Question

Let $\left[ x \right]$be the greatest integer less than or equal to x. Then, at which of the following point(s), the function$f(x) = x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right)$is discontinuous?
(A). $x = 2$
(B). $x = 0$
(C). $x = 1$
(D). $x = - 1$

Answer 1

To solve the question we need to check the continuity of function at each point given in the options. To check the discontinuity of the function f(x) at first we have to determine the LHL (Left Hand Limit) and RHL (Right Hand limit) of the particular points. When the point at which the LHL and RHL are equal, then the function f(x) is continuous at that particular point.

Complete step-by-step answer :
Now we will find out the LHL and RHL of the points given in the options.
We know for a point $x = a$the LHL and RHL are given by
$LHL = \mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(a - h)$ ………………………………………… (1)
And
$RHL = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(a + h)$ ……………………….…………. (2)
Given function is$f(x) = x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right)$ ……………………………………… (3)
Applying this formula let’s find out the LHL and RHL at$x = 2$.

LHL = \mathop {\lim }\limits_{x \to {2^ - }} f(x) \\\ = \mathop {\lim }\limits_{x \to {2^ - }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\\ $$ ………………………………………. (4) Now here we see that as$$x \to {2^ - }$$, the value of x approaches to 2 from left side and the value of x must lie between 1 and 2 that means $$1 \leqslant x \leqslant 2$$.Therefore $$\left[ x \right] = 1$$ now the eq. (4) reduces to

LHL = \mathop {\lim }\limits_{x \to {2^ - }} x\cos \left( {\pi \left( {x + 1} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (2 - h)\cos \left( {\pi \left( {2 - h + 1} \right)} \right) \\
= 2\cos 3\pi \\
= - 2 \\

$$………………………………... (5) And$$

RHL = \mathop {\lim }\limits_{x \to {2^ + }} f(x) \\
= \mathop {\lim }\limits_{x \to {2^ + }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

………………………………. (6) Now here we see that as$$x \to {2^ + }$$, the value of x approaches to 2 from right side and the value of x must lie between 2 and 3 that means $$2 \leqslant x \leqslant 3$$.Therefore $$\left[ x \right] = 2$$ now the eq. (6) reduces to

RHL = \mathop {\lim }\limits_{x \to {2^ + }} x\cos \left( {\pi \left( {x + 2} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (2 - h)\cos \left( {\pi \left( {2 - h + 2} \right)} \right) \\
= 2\cos 4\pi \\
= 2 \\

………………………………………… (7) Here $$LHL \ne RHL$$,hence f(x) is discontinuous at$$x = 2$$. Then for$$x = 0$$,

LHL = \mathop {\lim }\limits_{x \to {0^ - }} f(x) \\
= \mathop {\lim }\limits_{x \to {0^ - }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\
\\

………………………………………………. (8) Now here we see that as$$x \to {0^ - }$$, the value of x approaches to 0 from left side and the value of x must lie between -1 and 0 that means $$ - 1 \leqslant x \leqslant 0$$.Therefore $$\left[ x \right] = - 1$$ now the eq. (8) reduces to

LHL = \mathop {\lim }\limits_{x \to {0^ - }} x\cos \left( {\pi \left( {x - 1} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (0 - h)\cos \left( {\pi \left( {0 - h - 1} \right)} \right) \\
= 0 \times \cos \left( { - \pi } \right) \\
= 0 \\

$$…………………………………. (9)$$

RHL = \mathop {\lim }\limits_{x \to {0^ + }} f(x) \\
= \mathop {\lim }\limits_{x \to {0^ + }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

………………………………… (10) Now here we see that as$$x \to {0^ + }$$, the value of x approaches to 0 from right side and the value of x must lie between 0 and 1 that means $$0 \leqslant x \leqslant 1$$.Therefore $$\left[ x \right] = 0$$ now the eq. (10) reduces to

RHL = \mathop {\lim }\limits_{x \to {0^ + }} x\cos \left( {\pi \left( {x + 0} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (0 - h)\cos \left( {\pi \left( {0 - h} \right)} \right) \\
= 0 \times \cos 0 \\
= 0 \\

