QJEEM19Ma8318: JEE Main 2019 Mathematics - Class 12 JEEM19Ma8318 Solution: If ,x 4 , where [x] denotes the greatest integer function, then

Question

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If

f(x) = [x] - \left[ {{x \over 4}} \right]

,x

\in

4 , where [x] denotes the greatest integer function, then

A Both $\mathop {\lim }\limits_{x \to 4 - } f(x)$ and $\mathop {\lim }\limits_{x \to 4 + } f(x)$ exist but are not equal
B f is continuous at x = 4
C $\mathop {\lim }\limits_{x \to 4 + } f(x)$ exists but $\mathop {\lim }\limits_{x \to 4 - } f(x)$ does not exist
D $\mathop {\lim }\limits_{x \to 4 - } f(x)$ exists but $\mathop {\lim }\limits_{x \to 4 + } f(x)$ does not exist

Accepted Answer

f(x) = [x] - \left[ {{x \over 4}} \right]

Here check continuty at x = 4

LHL =

\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)

=

\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]

=

\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]

= 3 - 0 = 3

h

\to

0 means h > 0 (very small positive value)

So 4 - h = 3.something (means less than 4)

\therefore

[4 - h] = [3.something] = 3

and

\left[ {{{4 - h} \over 4}} \right]

= [0.something] = 0

RHL =

\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)

=

\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]

=

\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]

= 4 - 1 = 3

Here [4 + h] = [4.something] = 4

and

\left[ {{{4 + h} \over 4}} \right]

= [1.something] = 1

And f(4) =

\left[ 4 \right] - \left[ {{4 \over 4}} \right]

= 4 - 1 = 3

As LHL = RHL = f(4)

\therefore

f is continuous at x = 4.

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