1
JEE Main 2023 (Online) 25th January Morning Shift
+4
-1

The value of $$\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}$$ is :

A
$${3 \over {2\sqrt 2 }}$$
B
$${3 \over 2}(\sqrt 2 + 1)$$
C
$$3(\sqrt 2 + 1)$$
D
$${{\sqrt 2 + 1} \over 2}$$
2
JEE Main 2023 (Online) 24th January Evening Shift
+4
-1

The set of all values of $$a$$ for which $$\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0$$, where [$$\alpha$$] denotes the greatest integer less than or equal to $$\alpha$$ is equal to

A
$$[-7.5,-6.5]$$
B
$$(-7.5,-6.5]$$
C
$$[-7.5,-6.5)$$
D
$$(-7.5,-6.5)$$
3
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1

$$\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}$$ is equal to

A
$${{n(n + 1)} \over 2}$$
B
n
C
n$$^2$$ + n
D
n$$^2$$
4
JEE Main 2023 (Online) 24th January Morning Shift
+4
-1

Let $$f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right.$$

Then at $$x=0$$

A
$$f$$ is continuous but $$f'$$ is not continuous
B
$$f$$ and $$f'$$ both are continuous
C
$$f$$ is continuous but not differentiable
D
$$f'$$ is continuous but not differentiable
EXAM MAP
Medical
NEET