JEE Main 2021 (18 March Shift 1)

For the four circles $$M, N, O$$ and $$P$$, following four equations are given:
Circle M: $$x^2 + y^2 = 1$$
Circle N: $$x^2 + y^2 - 2x = 0$$
Circle O: $$x^2 + y^2 - 2x - 2y + 1 = 0$$
Circle P: $$x^2 + y^2 - 2y = 0$$
If the centre of circle M is joined with centre of circle N, further centre of circle N is joined with centre of circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a:

If $$\lim_{x \to 0} \frac{\sin^{-1}x - \tan^{-1}x}{3x^3}$$ is equal to $$L$$, then the value of $$(6L + 1)$$ is:

Let $$A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$$ and $$2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$$. If $$Tr(A)$$ denotes the sum of all diagonal elements of the matrix $$A$$, then Tr($$A$$) - Tr($$B$$) has value equal to:

Let $$\alpha, \beta, \gamma$$ be the real roots of the equation, $$x^3 + ax^2 + bx + c = 0$$, $$(a, b, c \in R$$ and $$a, b \neq 0)$$. If the system of equations (in $$u, v, w$$) given by $$\alpha u + \beta v + \gamma w = 0$$, $$\beta u + \gamma v + \alpha w = 0$$, $$\gamma u + \alpha v + \beta w = 0$$ has non-trivial solution, then the value of $$\frac{a^2}{b}$$ is:

The real valued function $$f(x) = \frac{\text{cosec}^{-1} x}{\sqrt{x - [x]}}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is defined for all $$x$$ belonging to:

If the functions are defined as $$f(x) = \sqrt{x}$$ and $$g(x) = \sqrt{1-x}$$, then what is the common domain of the following functions: $$f+g, f-g, f/g, g/f, g-f$$, where $$(f \pm g)(x) = f(x) \pm g(x)$$, $$(f/g)(x) = \frac{f(x)}{g(x)}$$:

If $$f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$$ is differentiable at every point of the domain, then the values of $$a$$ and $$b$$ are respectively:

The integral $$\int \frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{4x^2-4x+6}}dx$$ is equal to (where $$c$$ is a constant of integration):

The differential equation satisfied by the system of parabolas $$y^2 = 4a(x+a)$$ is:

A vector $$\vec{a}$$ has components $$3p$$ and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $$\vec{a}$$ has components $$p+1$$ and $$\sqrt{10}$$, then a value of $$p$$ is equal to: