1
JEE Main 2022 (Online) 25th June Evening Shift
+4
-1

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to

A
$${1 \over {12}}$$
B
$$-$$$${1 \over {18}}$$
C
$$-$$$${1 \over {12}}$$
D
$${1 \over {6}}$$
2
JEE Main 2022 (Online) 24th June Evening Shift
+4
-1

Let $$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & {,\,|x| < 1} \cr 1 & {,\,otherwise} \cr } } \right.$$

where [t] denotes greatest integer $$\le$$ t. If m is the number of points where $$f$$ is not continuous and n is the number of points where $$f$$ is not differentiable, then the ordered pair (m, n) is :

A
(3, 3)
B
(2, 4)
C
(2, 3)
D
(3, 4)
3
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
If $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ and $$\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$$ are the roots of the equation, ax2 + bx $$-$$ 4 = 0, then the ordered pair (a, b) is :
A
(1, $$-$$3)
B
($$-$$1, 3)
C
($$-$$1, $$-$$3)
D
(1, 3)
4
JEE Main 2021 (Online) 31st August Evening Shift
+4
-1
Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
A
f''(x) = 0 for all x $$\in$$ (0, 2)
B
f''(x) = 0 for some x $$\in$$ (0, 2)
C
f'(x) = 0 for some x $$\in$$ [0, 2]
D
f''(x) > 0 for all x $$\in$$ (0, 2)
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