NTA JEE Main 27th July 2021 Shift 1

Let $$\alpha, \beta$$ be two roots of the equation $$x^2 + (20)^{1/4}x + (5)^{1/2} = 0$$. Then $$\alpha^8 + \beta^8$$ is equal to:

Let $$C$$ be the set of all complex numbers. Let
$$S_1 = \{z \in C \mid |z - 3 - 2i|^2 = 8\}$$,
$$S_2 = \{z \in C \mid \text{Re}(z) \geq 5\}$$ and
$$S_3 = \{z \in C \mid |z - \bar{z}| \geq 8\}$$.
Then the number of elements in $$S_1 \cap S_2 \cap S_3$$ is equal to

If the coefficients of $$x^7$$ in $$\left(x^2 + \frac{1}{bx}\right)^{11}$$ and $$x^{-7}$$ in $$\left(x - \frac{1}{bx^2}\right)^{11}$$, $$b \neq 0$$, are equal, then the value of $$b$$ is equal to:

If $$\sin \theta + \cos \theta = \frac{1}{2}$$, then $$16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$$ is equal to:

Two tangents are drawn from the point $$P(-1, 1)$$ to the circle $$x^2 + y^2 - 2x - 6y + 6 = 0$$. If these tangents touch the circle at points $$A$$ and $$B$$, and if $$D$$ is a point on the circle such that length of the segments $$AB$$ and $$AD$$ are equal, then the area of the triangle $$ABD$$ is equal to:

Let $$P$$ and $$Q$$ be two distinct points on a circle which has center at $$C(2, 3)$$ and which passes through origin $$O$$. If $$OC$$ is perpendicular to both the line segments $$CP$$ and $$CQ$$, then the set $$\{P, Q\}$$ is equal to

Let
$$A = \{(x, y) \in R \times R \mid 2x^2 + 2y^2 - 2x - 2y = 1\}$$
$$B = \{(x, y) \in R \times R \mid 4x^2 + 4y^2 - 16y + 7 = 0\}$$ and
$$C = \{(x, y) \in R \times R \mid x^2 + y^2 - 4x - 2y + 5 \leq r^2\}$$.
Then the minimum value of $$|r|$$ such that $$A \cup B \subseteq C$$ is equal to

A ray of light through $$(2, 1)$$ is reflected at a point $$P$$ on the $$y$$-axis and then passes through the point $$(5, 3)$$. If this reflected ray is the directrix of an ellipse with eccentricity $$\frac{1}{3}$$ and the distance of the nearer focus from this directrix is $$\frac{8}{\sqrt{53}}$$, then the equation of the other directrix can be:

Let $$f : R \rightarrow R$$ be a function such that $$f(2) = 4$$ and $$f'(2) = 1$$. Then, the value of $$\lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2}$$ is equal to:

The compound statement $$(P \vee Q) \wedge (\sim P) \Rightarrow Q$$ equivalent to: