1
JEE Main 2020 (Online) 6th September Evening Slot
+4
-1
Let f : R $$\to$$ R be a function defined by
f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable. Then :
A
{0, 1}
B
{0}
C
$$\phi$$(an empty set)
D
{1}
2
JEE Main 2020 (Online) 6th September Evening Slot
+4
-1
For all twice differentiable functions f : R $$\to$$ R,
with f(0) = f(1) = f'(0) = 0
A
f''(x) $$\ne$$ 0, at every point x $$\in$$ (0, 1)
B
f''(x) = 0, for some x $$\in$$ (0, 1)
C
f''(0) = 0
D
f''(x) = 0, at every point x $$\in$$ (0, 1)
3
JEE Main 2020 (Online) 6th September Morning Slot
+4
-1
$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$
A
is equal to 0
B
is equal to $${1 \over 2}$$
C
does not exist
D
is equal to $$- {1 \over 2}$$
4
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.
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