🎓 JEE MAIN📅 Year: 2023📚 Mathematics🏷 Area enclosed between the curves Definite Integration
3
Let $\Delta$ be the area of the region $\{(x,y)\in\mathbb{R}^{2}:\ x^{2}+y^{2}\le 21,\ y^{2}\le 4x,\ x\ge 1\}$.
Then $\dfrac{1}{2}\Big(\Delta-21\sin^{-1}\!\dfrac{2}{\sqrt{7}}\Big)$ is equal to:
The value of $\dfrac{e^{-\pi/4}+\displaystyle\int_{0}^{\pi/4} e^{-x}\tan^{50}x\,dx}{\displaystyle\int_{0}^{\pi/4} e^{-x}\big(\tan^{49}x+\tan^{51}x\big)\,dx}$ is:
For a triangle $ABC$,
$\overrightarrow{AB}=-2\hat i+\hat j+3\hat k$
$\overrightarrow{CB}=\alpha\hat i+\beta\hat j+\gamma\hat k$
$\overrightarrow{CA}=4\hat i+3\hat j+\delta\hat k$
If $\delta>0$ and the area of the triangle $ABC$ is $5\sqrt{6}$, then $\overrightarrow{CB}\cdot\overrightarrow{CA}$ is equal to:
Let $[x]$ denote the greatest integer $\le x$. Consider the function
$$f(x)=\max\{x^{2},\,1+[x]\}.$$
Then the value of the integral $\displaystyle \int_{0}^{2} f(x)\,dx$ i
Let $x=x(y)$ be the solution of the differential equation
$2(y+2)\log_e(y+2)\,dx+\big(x+4-2\log_e(y+2)\big)\,dy=0,\quad y>-1$
with $x\big(e^{4}-2\big)=1$. Then $x\big(e^{9}-2\big)$ is equal to:
🎓 JEE MAIN📅 Year: 2023📚 Mathematics🏷 Area enclosed between the curves Definite Integration
2
Let
$$A=\{(x,y)\in\mathbb{R}^{2}:\ y\ge 0,\ 2x\le y\le \sqrt{4-(x-1)^{2}}\}$$
and
$$B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ 0\le y\le \min\{2x,\ \sqrt{4-(x-1)^{2}}\}\}.$$
Then the ratio of the area of $A$ to the area of $B$ is
If $\displaystyle \int_{0}^{1} \frac{1}{(5+2x-2x^2)\,(1+e^{\,2-4x})}\,dx=\frac{1}{\alpha}\log_e\!\left(\frac{\alpha+1}{\beta}\right),\ \alpha,\beta>0,$ then $\alpha^4-\beta^4$ is equal to: