Let [t] denote the greatest integer $\le$ t. The number of points where the function $$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1],x \in ( - 2,2)$$ is not continuous is _____________.
Answer (integer)
2
Solution
$$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$$<br><br>$$f\left( x \right) = \left\{ {\matrix{
{ - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr
{ - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr
{\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr
{\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr
} } \right.$$
<br><br>$\therefore$ at x = -1, $\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$ and $\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$
<br><br>Hence continuous at x = –1
<br><br>Similarly check at x = 0,
<br><br>$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$ and $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$
<br><br>So, f(x) discontinuous
and at x = 0
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$ and $$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$$
<br><br>So, f(x) discontinuous
and at x = 1
<br><br>Hence 2 points of discontinuity.
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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