WB JEE 2022 | Application of Derivatives Question 28 | Mathematics

WB JEE 2022

MCQ (More than One Correct Answer)

-0

From a balloon rising vertically with uniform velocity v ft/sec a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after 4 sec is [g = 32 ft/sec²]

220 ft

240 ft

256 ft

260 ft

WB JEE 2021

MCQ (More than One Correct Answer)

-0

The greatest and least value of $$f(x) = {\tan ^{ - 1}} - {1 \over 2}\,ln \,x\,on\,\left[ {{1 \over {\sqrt 3 }},\sqrt 3 } \right]$$ are

$${f_{\min }} = \sqrt 3 - 1$$

$${f_{\max }} = {\pi \over 6} + {1 \over 4}\ln 3$$

$${f_{\min }} = {\pi \over 3} - {1 \over 4}\ln 3$$

$${f_{\max }} = {\pi \over {12}} + \ln 5$$

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

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