If x $$=$$ 3 tan t and y $$=$$ 3 sec t, then the value of {{{d^2}y} \over {d{x^2}}} at t = {\pi \over 4}, is :

For x > 1, if (2x)2y = 4e2x$$-$$2y, then (1 + loge 2x)2 {{dy} \over {dx}} is equal to :

If y = y(x) is the solution of the differential equation, x$$dy\frac{d}{x}$$ + 2y = x2, satisfying y(1) = 1, then y(1\over2) is equal to :

The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x $$\ne$$ 0) is : (where c is a constant of integration)

Consider the differential equation, {y^2}dx + \left( {x - {1 \over y}} \right)dy = 0, If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Let y = y(x) be the solution of the differential equation, {{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x, x \in \left( { - {\pi \over 2},{\pi \over 2}} \right), such that y(0) = 1. Then :

If y = y(x) is the solution of the differential equation {{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x, x \in \left( { - {\pi \over 2},{\pi \over 2}} \right), such that y (0) = 0, then y\left( { - {\pi \over 4}} \right) is equal to :

If \cos x{{dy} \over {dx}} - y\sin x = 6x, (0 < x < {\pi \over 2}) and y\left( {{\pi \over 3}} \right) = 0 then y\left( {{\pi \over 6}} \right) is equal to :-

The solution of the differential equation x{{dy} \over {dx}} + 2y = x2 (x $$\ne$$ 0) with y(1) = 1, is :

Let y = y(x) be the solution of the differential equation, {({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1 such that y(0) = 0. If $$\sqrt{a}y(1)$$ = $$\pi \frac{3}{2}$$ , then the value of 'a' is :

If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as {{{x^2} - 2y} \over x}, then the curve also passes through the point :

Let y = y(x) be the solution of the differential equation, x{{dy} \over {dx}} + y = x loge x, (x > 1). If 2y(2) = loge 4 $$-$$ 1, then y(e) is equal to :

The solution of the differential equation, {{dy} \over {dx}} = (x – y)2, when y(1) = 1, is :

If y(x) is the solution of the differential equation {{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\, where y\left( 1 \right) = {1 \over 2}{e^{ - 2}}, then

Let f be a differentiable function such that f '(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}, (x > 0) and f(1) $$\ne$$ 4. Then $$\underset{x\to 0^{\prime }}{\lim }$$ xf\left( {{1 \over x}} \right) :

The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :

If {{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right) and y\left( {{\pi \over 4}} \right) = {4 \over 3}, then y\left( { - {\pi \over 4}} \right) equals -

Let f : [0,1] $$\to$$ R be such that f(xy) = f(x).f(y), for all x, y $$\in$$ [0, 1], and f(0) $$\ne$$ 0. If y = y(x) satiesfies the differential equation, {{dy} \over {dx}} = f(x) with y(0) = 1, then y\left( {{1 \over 4}} \right) + y\left( {{3 \over 4}} \right) is equal to :

Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true?

A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60o with the line x + y = 0. Then an equation of the line L is :