If $${{dy} \over {dx}} + 2y\tan x = \sin x,\,0

Let the solution curve $$y=f(x)$$ of the differential equation $$\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$$, $$x\in (-1,1)$$ pass through the origin. Then $$\underset{-\frac{\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f(x)dx$$ is equal to

Let $$y=y_1(x)$$ and $$y=y_2(x)$$ be two distinct solutions of the differential equation $$\frac{dy}{dx}=x+y$$, with $$y_1(0)=0$$ and $$y_2(0)=1$$ respectively. Then, the number of points of intersection of $$y=y_1(x)$$ and $$y=y_2(x)$$ is

Let the solution curve of the differential equation $$x\mathrm{d}y=\left(\sqrt{x^2+y^2}+y\right)\mathrm{d}x,x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

If $$y=y(x),x\in (0,\pi /2)$$ be the solution curve of the differential equation $$\left({\sin }^{2}2x\right)\frac{dy}{dx}+\left(8{\sin }^{2}2x+2\sin 4x\right)y=2{\mathrm{e}}^{-4x}(2\sin 2x+\cos 2x)$$, with $$y(\pi /4)={\mathrm{e}}^{-\pi }$$, then $$y(\pi /6)$$ is equal to :

Let $$y=y(x)$$ be the solution curve of the differential equation $$\frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{1/2}$$, $$x>1$$ passing through the point $$\left(2,\sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}y(8)$$ is equal to :

The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :

Let the solution curve $$y=y(x)$$ of the differential equation $$\left(1+{\mathrm{e}}^{2x}\right)\left(\frac{\mathrm{d}y}{\mathrm{d}x}+y\right)=1$$ pass through the point $$\left(0,\frac{\pi }{2}\right)$$. Then, $$\underset{x\to \infty }{\lim }{\mathrm{e}}^{x}y(x)$$ is equal to :

If the solution curve of the differential equation $$\frac{dy}{dx}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2),\mathrm{k}>0$$, then

Let $$y=y(x)$$ be the solution curve of the differential equation $$\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{(x+3)}{x+1},x>-1$$, which passes through the point $$(0,1)$$. Then $$y(1)$$ is equal to :

The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle {\pi \over 4} at the origin, is equal to :

The distance of the origin from the centroid of the triangle whose two sides have the equations $$x-2y+1=0$$ and $$2x-y-1=0$$ and whose orthocenter is \left( {{7 \over 3},{7 \over 3}} \right) is :

Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $$-$$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :

In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($$\alpha$$, $$\beta$$) is the centroid of $$\Delta$$ABC, then 15($$\alpha$$ + $$\beta$$) is equal to :

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $$\Delta$$PQR is :

Let the area of the triangle with vertices A(1, $$\alpha$$), B($$\alpha$$, 0) and C(0, $$\alpha$$) be 4 sq. units. If the points ($$\alpha$$, $$-$$$$\alpha$$), ($$-$$$$\alpha$$, $$\alpha$$) and ($$\alpha$$2, $$\beta$$) are collinear, then $$\beta$$ is equal to :

Let $$\alpha$$1, $$\alpha$$2 ($$\alpha$$1 2) be the values of $$\alpha$$ fo the points ($$\alpha$$, $$-$$3), (2, 0) and (1, $$\alpha$$) to be collinear. Then the equation of the line, passing through ($$\alpha$$1, $$\alpha$$2) and making an angle of {\pi \over 3} with the positive direction of the x-axis, is :

A line, with the slope greater than one, passes through the point $$A(4,3)$$ and intersects the line $$x-y-2=0$$ at the point B. If the length of the line segment $$AB$$ is $$\frac{\sqrt{29}}{3}$$, then $$B$$ also lies on the line :

Let the point $$P(\alpha ,\beta )$$ be at a unit distance from each of the two lines $$L_1:3x-4y+12=0$$, and $$L_2:8x+6y+11=0$$. If $$P$$ lies below $$L_1$$ and above $$L_2$$, then $$100(\alpha +\beta )$$ is equal to :

A point $$P$$ moves so that the sum of squares of its distances from the points $$(1,2)$$ and $$(-2,1)$$ is 14. Let $$f(x,y)=0$$ be the locus of $$\mathrm{P}$$, which intersects the $$x$$-axis at the points $$\mathrm{A}$$, $$\mathrm{B}$$ and the $$y$$-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :