JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 22)
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
2
Explanation
$$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$$
$$f\left( x \right) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.$$
$$ \therefore $$ at x = -1, $$\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$$ and $$\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$$
Hence continuous at x = –1
Similarly check at x = 0,
$$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$$ and $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$
So, f(x) discontinuous and at x = 0
$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$ and $$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$$
So, f(x) discontinuous and at x = 1
Hence 2 points of discontinuity.
$$f\left( x \right) = \left\{ {\matrix{ { - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr { - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr {\sin {\pi \over 3} + 1,} & {0 \le x < 1} \cr {\left| {{x^2} - 1} \right| + {1 \over {\sqrt 2 }} - 2,} & {1 \le x < 2} \cr } } \right.$$
$$ \therefore $$ at x = -1, $$\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 1$$ and $$\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 1$$
Hence continuous at x = –1
Similarly check at x = 0,
$$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1$$ and $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$
So, f(x) discontinuous and at x = 0
$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1 + {{\sqrt 3 } \over 2}$$ and $$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {1 \over {\sqrt 2 }} - 2$$
So, f(x) discontinuous and at x = 1
Hence 2 points of discontinuity.
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