NTA JEE Main 31st August 2021 Shift 1

If $$a_r = \cos\frac{2r\pi}{9} + i\sin\frac{2r\pi}{9}$$, $$r = 1, 2, 3, \ldots$$, $$i = \sqrt{-1}$$, then the determinant $$\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix}$$ is equal to:

If the following system of linear equations
$$2x + y + z = 5$$
$$x - y + z = 3$$
$$x + y + az = b$$
has no solution, then:

The function $$f(x) = |x^2 - 2x - 3| \cdot e^{9x^2-12x+4}$$ is not differentiable at exactly:

If the function $$f(x) = \begin{cases} \frac{1}{x}\log_e\left(\frac{1+\frac{x}{b}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2+1}-1}, & x > 0 \end{cases}$$ is continuous at $$x = 0$$, then $$\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$$ is equal to:

The number of real roots of the equation $$e^{4x} + 2e^{3x} - e^x - 6 = 0$$ is:

The integral $$\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} dx$$ is equal to: (where $$C$$ is a constant of integration)

Let $$f$$ be a non-negative function in $$[0, 1]$$ and twice differentiable in $$(0, 1)$$. If
$$\int_0^x \sqrt{1 - (f'(t))^2} dt = \int_0^x f(t) dt$$, $$0 \leq x \leq 1$$ and $$f(0) = 0$$, then $$\lim_{x \to 0} \frac{1}{x^2} \int_0^x f(t) dt$$:

If $$\frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y}$$, $$y(0) = 1$$, then $$y(1)$$ is equal to:

Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|$$ and the angle between $$\vec{a}$$ and $$\vec{b}$$ is 60°. If $$\frac{1}{8}\vec{a}$$ is a unit vector, then $$|\vec{b}|$$ is equal to:

Let the equation of the plane, that passes through the point $$(1, 4, -3)$$ and contains the line of intersection of the planes $$3x - 2y + 4z - 7 = 0$$ and $$x + 5y - 2z + 9 = 0$$, be $$\alpha x + \beta y + \gamma z + 3 = 0$$, then $$\alpha + \beta + \gamma$$ is equal to: