Let $$y=y(x)$$ be the solution of the differential equation $$(1+y^2)e^{\tan x}dx+{\cos }^{2}x(1+e^{2\tan x})dy=0,y(0)=1$$. Then $$y\left(\frac{\pi }{4}\right)$$ is equal to

If $$y=y(x)$$ is the solution of the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x}+2y=\sin (2x),y(0)=\frac{3}{4}$$, then $$y\left(\frac{\pi }{8}\right)$$ is equal to :

The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :

Suppose the solution of the differential equation $$\frac{dy}{dx}=\frac{(2+\alpha )x-\beta y+2}{\beta x-2\alpha y-(\beta \gamma -4\alpha )}$$ represents a circle passing through origin. Then the radius of this circle is :

Let $$y=y(x)$$ be the solution of the differential equation $$\left(2x{\log }_{e}x\right)\frac{dy}{dx}+2y=\frac{3}{x}{\log }_{e}x,x>0$$ and $$y\left(e^{-1}\right)=0$$. Then, $$y(e)$$ is equal to

Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$$, $$y(1)=0$$. Then $$y(0)$$ is

The portion of the line $4 x+5 y=20$ in the first quadrant is trisected by the lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ passing through the origin. The tangent of an angle between the lines ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ is :

Let $$\mathrm{R}$$ be the interior region between the lines $$3x-y+1=0$$ and $$x+2y-5=0$$ containing the origin. The set of all values of $$a$$, for which the points $$\left(a^2,a+1\right)$$ lie in $$R$$, is :

Let $$A(a,b),B(3,4)$$ and $$C(-6,-8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2a+3,7b+5)$$ from the line $$2x+3y-4=0$$ measured parallel to the line $$x-2y-1=0$$ is

Let $$\alpha ,\beta ,\gamma ,\delta \in \mathbb{Z}$$ and let $$A(\alpha ,\beta ),B(1,0),C(\gamma ,\delta )$$ and $$D(1,2)$$ be the vertices of a parallelogram $$\mathrm{ABCD}$$. If $$AB=\sqrt{10}$$ and the points $$\mathrm{A}$$ and $$\mathrm{C}$$ lie on the line $$3y=2x+1$$, then $$2(\alpha +\beta +\gamma +\delta )$$ is equal to

In a $$△ABC$$, suppose $$y=x$$ is the equation of the bisector of the angle $$B$$ and the equation of the side $$AC$$ is $$2x-y=2$$. If $$2AB=BC$$ and the points $$A$$ and $$B$$ are respectively $$(4,6)$$ and $$(\alpha ,\beta )$$, then $$\alpha +2\beta$$ is equal to

Let $$\mathrm{A}$$ be the point of intersection of the lines $$3x+2y=14,5x-y=6$$ and $$\mathrm{B}$$ be the point of intersection of the lines $$4x+3y=8,6x+y=5$$. The distance of the point $$P(5,-2)$$ from the line $$\mathrm{AB}$$ is

The distance of the point $$(2,3)$$ from the line $$2x-3y+28=0$$, measured parallel to the line $$\sqrt{3}x-y+1=0$$, is equal to

If $$x^2-y^2+2hxy+2gx+2fy+c=0$$ is the locus of a point, which moves such that it is always equidistant from the lines $$x+2y+7=0$$ and $$2x-y+8=0$$, then the value of $$g+c+h-f$$ equals

A line passing through the point $$\mathrm{A}(9,0)$$ makes an angle of $$30^{\circ }$$ with the positive direction of $$x$$-axis. If this line is rotated about A through an angle of $$15^{\circ }$$ in the clockwise direction, then its equation in the new position is :

A variable line $$\mathrm{L}$$ passes through the point $$(3,5)$$ and intersects the positive coordinate axes at the points $$\mathrm{A}$$ and $$\mathrm{B}$$. The minimum area of the triangle $$\mathrm{OAB}$$, where $$\mathrm{O}$$ is the origin, is :

A ray of light coming from the point $$\mathrm{P}(1,2)$$ gets reflected from the point $$\mathrm{Q}$$ on the $$x$$-axis and then passes through the point $$R(4,3)$$. If the point $$S(h,k)$$ is such that $$PQRS$$ is a parallelogram, then $$h{k}^2$$ is equal to:

The vertices of a triangle are $$\mathrm{A}(-1,3),\mathrm{B}(-2,2)$$ and $$\mathrm{C}(3,-1)$$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

If the line segment joining the points $$(5,2)$$ and $$(2,a)$$ subtends an angle $$\frac{\pi }{4}$$ at the origin, then the absolute value of the product of all possible values of $a$ is :

The equations of two sides $$\mathrm{AB}$$ and $$\mathrm{AC}$$ of a triangle $$\mathrm{ABC}$$ are $$4x+y=14$$ and $$3x-2y=5$$, respectively. The point $$\left(2,-\frac{4}{3}\right)$$ divides the third side $$\mathrm{BC}$$ internally in the ratio $$2:1$$, the equation of the side $$\mathrm{BC}$$ is