…………………………………. (11) Here $$LHL = RHL$$,hence f(x) is continuous at $$x = 0$$. Then for$$x = 1$$,

LHL = \mathop {\lim }\limits_{x \to {1^ - }} f(x) \\
= \mathop {\lim }\limits_{x \to {1^ - }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

…………………………. (12) Now here we see that as$$x \to {1^ - }$$, the value of x approaches to 1 from left side and the value of x must lie between 0 and 1 that means $$0 \leqslant x \leqslant 1$$.Therefore $$\left[ x \right] = 0$$ now the eq. (12) reduces to For $$x = 1$$

LHL = \mathop {\lim }\limits_{x \to {1^ - }} x\cos \left( {\pi \left( {x + 0} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (1 - h)\cos \left( {\pi \left( {1 - h} \right)} \right) \\
= 1 \times \cos \pi \\
= \cos \pi \\
= - 1 \\

$$…………………………………….. (13)$$

RHL = \mathop {\lim }\limits_{x \to {1^ + }} f(x) \\
= \mathop {\lim }\limits_{x \to {1^ + }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

……………………………………….. (14) Now here we see that as$$x \to {1^ + }$$, the value of x approaches to 1 from right side and the value of x must lie between 0 and 1 that means $$1 \leqslant x \leqslant 2$$.Therefore $$\left[ x \right] = 1$$ now the eq. (14) reduces to

RHL = \mathop {\lim }\limits_{x \to {1^ + }} x\cos \left( {\pi \left( {x + 1} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} (1 - h)\cos \left( {\pi \left( {1 - h + 1} \right)} \right) \\
= 1 \times \cos 2\pi \\
= 1 \\

…………………………….. (15) Here $$LHL \ne RHL$$,hence f(x) is discontinuous at $$x = 1$$. For$$x = - 1$$,

LHL = \mathop {\lim }\limits_{x \to - {1^ - }} f(x) \\
= \mathop {\lim }\limits_{x \to - {1^ - }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

……………………………………….. (16) Now here we see that as$$x \to - {1^ - }$$, the value of x approaches to -1 from left side and the value of x must lie between 0 and 1 that means $$ - 2 \leqslant x \leqslant - 1$$.Therefore $$\left[ x \right] = - 2$$ now the eq. (16) reduces to

LHL = \mathop {\lim }\limits_{x \to - {1^ - }} x\cos \left( {\pi \left( {x - 2} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} ( - 1 - h)\cos \left( {\pi \left( { - 1 - h - 2} \right)} \right) \\
= - 1 \times \cos ( - 3\pi ) \\
= - \cos 3\pi \\
= 1 \\

$$……………………………….. (17)$$

RHL = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) \\
= \mathop {\lim }\limits_{x \to - {1^ + }} x\cos \left( {\pi \left( {x + \left[ x \right]} \right)} \right) \\

…………………….. (18) Now here we see that as$$x \to - {1^ + }$$, the value of x approaches to -1from right side and the value of x must lie between -1 and 0 that means $$ - 1 \leqslant x \leqslant 0$$.Therefore $$\left[ x \right] = - 1$$ now the eq. (18) reduces to

RHL = \mathop {\lim }\limits_{x \to - {1^ + }} x\cos \left( {\pi \left( {x - 1} \right)} \right) \\
= \mathop {\lim }\limits_{h \to 0} ( - 1 - h)\cos \left( {\pi \left( { - 1 - h - 1} \right)} \right) \\
= - 1 \times \cos ( - 2\pi ) \\
= - 1 \\

…………………………….. (19) Here $$LHL \ne RHL$$,hence f(x) is discontinuous at $$x = - 1$$. Therefore the options (A), (C) and (D) are correct. **Note** : The statement$$x \to {a^ - }$$, the value of x approaches to a from left side that means x is a number less than a and very very close to a and right hand limits of a function at a given point. Therefore $$x \to {a^ - }$$is equivalent to$$x = a - h$$. Similarly the statement$$x \to {a^ + }$$, the value of x approaches a from the right hand side that means x is a number less than a and very very close to a. Therefore $$x \to {a^ + }$$is equivalent to$$x = a + h$$.