For the following questions answer them individually
$$\tan^{-1}\frac{1+\sqrt{3}}{3+\sqrt{3}} + \sec^{-1}\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}} =$$
The equation $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has:
Let $$f(x) = \begin{cases} x^2\sin\frac{1}{x}; & x \neq 0 \\ 0; & x = 0 \end{cases}$$, then at $$x = 0$$
The area enclosed between the curves $$y^2 + 4x = 4$$ and $$y - 2x = 2$$ is
Let $$y = y(x)$$ be the solution of the differential equation $$x^3 dy + (xy - 1)dx = 0$$, $$x > 0$$, $$y\left(\frac{1}{2}\right) = 3 - e$$. Then $$y(1)$$ is equal to
Let $$\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$$, $$\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{v} \cdot \vec{w} = 2$$ and $$\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$$, then $$\vec{u} \cdot \vec{w}$$ is equal to
Let $$PQR$$ be a triangle. The points $$A, B$$ and $$C$$ are on the sides $$QR, RP$$ and $$PQ$$ respectively such that $$\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$$. Then $$\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$$ is equal to
The distance of the point $$(7, -3, -4)$$ from the plane containing the points $$(2, -3, 1)$$, $$(-1, 1, -2)$$ and $$(3, -4, 2)$$ is equal to:
The distance of the point $$(-1, 9, -16)$$ from the plane $$2x + 3y - z = 5$$ measured parallel to the line $$\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$$ is
Let $$\Omega$$ be the sample space and $$A \subseteq \Omega$$ be an event. Given below are two statements:
(S1): If $$P(A) = 0$$, then $$A = \phi$$
(S2): If $$P(A) = 1$$, then $$A = \Omega$$
Then
