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

Let $x=x(y)$ be the solution of the differential equation $2(y+2){\log }_e(y+2)dx+\left(x+4-2{\log }_e(y+2)\right)dy=0,y>-1$ with $x\left(e^4-2\right)=1$. Then $x\left(e^9-2\right)$ is equal to :

Let $$y=y_1(x)$$ and $$y=y_2(x)$$ be the solution curves of the differential equation $$\frac{dy}{dx}=y+7$$ with initial conditions $$y_1(0)=0$$ and $$y_2(0)=1$$ respectively. Then the curves $$y=y_1(x)$$ and $$y=y_2(x)$$ intersect at

Let $$y=y(x),y>0$$, be a solution curve of the differential equation $$\left(1+x^2\right)\mathrm{d}y=y(x-y)\mathrm{d}x$$. If $$y(0)=1$$ and $$y(2\sqrt{2})=\beta$$, then

Let $$y=y(x)$$ be the solution of the differential equation $$\frac{dy}{dx}+\frac{5}{x\left(x^5+1\right)}y=\frac{{\left(x^5+1\right)}^2}{x^7},x>0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :

Let $$y=y(x)$$ be a solution curve of the differential equation. $$\left(1-x^2{y}^2\right)dx=ydx+xdy$$. If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\alpha$$, then a value of $$\alpha$$ is :

Let $$f$$ be a differentiable function such that $$x^{2}f(x)-x=4\underset{0}{\overset{x}{\int }}tf(t)dt$$, f(1) = {2 \over 3}. Then $$18f(3)$$ is equal to :

If the solution curve $$f(x,y)=0$$ of the differential equation $$\left(1+{\log }_{e}x\right)\frac{dx}{dy}-x{\log }_{e}x=e^y,x>0$$, passes through the points $$(1,0)$$ and $$(\alpha ,2)$$, then $${\alpha }^{\alpha }$$ is equal to :

The combined equation of the two lines $$ax+by+c=0$$ and $$a^{\prime }x+b^{\prime }y+c^{\prime }=0$$ can be written as $$(ax+by+c)(a^{\prime }x+b^{\prime }y+c^{\prime })=0$$. The equation of the angle bisectors of the lines represented by the equation $$2x^2+xy-3y^2=0$$ is :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\alpha ,\beta )$$, then the quadratic equation whose roots are $$\alpha +4\beta$$ and $$4\alpha +\beta$$, is :

Let $$B$$ and $$C$$ be the two points on the line $$y+x=0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y-2x=2$$ such that $$△ABC$$ is an equilateral triangle. Then, the area of the $$△ABC$$ is :

A light ray emits from the origin making an angle 30$${}^{\circ }$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :

If $(\alpha ,\beta )$ is the orthocenter of the triangle $\mathrm{ABC}$ with vertices $A(3,-7), B(-1,2)$ and $C(4,5)$, then $9\alpha -6\beta +60$ is equal to :

Let $$(\alpha ,\beta )$$ be the centroid of the triangle formed by the lines $$15x-y=82,6x-5y=-4$$ and $$9x+4y=17$$. Then $$\alpha +2\beta$$ and $$2\alpha -\beta$$ are the roots of the equation :

If the point $$\left(\alpha ,\frac{7\sqrt{3}}{3}\right)$$ lies on the curve traced by the mid-points of the line segments of the lines $$x\cos \theta +y\sin \theta =7,\theta \in \left(0,\frac{\pi }{2}\right)$$ between the co-ordinates axes, then $$\alpha$$ is equal to :

Let $$C(\alpha ,\beta )$$ be the circumcenter of the triangle formed by the lines $$4x+3y=69$$ $$4y-3x=17$$, and $$x+7y=61$$. Then $$(\alpha -\beta {)}^2+\alpha +\beta$$ is equal to :

The straight lines $${\mathrm{l}}_1$$ and $${\mathrm{l}}_2$$ pass through the origin and trisect the line segment of the line L : $$9x+5y=45$$ between the axes. If $${\mathrm{m}}_1$$ and $${\mathrm{m}}_2$$ are the slopes of the lines $${\mathrm{l}}_1$$ and $${\mathrm{l}}_2$$, then the point of intersection of the line $$\mathrm{y}=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)x$$ with L lies on :

If two vectors $$\overrightarrow{P}=\hat{i}+2m\hat{j}+m\hat{k}$$ and $$\overrightarrow{Q}=4\hat{i}-2\hat{j}+m\hat{k}$$ are perpendicular to each other. Then, the value of m will be :

A vector in $x-y$ plane makes an angle of $30^{\circ }$ with $y$-axis. The magnitude of $\mathrm{y}$-component of vector is $2\sqrt{3}$. The magnitude of $x$-component of the vector will be :

When vector $$\vec{A}=2\hat{i}+3\hat{j}+2\hat{k}$$ is subtracted from vector $$\overrightarrow{\mathrm{B}}$$, it gives a vector equal to $$2\hat{j}$$. Then the magnitude of vector $$\overrightarrow{\mathrm{B}}$$ will be :