1
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Let S be the set of all functions ƒ : [0,1] $$\to$$ R, which are continuous on [0,1] and differentiable on (0,1). Then for every ƒ in S, there exists a c $$\in$$ (0,1), depending on ƒ, such that
A
$$\left| {f(c) - f(1)} \right| < \left| {f'(c)} \right|$$
B
$$\left| {f(c) + f(1)} \right| < \left( {1 + c} \right)\left| {f'(c)} \right|$$
C
$$\left| {f(c) - f(1)} \right| < \left( {1 - c} \right)\left| {f'(c)} \right|$$
D
None
2
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
$$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to
A
e
B
e2
C
$${1 \over {{e^2}}}$$
D
$${1 \over e}$$
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
$$\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$$ is :
A
6
B
1
C
3
D
2
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$\in$$ R. If f(x) attains maximum value at $$\alpha$$ and g(x) attains minimum value at $$\beta$$, then $$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :
A
$${1 \over 2}$$
B
$$-{1 \over 2}$$
C
$${3 \over 2}$$
D
$$-{3 \over 2}$$
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