JEE Main 2019 (Online) 12th January Morning Slot | Differential Equations Question 168 | Mathematics

Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x log_e x, (x > 1). If 2y(2) = log_e 4 $$-$$ 1, then y(e) is equal to :

$$ - {e \over 2}$$

$$ - {{{e^2}} \over 2}$$

$${{{e^2}} \over 4}$$

$${e \over 4}$$

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

The solution of the differential equation,

$${{dy} \over {dx}}$$ = (x – y)², when y(1) = 1, is :

$$-$$ log_e $$\left| {{{1 + x - y} \over {1 - x + y}}} \right|$$ = x + y $$-$$ 2

log_e $$\left| {{{2 - x} \over {2 - y}}} \right|$$ = x $$-$$ y

log_e $$\left| {{{2 - y} \over {2 - x}}} \right|$$ = 2(y $$-$$ 1)

$$-$$ log_e $$\left| {{{1 - x + y} \over {1 + x - y}}} \right|$$ = 2(x $$-$$ 1)

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then

y(log_e2) = log_e4

y(x) is decreasing in (0, 1)

y(log_e2) = $${{{{\log }_e}2} \over 4}$$

y(x) is decreasing in $$\left( {{1 \over 2},1} \right)$$

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Let f be a differentiable function such that f '(x) = 7 - $${3 \over 4}{{f\left( x \right)} \over x},$$ (x > 0) and f(1) $$ \ne $$ 4. Then $$\mathop {\lim }\limits_{x \to 0'} \,$$ xf$$\left( {{1 \over x}} \right)$$ :

does not exist

exists and equals $${4 \over 7}$$

exists and equals 4

exists and equals 0

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