Competition Level Test: Vector Algebra- 1
30 Questions MCQ Test Mathematics For JEE | Competition Level Test: Vector Algebra- 1
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The degree of the differential equation satisfying 
Putting x = sin A and y = sin B in the given relation, we get
cos A + cos B = a(sin A - sin B)
Clearly, it is a differential equation of degree one
The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of
Again differentiating w.r.t. x, we get
Differential equation of the family of circles touching the line y = 2 at (0, 2) is
Options A, B, C – all have a family of circles with centre (0,2).
But according to given conditions (0,2) is a point on the circle.
So, none of the options are correct.
The differential equation of all parabolas whose axis are parallel to the y-axis is
The equation of a member of the family of parabolas having axis parallel toy-axis is
y — Ax2 + Bx+ C ......(1)
where A, B, and C are arbitrary constants
Differentiating equation (1) w.r.t. x, we get 
......(2)
which on again differentiating w.r.t, x gives 
....(3)
Differentiating (3) w.r.t. x, we get 
The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is
If (0, k) be the centre ony-axis then its radius will be k as it passes through origin. Hence its equation is
The differential equation whose general solution is given by, y = 
, where c1, c2, c3, c4, c5 are arbitrary constants, is
where l, m, n are arbitrary constant
The equation of the curves through the point (1, 0) and whose slope is 
The solution of the equation log(dy/dx) = ax + by is
Solution of differential equation dy – sin x sin ydx = 0 is
Putting u = x - y, we get du/dx = 1 - dy/dx. The given equation can be written as 1 - du/dx = cos u
The general solution of the differential equation 
The solutions of (x + y + 1) dy = dx is
Putting x + y + 1 = u, we have du = dx + dy and the given equation reduces to u(du - dx) = dx
The slope of the tangent at (x, y) to a curve passing through 
then the equation of the curve is
The solution of (x2 + xy)dy = (x2 + y2)dx is
The solution of (y + x + 5)dy = (y – x + 1) dx is
The intersection of y – x + 1 = 0 and y + x + 5 = 0 is (-2, -3).
Put x = X - 2, y = Y - 3
The given equation reduces to 
The slope of the tangent at (x, y) to a curve passing through a point (2, 1) is 
then the equation of the curve is
The solution of 
satisfying y(1) = 1 is given by
Rewritting the given equation as
Hence y2 = x(1 + x) - 1 which represents a system of hyperbola.
The given differential equation can be written as 
which is linear differential equation of first order.
A function y = f(x) satisfies 
then f(x) is
The curve satisfying the equation 
and passing through the point (4, - 2) is
The solution of the differential equation 
Which of the following is not the differential equation of family of curves whose tangent from an angle of π/4 with the hyperbola xy = c2?
Tangent to a curve intercepts the y-axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is
The equation of the tangent at the point
The coordinates of the point 
which is the required differential equation to the curve at y =f(x).
The curve for which the normal at any point (x, y) and the line joining the origin to that point from an isosceles triangle with the x-axis as base is
t is given that the triangle OPC is an isosceles triangle.
On integration, we get x2-y2 = C, which is a rectangular hyperbola.
A normal at P(x, y) on a curve meets the x-axis at Q and N is the foot of the ordinate at P. If NQ = 
then the equation of curve given that it passes through the point (3,1) is
Let the equation of the curve be y = f (x).
The equation of the curve passing through (2, 7/2) and having gradient 
A normal at any point (x, y) to the curve y = f(x) cuts a triangle of unit area with the axis, the differential equation of the curve is
Equation of normal at point p is
Y – y = (dx/dy)(X – x)
The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is
Equation to the family of parabolas is (y - k)2 = 4a(x - h).
Hence the order is 2 and the degree is 1.
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