For the following questions answer them individually
Let $$f(x) = ax^3 + bx^2 + cx + 41$$ be such that $$f(1) = 40$$, $$f'(1) = 2$$ and $$f''(1) = 4$$. Then $$a^2 + b^2 + c^2$$ is equal to:
A variable line $$L$$ passes through the point $$(3, 5)$$ and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle OAB, where O is the origin, is :
Let $$\int \frac{2 - \tan x}{3 + \tan x} dx = \frac{1}{2}(\alpha x + \log_e|\beta \sin x + \gamma \cos x|) + C$$, where $$C$$ is the constant of integration. Then $$\alpha + \frac{\gamma}{\beta}$$ is equal to :
The parabola $$y^2 = 4x$$ divides the area of the circle $$x^2 + y^2 = 5$$ in two parts. The area of the smaller part is equal to:
The solution curve, of the differential equation $$2y\frac{dy}{dx} + 3 = 5\frac{dy}{dx}$$, passing through the point $$(0, 1)$$ is a conic, whose vertex lies on the line:
The solution of the differential equation $$(x^2 + y^2)dx - 5xy \, dy = 0$$, $$y(1) = 0$$, is :
Let three vectors $$\vec{a} = \alpha\hat{i} + 4\hat{j} + 2\hat{k}$$, $$\vec{b} = 5\hat{i} + 3\hat{j} + 4\hat{k}$$, $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$ form a triangle such that $$\vec{c} = \vec{a} - \vec{b}$$ and the area of the triangle is $$5\sqrt{6}$$. If $$\alpha$$ is a positive real number, then $$|\vec{c}|^2$$ is equal to:
Let $$\vec{OA} = 2\vec{a}$$, $$\vec{OB} = 6\vec{a} + 5\vec{b}$$ and $$\vec{OC} = 3\vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\vec{OA}$$ and $$\vec{OC}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to :
Let the line L intersect the lines $$x - 2 = -y = z - 1$$, $$2(x + 1) = 2(y - 1) = z + 1$$ and be parallel to the line $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$. Then which of the following points lies on L?
The shortest distance between the lines $$\frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5}$$ and $$\frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1}$$ is:
