JEE MAIN 2022 Previous Year Questions (PYQs) – Page 2 of 31

$\int_{0}^{20\pi} (|\sin x| + |\cos x|)^{2} \, dx$ is equal to:

Let the solution curve $y=f(x)$ of the differential equation $\dfrac{dy}{dx}+\dfrac{xy}{x^{2}-1}=\dfrac{x^{4}+2x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$, pass through the origin. Then $\displaystyle \int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x)\,dx$ is equal to:

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4x-6=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}+2y-7=0$. If $PQ$ is a diameter of the circle $x^{2}+y^{2}+2ax+2by+c=0$, then the value of $(a+b-c)$ is _________. (A)

If the line $x-1=0$ is a directrix of the hyperbola $kx^{2}-y^{2}=6$, then the hyperbola passes through the point:

$ \text{If } 0 < x < \tfrac{1}{\sqrt{2}} \text{ and } \tfrac{\sin^{-1}x}{\alpha} = \tfrac{\cos^{-1}x}{\beta}, \text{ then the value of } \sin!\left(\tfrac{2\pi\alpha}{\alpha+\beta}\right) \text{ is :}$

$ \text{The integral } \int \dfrac{\left(1-\tfrac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{1+\tfrac{2}{\sqrt{3}}\sin 2x},dx \text{ is equal to :}$

$ \text{The area bounded by the curves } y=\lvert x^{2}-1\rvert \text{ and } y=1 \text{ is :}$

$ \text{Let } R_1 \text{ and } R_2 \text{ be two relations defined on } \mathbb{R} \text{ by } a R_1 b \Leftrightarrow ab \ge 0 \text{ and } aR_2b \Leftrightarrow a \ge b. \text{ Then,}$

$ \text{Let } f,g : \mathbb{N} - \{1\} \to \mathbb{N} \text{ be functions defined by } f(a) = \alpha, \text{ where } \alpha \text{ is the maximum of the powers of those primes } p \text{ such that } p^\alpha \text{ divides } a, \text{ and } g(a) = a+1, \text{ for all } a \in \mathbb{N} - \{1\}. \text{ Then, the function } f+g \text{ is} $

$ \text{Let the minimum value } v_{0} \text{ of } v=\lvert z\rvert^{2}+\lvert z-3\rvert^{2}+\lvert z-6i\rvert^{2},\ z\in\mathbb{C} \text{ be attained at } z=z_{0}. \text{ Then } \lvert 2z_{0}^{2}-\overline{z_{0}}^{\,3}+3\rvert^{2}+v_{0}^{2} \text{ is equal to:} $