frac{x}{2}]$
Domain: $- 10 < x < 10$
$[\cdot]$ represents the Greatest Integer Function (GIF).

$P r o p er t i eso f G r e a t es t I n t e g er F u n c t i o n$

The function $[t]$ jumps at every integer value of $t$ .
Therefore, $[t]$ is discontinuous for all $t \in Z$ .
For $f (x)$ , the critical points occur when the inner term $\frac{x}{2}$ is an integer.

$I d e n t i f y in g C r i t i c a l P o in t s$

Let $\frac{x}{2} = k$ , where $k \in Z$ .
This implies $x = 2 k$ .
These are the only points where $f (x)$ might be discontinuous.

$A ppl y in g D o main C o n s t r ain t s$

Given domain: $- 10 < x < 10$
Substitute $x = 2 k$ : $- 10 < 2 k < 10$
Divide by $2$ : $- 5 < k < 5$

$L i s t in g P o t e n t ia l P o in t so f D i sco n t in u i t y$

Possible integers for $k$ : ${- 4, - 3, - 2, - 1, 0, 1, 2, 3, 4}$
Corresponding $x$ values ( $x = 2 k$ ):
$x \in {0, \pm 2, \pm 4, \pm 6, \pm 8}$

$C h ec k in g C o n t in u i t y a t x = 0$

Let's first check the behavior at $x = 0$ .
Function value: $f (0) = 0 \cdot [\frac{0}{2}] = 0 \cdot 0 = 0$ .

$L e f t H an d L imi t a t x = 0$

$LHL = lim_{x \to 0^{-}} x \cdot [\frac{x}{2}]$
Let $x = - h$ where $h \to 0^{+}$
$LHL = lim_{h \to 0^{+}} (- h) \cdot [\frac{- h}{2}]$
Since $- 0.5 < \frac{- h}{2} < 0$ , $[\frac{- h}{2}] = - 1$ .
$LHL = lim_{h \to 0^{+}} (- h) (- 1) = 0$ .

$R i g h t H an d L imi t a t x = 0$

$RHL = lim_{x \to 0^{+}} x \cdot [\frac{x}{2}]$
Let $x = h$ where $h \to 0^{+}$
$RHL = lim_{h \to 0^{+}} (h) \cdot [\frac{h}{2}]$
Since $0 < \frac{h}{2} < 0.5$ , $[\frac{h}{2}] = 0$ .
$RHL = lim_{h \to 0^{+}} (h) (0) = 0$ .
Conclusion: $LHL = RHL = f (0) = 0 ⟹$ Continuous at $x = 0$ .

$G e n er a l C a se : x = 2 k (k \neq = 0)$

Now consider $x = 2 k$ where $k \in {- 4, - 3, - 2, - 1, 1, 2, 3, 4}$ .
Function value: $f (2 k) = 2 k \cdot [\frac{2 k}{2}]$
$f (2 k) = 2 k \cdot [k] = 2 k \cdot k = 2 k^{2}$ .

$L e f t H an d L imi t a t x = 2 k$

$LHL = lim_{h \to 0^{+}} (2 k - h) \cdot [\frac{2 k - h}{2}]$
$LHL = lim_{h \to 0^{+}} (2 k - h) \cdot [k - \frac{h}{2}]$
Since $h$ is small and positive, $k - \frac{h}{2}$ is slightly less than $k$ .
Thus, $[k - \frac{h}{2}] = k - 1$ .
$LHL = 2 k (k - 1) = 2 k^{2} - 2 k$ .

$R i g h t H an d L imi t a t x = 2 k$

$RHL = lim_{h \to 0^{+}} (2 k + h) \cdot [\frac{2 k + h}{2}]$
$RHL = lim_{h \to 0^{+}} (2 k + h) \cdot [k + \frac{h}{2}]$
Since $h$ is small and positive, $k + \frac{h}{2}$ is slightly more than $k$ .
Thus, $[k + \frac{h}{2}] = k$ .
$RHL = 2 k (k) = 2 k^{2}$ .

$C o n d i t i o n f or C o n t in u i t y$

For continuity at $x = 2 k$ , we must have $LHL = RHL$ .
$2 k^{2} - 2 k = 2 k^{2}$
$- 2 k = 0 ⟹ k = 0$
But we assumed $k \neq = 0$ .
Therefore, $f (x)$ is discontinuous at all $x = 2 k$ for $k \neq = 0$ .

$F ina l C o u n t o f D i sco n t in u i t i es$

The points of discontinuity are $x = \pm 2, \pm 4, \pm 6, \pm 8$ .
Counting them:
Positive: $2, 4, 6, 8$ ( $4$ points)
Negative: $- 2, - 4, - 6, - 8$ ( $4$ points)
Total number of points of discontinuity = $4 + 4 = 8$ .

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Solution Diagram

Analyzing the Setup

The function under investigation is

f (x) = x \cdot [\frac{x}{2}]

within the domain

- 10 < x < 10

. The Greatest Integer Function (GIF), denoted by

[\cdot]

, is inherently discontinuous at every integer value of its argument.

Consequently,

f (x)

is potentially discontinuous whenever the argument

\frac{x}{2}

is an integer. Let

\frac{x}{2} = k

, where

k \in Z

This yields the critical points

x = 2 k

. Given the domain

- 10 < x < 10

, we find the range of

k

by solving

- 10 < 2 k < 10

, which simplifies to

- 5 < k < 5

Thus, our set of critical points corresponds to

k \in {- 4, - 3, - 2, - 1, 0, 1, 2, 3, 4}

. We must evaluate the continuity of

f (x)

at each of these nine points.

The Anomaly at the Origin

We first examine the point

x = 0

(where

k = 0

). The function value is

f (0) = 0 \cdot [0] = 0

To determine continuity, we evaluate the limits. For the Left Hand Limit (LHL) as

x \to 0^{-}

, we let

x = - h

for a small

h > 0

h \to 0^{+} lim (- h) \cdot [\frac{- h}{2}] = h \to 0^{+} lim (- h) (- 1) = 0

For the Right Hand Limit (RHL) as

x \to 0^{+}

, we let

x = h

h \to 0^{+} lim (h) \cdot [\frac{h}{2}] = h \to 0^{+} lim (h) (0) = 0

Since

LHL = RHL = f (0) = 0

, the function is continuous at $x = 0$ .

The General Case of Discontinuity

Now, consider the non-zero integers

k \in {- 4, - 3, - 2, - 1, 1, 2, 3, 4}

. The function value at these points is

f (2 k) = 2 k \cdot [k] = 2 k^{2}

For the LHL, we approach

x = 2 k

from the left by setting

x = 2 k - h

h \to 0^{+} lim (2 k - h) \cdot [\frac{2 k - h}{2}] = h \to 0^{+} lim (2 k - h) \cdot [k - \frac{h}{2}] = 2 k (k - 1) = 2 k^{2} - 2 k

For the RHL, we approach

x = 2 k

from the right by setting

x = 2 k + h

h \to 0^{+} lim (2 k + h) \cdot [\frac{2 k + h}{2}] = h \to 0^{+} lim (2 k + h) \cdot [k + \frac{h}{2}] = 2 k (k) = 2 k^{2}

For continuity, we require

LHL = RHL

, which implies

2 k^{2} - 2 k = 2 k^{2}

. This simplifies to

- 2 k = 0

, or

k = 0

Final Calculation

Since we are analyzing $k eq 0$, the condition for continuity is never satisfied for these points. Therefore, the function is discontinuous at all

x = 2 k

where

k \in {- 4, - 3, - 2, - 1, 1, 2, 3, 4}

Counting these points, we have four values on the negative side and four values on the positive side. The total number of points of discontinuity is 8.

Common Traps & FAQs

Wait, why is

x = 0

continuous when all other

2 k

points are discontinuous?

Great question! Continuity requires the LHL, RHL, and function value to be equal. At

x = 0

f (0) = 0

. The LHL approaches

0

from the negative side, and the RHL approaches

0

from the positive side. Because the multiplier

x

is zero, it 'crushes' the jump of the GIF, making the function continuous at the origin. It is the only point where this happens!

Why does the LHL at

x = 2 k

give

[k - \frac{h}{2}] = k - 1

Think of the number line. If

k

is an integer,

k - tiny

is a number like

4.999

. The greatest integer of

4.999

4

, which is

k - 1

. This is the fundamental behavior of the GIF: it always rounds down to the nearest integer below the value.

Can I just graph this to find the answer?

Visualizing is a powerful tool, but in JEE Advanced, you must back it up with algebra. The graph helps you identify the candidates, but the limit calculation is the proof. Always use the limit definition to be absolutely certain, especially when the function involves a product of a variable and a GIF.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = x \cdot [\frac{x}{2}]$ , for $- 10 < x < 10$ , where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

Enter Numerical Value:

Visualized Solution

$A na l y z in g t h e F u n c t i o n f (x)$

$P r o p er t i eso f G r e a t es t I n t e g er F u n c t i o n$

$I d e n t i f y in g C r i t i c a l P o in t s$

$A ppl y in g D o main C o n s t r ain t s$

$L i s t in g P o t e n t ia l P o in t so f D i sco n t in u i t y$

$C h ec k in g C o n t in u i t y a t x = 0$

$L e f t H an d L imi t a t x = 0$

$R i g h t H an d L imi t a t x = 0$

$G e n er a l C a se : x = 2 k (k \neq = 0)$

$L e f t H an d L imi t a t x = 2 k$

$R i g h t H an d L imi t a t x = 2 k$

$C o n d i t i o n f or C o n t in u i t y$

$F ina l C o u n t o f D i sco n t in u i t i es$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Anomaly at the Origin

The General Case of Discontinuity

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingtheFunctionf(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PropertiesofGreatestIntegerFunction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IdentifyingCriticalPoints

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingDomainConstraints

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ListingPotentialPointsofDiscontinuity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CheckingContinuityatx=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LeftHandLimitatx=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RightHandLimitatx=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } GeneralCase:x=2k(k=0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LeftHandLimitatx=2k

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RightHandLimitatx=2k

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConditionforContinuity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCountofDiscontinuities

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Anomaly at the Origin

The General Case of Discontinuity

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [t] denote the greatest integer ≤t and limx→0​x[x4​]=A. Then the function, f(x)=[x2]sin(πx) is discontinuous, when x is equal to :

$A na l y z in g t h e F u n c t i o n f (x)$

$P r o p er t i eso f G r e a t es t I n t e g er F u n c t i o n$

$I d e n t i f y in g C r i t i c a l P o in t s$

$A ppl y in g D o main C o n s t r ain t s$

$L i s t in g P o t e n t ia l P o in t so f D i sco n t in u i t y$

$C h ec k in g C o n t in u i t y a t x = 0$

$L e f t H an d L imi t a t x = 0$

$R i g h t H an d L imi t a t x = 0$

$G e n er a l C a se : x = 2 k (k \neq = 0)$

$L e f t H an d L imi t a t x = 2 k$

$R i g h t H an d L imi t a t x = 2 k$

$C o n d i t i o n f or C o n t in u i t y$

$F ina l C o u n t o f D i sco n t in u i t i es$

The number of points of discontinuity of the function $f (x) = [\frac{x ^{2}}{2}] - [x], x \in [0, 4]$ , where $[\cdot]$ denotes the greatest integer function is ________.

Let $[x]$ be the greatest integer $\leq x$ . Then the number of points in the interval $(- 2, 1)$ where the function $f (x) = ∣ [x] ∣ + x - [x]$ is discontinuous, is _____.

The function $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , $[.]$ denotes the greatest integer function, is discontinuous at

The function $f (x) = [x]^{2} - [x^{2}]$ (where $[y]$ is the greatest integer less than or equal to $y$ ), is discontinuous at

Let $f : [- 1, 2] \to R$ be given by $f (x) = 2 x^{2} + x + [x^{2}] - [x]$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . The number of points, where $f$ is not continuous, is :

Let $[t]$ denote the greatest integer $\leq t$ . The number of points where the function $f (x) = [x] ∣ x^{2} - 1∣ + sin (\frac{π}{[ x ] + 3}) - [x + 1]$ , $x \in (- 2, 2)$ is not continuous is .

Let $[t]$ denotes the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin π x$ is discontinuous, when $x$ is equal to

Let $[t]$ denote the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin (π x)$ is discontinuous, when $x$ is equal to :