JEE Main 2020 (Online) 4th September Evening Slot | Differential Equations Question 93 | Mathematics

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

The solution of the differential equation

$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:

(where c is a constant of integration)

$$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$$

$$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$$

x – log_e(y+3x) = C

x – 2log_e(y+3x) = C

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation,
xy'- y = x²(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to

$$1 + {\pi \over 2}$$

$$2 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$

$$2 + {\pi \over 2}$$

$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-1

Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x) $$ \ge $$ 1 and f''(x) $$ \ge $$ 4, for all x $$ \in $$ (1, 6), then :

f(5) $$ \le $$ 10

f(5) + f'(5) $$ \ge $$ 28

f(5) + f'(5) $$ \le $$ 26

f'(5) + f''(5) $$ \le $$ 20

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

The solution curve of the differential equation,

(1 + e^-x)(1 + y²)$${{dy} \over {dx}}$$ = y²,

which passes through the point (0, 1), is :

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$

y² + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$

y² = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$