1
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
Let f : R $$\to$$ R be a function defined as

$$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \hfill \cr b,\,if\,x\, = 0 \hfill \cr {{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \hfill \cr} \right.$$

If f is continuous at x = 0, then the value of a + b is equal to :
A
$$-$$3
B
$$-$$2
C
$$- {5 \over 2}$$
D
$$- {3 \over 2}$$
2
JEE Main 2021 (Online) 18th March Morning Shift
+4
-1
If $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$$ is equal to L, then the value of (6L + 1) is
A
$${1 \over 6}$$
B
$${1 \over 2}$$
C
6
D
2
3
JEE Main 2021 (Online) 18th March Morning Shift
+4
-1
If $$f(x) = \left\{ {\matrix{ {{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr {a{x^2} + b} & {;\,|x|\, < 1} \cr } } \right.$$ is differentiable at every point of the domain, then the values of a and b are respectively :
A
$${1 \over 2},{1 \over 2}$$
B
$${1 \over 2}, - {3 \over 2}$$
C
$${5 \over 2}, - {3 \over 2}$$
D
$$- {1 \over 2},{3 \over 2}$$
4
JEE Main 2021 (Online) 17th March Evening Shift
+4
-1
The value of the limit

$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ is equal to :
A
0
B
$$-$$$${1 \over 2}$$
C
$${1 \over 4}$$
D
$$-$$$${1 \over 4}$$
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