1
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
2
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
3
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$ is equal to :
A
$${\pi ^2}$$
B
$$2{\pi ^2}$$
C
$$4{\pi ^2}$$
D
$$4\pi$$
4
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is :
A
$$\left( {1,{1 \over 2}} \right)$$
B
$$\left( {1, - {1 \over 2}} \right)$$
C
$$\left( { - 1,{1 \over 2}} \right)$$
D
$$\left( { - 1, - {1 \over 2}} \right)$$
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