1
JEE Main 2024 (Online) 31st January Evening Shift
+4
-1

The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\circ} \mathrm{F}$$ and it decreases continuously as per the differential equation $$\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\circ} \mathrm{F}$$, then $$T(45)$$ is equal to

A
90$$^\circ$$ F
B
85$$^\circ$$ F
C
80$$^\circ$$ F
D
95$$^\circ$$ F
2
JEE Main 2024 (Online) 31st January Morning Shift
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \frac{\pi}{2}\right)$$ satisfying the condition $$y\left(\frac{\pi}{4}\right)=2$$. Then, $$y\left(\frac{\pi}{3}\right)$$ is

A
$$\sqrt{3}\left(2+\log _e 3\right)$$
B
$$\sqrt{3}\left(1+2 \log _e 3\right)$$
C
$$\sqrt{3}\left(2+\log _e \sqrt{3}\right)$$
D
$$\frac{\sqrt{3}}{2}\left(2+\log _e 3\right)$$
3
JEE Main 2024 (Online) 31st January Morning Shift
+4
-1

The solution curve of the differential equation $$y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0$$ passing through the point $$(e, 1)$$ is

A
$$\left|\log _e \frac{y}{x}\right|=y^2$$
B
$$\left|\log _e \frac{y}{x}\right|=x$$
C
$$\left|\log _e \frac{x}{y}\right|=y$$
D
$$2\left|\log _e \frac{x}{y}\right|=y+1$$
4
JEE Main 2024 (Online) 30th January Morning Shift
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$\sec x \mathrm{~d} y+\{2(1-x) \tan x+x(2-x)\} \mathrm{d} x=0$$ such that $$y(0)=2$$. Then $$y(2)$$ is equal to:

A
$$2\{\sin (2)+1\}$$
B
2
C
1
D
$$2\{1-\sin (2)\}$$
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