1
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)
2
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)
3
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
The function

$$f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}}$$ is not differentiable at exactly :
A
four points
B
three points
C
two points
D
one point
4
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
If the function
$$f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$$ is continuous

at x = 0, then $${1 \over a} + {1 \over b} + {4 \over k}$$ is equal to :
A
$$-$$5
B
5
C
$$-$$4
D
4
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