JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 18)
Let ƒ : [–1,3] $$ \to $$ R be defined as
$$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$$
where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at:
$$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$$
where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at:
only three points
four or more points
only two points
only one point
Explanation
$$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$
= $$ = \left\{ {\matrix{ { - x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {2x,} & {1 \le x < 2} \cr {x + 2,} & {2 \le x < 3} \cr {6,} & {x = 3} \cr } } \right.$$
f(-1) = $$\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1} \right)$$ = -( -1) - 1 = 0
f(-1+) = $$\mathop {\lim }\limits_{x \to - {1^ + }} \left( { - x - 1} \right)$$ = 0
$$ \therefore $$ f(x) is continuous at x = - 1
f(0-) =$$\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)$$
= -(0) - 1 = - 1
f(0) = $$\mathop {\lim }\limits_{x \to 0} \left( x \right)$$ = 0
f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)$$ = 0
$$ \therefore $$ f(x) is discontinuous at x = 0
f(1-) = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)$$ = 1
f(1) = $$\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$$ = 2
f(1+) = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$$ = 2
$$ \therefore $$ f(x) is discontinuous at x = 1
f(3-) = $$\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$$ = 3 + 2 = 5
f(3) = 6
$$ \therefore $$ f(x) is discontinuous at x = 3
So, f(x) is discontinuous at x = {0, 1, 3}.
= $$ = \left\{ {\matrix{ { - x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {2x,} & {1 \le x < 2} \cr {x + 2,} & {2 \le x < 3} \cr {6,} & {x = 3} \cr } } \right.$$
f(-1) = $$\mathop {\lim }\limits_{x \to - 1} \left( { - x - 1} \right)$$ = -( -1) - 1 = 0
f(-1+) = $$\mathop {\lim }\limits_{x \to - {1^ + }} \left( { - x - 1} \right)$$ = 0
$$ \therefore $$ f(x) is continuous at x = - 1
f(0-) =$$\mathop {\lim }\limits_{x \to {0^ - }} \left( { - x - 1} \right)$$
= -(0) - 1 = - 1
f(0) = $$\mathop {\lim }\limits_{x \to 0} \left( x \right)$$ = 0
f(0+) = $$\mathop {\lim }\limits_{x \to {0^ + }} \left( x \right)$$ = 0
$$ \therefore $$ f(x) is discontinuous at x = 0
f(1-) = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( x \right)$$ = 1
f(1) = $$\mathop {\lim }\limits_{x \to 1} \left( {2x} \right)$$ = 2
f(1+) = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {2x} \right)$$ = 2
$$ \therefore $$ f(x) is discontinuous at x = 1
f(3-) = $$\mathop {\lim }\limits_{x \to {3^ - }} \left( {x + 2} \right)$$ = 3 + 2 = 5
f(3) = 6
$$ \therefore $$ f(x) is discontinuous at x = 3
So, f(x) is discontinuous at x = {0, 1, 3}.
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