WB JEE 2020 | Application of Derivatives Question 38 | Mathematics

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WB JEE 2020

MCQ (More than One Correct Answer)

-0

A particle is projected vertically upwards. If it has to stay above the ground for 12 sec, then

velocity of projection is 192 ft/sec

greatest height attained is 600 ft

velocity of projection is 196 ft/sec

greatest height attained is 576 ft

WB JEE 2020

MCQ (More than One Correct Answer)

-0

Tangent is drawn at any point P(x, y) on a curve, which passes through (1, 1). The tangent cuts X-axis and Y-axis at A and B respectively. If AP : BP = 3 : 1, then

The differential equation of the curve is $$3x{{dy} \over {dx}} + y = 0$$

the differential equation of the curve is $$3x{{dy} \over {dx}} - y = 0$$

the curve passes through $$\left( {{1 \over 8},2} \right)$$

the normal at (1, 1) is x + 3y = 4

WB JEE 2019

MCQ (More than One Correct Answer)

-0

Two particles A and B move from rest along a straight line with constant accelerations f and h, respectively. If A takes m seconds more than B and describes n units more than that of B acquiring the same speed, then

$$(f + h){m^2} = fhn$$

$$(f - h){m^2} = fhn$$

$$(h - f)n = {1 \over 2}fh{m^2}$$

$${1 \over 2}(f + h)n = fh{m^2}$$

WB JEE 2018

MCQ (More than One Correct Answer)

-0

A particle is in motion along a curve 12y = x³. The rate of change of its ordinate exceeds that of abscissa in

$$-$$ 2 < x < 2

x = $$ \pm $$ 2

x < $$-$$2

x > 2

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