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# Test: Vector Algebra (Competition Level) - 1

## 30 Questions MCQ Test Mathematics for JEE Mains | Test: Vector Algebra (Competition Level) - 1

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This mock test of Test: Vector Algebra (Competition Level) - 1 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Vector Algebra (Competition Level) - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Vector Algebra (Competition Level) - 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Vector Algebra (Competition Level) - 1 exercise for a better result in the exam. You can find other Test: Vector Algebra (Competition Level) - 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### The degree of the differential equation satisfying Solution:

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

QUESTION: 2

### The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of

Solution:  Again differentiating w.r.t. x, we get QUESTION: 3

### Differential equation of the family of circles touching the line y = 2 at (0, 2) is

Solution:

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.

QUESTION: 4

The differential equation of all parabolas whose axis are parallel to the y-axis is

Solution:

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 QUESTION: 5

The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is

Solution:

If (0, k) be the centre ony-axis then its radius will be k as it passes through origin. Hence its equation is   QUESTION: 6

The differential equation whose general solution is given by, y = , where c1, c2, c3, c4, c5 are arbitrary constants, is

Solution: where l, m, n are arbitrary constant   QUESTION: 7

The equation of the curves through the point (1, 0) and whose slope is Solution:  QUESTION: 8

The solution of the equation log(dy/dx) = ax + by is

Solution: QUESTION: 9

Solution of differential equation dy – sin x sin ydx = 0 is

Solution:   QUESTION: 10

The solution of the equation Solution:

Putting u = x - y, we get du/dx = 1 - dy/dx. The given equation can be written as 1 - du/dx = cos u QUESTION: 11 Solution:  QUESTION: 12

The general solution of the differential equation Solution:  QUESTION: 13

The solutions of (x + y + 1) dy = dx is

Solution:

Putting x + y + 1 = u, we have du = dx + dy and the given equation reduces to u(du - dx) = dx QUESTION: 14

The slope of the tangent at (x, y) to a curve passing through then the equation of the curve is

Solution:   QUESTION: 15

The solution of (x2 + xy)dy = (x2 + y2)dx is

Solution:    QUESTION: 16

The solution of (y + x + 5)dy = (y – x + 1) dx is

Solution:

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   QUESTION: 17

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

Solution:    QUESTION: 18

The solution of satisfying y(1) = 1 is given by

Solution:

Rewritting the given equation as    Hence y2 = x(1 + x) - 1 which represents a system of hyperbola.

QUESTION: 19

Solution of the equation Solution:

The given differential equation can be written as which is linear differential equation of first order.    QUESTION: 20

A function y = f(x) satisfies then f(x) is

Solution:  QUESTION: 21

The general solution of the equation Solution:  QUESTION: 22

The curve satisfying the equation and passing through the point (4, - 2) is

Solution:    QUESTION: 23

The solution of the differential equation Solution:  QUESTION: 24

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

Solution:   QUESTION: 25

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

Solution: 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).

QUESTION: 26

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

Solution: t is given that the triangle OPC is an isosceles triangle. On integration, we get x2-y2 = C, which is a rectangular hyperbola.

QUESTION: 27

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

Solution:

Let the equation of the  curve be y = f (x).    QUESTION: 28

The equation of the curve passing through (2, 7/2) and having gradient Solution:  QUESTION: 29

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

Solution: Equation of normal at point p is
Y – y = (dx/dy)(X – x) QUESTION: 30

The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is

Solution:

Equation to the family of parabolas is (y - k)2 = 4a(x - h).  Hence the order is 2 and the degree is 1.