1
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$
A
equals $$\pi$$ + 1
B
equals 0
C
does not exist
D
equals $$\pi$$
2
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
Let f : ($$-$$1, 1) $$\to$$ R be a function defined by f(x) = max $$\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly -
A
one element
B
three elements
C
five elements
D
two elements
3
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
For each t $$\in$$ R , let [t] be the greatest integer less than or equal to t

Then  $$\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$$
A
equals $$-$$ 1
B
equals 1
C
equals 0
D
does not exist
4
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
Let  $$f\left( x \right) = \left\{ {\matrix{ {\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr {8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr } } \right.$$

Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S
A
equals $$\left\{ { - 2, - 1,1,2} \right\}$$
B
equals $$\left\{ { - 2, - 1,0,1,2} \right\}$$
C
equals $$\left\{ { - 2,2} \right\}$$
D
is an empty set
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