4x2=y
y=8x+12
4x2=8x+12
x2−2x−3=0
x=−1,3
A=∫−13(8x+12−4x2)dx
A=28x2+12x−34x33−1=(4(9)+36−36)−(4−12+34)=36+8−34
=44−34=3132−4=3128
JEE Main 2020 — Mathematics Calculus
The area (in sq. units) of the region (x,y)∈R2∣4x2≤y≤8x+12 is
Held on 7 Jan 2020 · Verified 6 Jul 2026.
3125
3128
3124
3127
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in \mathbf{R}$. Consider the following two statements : (I) $f(x)$ is discontinous at $x=1$. (II) $f(x)$ is continous at $x=-1$. Then,
The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$ is:
The value of ∫₀¹ x·eˣ dx is:
Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
Work through every JEE Main Calculus PYQ, year by year.