JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 17)
Let [t] denote the greatest integer $$ \le $$ t
and $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A$$.
Then the function, f(x) = [x2]sin($$\pi $$x) is discontinuous, when x is equal to :
Then the function, f(x) = [x2]sin($$\pi $$x) is discontinuous, when x is equal to :
$$\sqrt {A + 1} $$
$$\sqrt {A + 5} $$
$$\sqrt {A + 21} $$
$$\sqrt {A} $$
Explanation
A = $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$$
= $$\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$$
= 4
Now, when x = $$\sqrt {A + 1} $$ = $$\sqrt 5 $$, f(x) = [x2]sin($$\pi $$x) is discontinuous at this non integer point.
But at x = 2, 3 and 5, f(x) is continuous.
= $$\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$$
= 4
Now, when x = $$\sqrt {A + 1} $$ = $$\sqrt 5 $$, f(x) = [x2]sin($$\pi $$x) is discontinuous at this non integer point.
But at x = 2, 3 and 5, f(x) is continuous.
Comments (0)
