1
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
$$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$
A
is equal to 0.
B
is equal to $$\sqrt e$$.
C
is equal to 1.
D
does not exist.
2
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If the function
$$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ is
twice differentiable, then the ordered pair (k1, k2) is equal to :
A
$$\left( {{1 \over 2},-1} \right)$$
B
(1, 1)
C
(1, 0)
D
$$\left( {{1 \over 2},1} \right)$$
3
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If $$\alpha$$ is positive root of the equation, p(x) = x2 - x - 2 = 0, then

$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to :
A
$${1 \over \sqrt2}$$
B
$${1 \over 2}$$
C
$${3 \over \sqrt2}$$
D
$${3 \over 2}$$
4
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
Let $$f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$$ be a differentiable function such that f(1) = e and
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :
A
$${1 \over e}$$
B
e
C
$${1 \over 2e}$$
D
2e
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