JEE Main 2021 (Online) 25th February Morning Shift | Differential Equations Question 87 | Mathematics

JEE Main 2021 (Online) 25th February Morning Shift

MCQ (Single Correct Answer)

-1

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $${{{x^2} - 4x + y + 8} \over {x - 2}}$$, then this curve also passes through the point :

(4, 4)

(5, 5)

(5, 4)

(4, 5)

JEE Main 2021 (Online) 24th February Morning Shift

MCQ (Single Correct Answer)

-1

The population P = P(t) at time 't' of a certain species follows the differential equation
$${{dP} \over {dt}}$$ = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :

$${\log _e}18$$

$${1 \over 2}{\log _e}18$$

2$${\log _e}18$$

$${\log _e}9$$

JEE Main 2020 (Online) 6th September Evening Slot

MCQ (Single Correct Answer)

-1

If $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$ is the solution

of the differential equation,
$${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$$,

$$0 < x < {\pi \over 2}$$, then the function p(x) is equal to :

cot x

sec x

tan x

cosec x

JEE Main 2020 (Online) 6th September Morning Slot

MCQ (Single Correct Answer)

-1

The general solution of the differential equation
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :

(where C is a constant of integration)

$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$

$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$

$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$

$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$

PREVIOUS NEXT