1
JEE Main 2024 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The area of the region enclosed by the parabolas $$y=4 x-x^2$$ and $$3 y=(x-4)^2$$ is equal to :

A
$$\frac{32}{9}$$
B
$$\frac{14}{3}$$
C
4
D
6
2
JEE Main 2024 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f, g:(0, \infty) \rightarrow \mathbb{R}$$ be two functions defined by $$f(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d t$$ and $$g(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d t$$. Then, the value of $$9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)$$ is equal to :

A
10
B
9
C
8
D
6
3
JEE Main 2024 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$$ be strictly increasing function such that $$\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$$ is equal to

A
0
B
4
C
1
D
7/5
4
JEE Main 2024 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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
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