1
WB JEE 2020
MCQ (Single Correct Answer)
+2
-0.5
Change Language
Let $$y = {1 \over {1 + x + lnx}}$$, then
A
$$x{{dy} \over {dx}} + y = x$$
B
$$x{{dy} \over {dx}} = y(y\ln x - 1)$$
C
$${x^2}{{dy} \over {dx}} = {y^2} + 1 - {x^2}$$
D
$$x{\left( {{{dy} \over {dx}}} \right)^2} = y - x$$
2
WB JEE 2020
MCQ (Single Correct Answer)
+2
-0.5
Change Language
Consider the curve $$y = b{e^{ - x/a}}$$, where a and b are non-zero real numbers. Then
A
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (0, 0)
B
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve, where the curve crosses the axis of y
C
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (a, 0)
D
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (2a, 0)
3
WB JEE 2020
MCQ (More than One Correct Answer)
+2
-0
Change Language
The area of the figure bounded by the parabola $$x = - 2{y^2},\,x = 1 - 3{y^2}$$ is
A
$${1 \over 3}$$sq unit
B
$${4 \over 3}$$sq unit
C
1 sq unit
D
2 sq units
4
WB JEE 2020
MCQ (More than One Correct Answer)
+2
-0
Change Language
A particle is projected vertically upwards. If it has to stay above the ground for 12 sec, then
A
velocity of projection is 192 ft/sec
B
greatest height attained is 600 ft
C
velocity of projection is 196 ft/sec
D
greatest height attained is 576 ft
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