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
JEE Main Papers
2023
2021
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12