1
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let a function ƒ : [0, 5] $$ \to $$ R be continuous, ƒ(1) = 3 and F be defined as :

$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$

Then for the function F, the point x = 1 is :
A
a point of inflection.
B
a point of local maxima.
C
a point of local minima.
D
not a critical point.
2
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :
A
$$\sqrt 2 e$$
B
$${1 \over 2}\sqrt 3 e$$
C
$${e \over {\sqrt 2 }}$$
D
$$\sqrt 3 e$$
3
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C$$

where C is a constant of integration, then the ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to :
A
(–1, 1 – tan$$\theta $$)
B
(1, 1 + tan$$\theta $$)
C
(–1, 1 + tan$$\theta $$)
D
(1, 1 – tan$$\theta $$)
4
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be :
A
$$\sqrt {10} $$
B
$$\sqrt {7} $$
C
$$\sqrt {{{17} \over 2}} $$
D
$$\sqrt {8} $$
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