The point where the line x - 12 = y 

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The differential equation of family of circles whose centre lies on x-axis, is

  • d2ydx2 + dydx2 + 1 = 0

  • yd2ydx2 + dydx2 - 1 = 0

  • yd2ydx2 - dydx2 - 1 = 0

  • yd2ydx2 + dydx2 + 1 = 0


22.

The solution of the differential equation y1 + logxdydx - xlogx = 0 is

  • x log(x) = y + c

  • x log(x) = yc

  • y(1 + log(x) = c

  • log(x) - y = c


23.

The order of the differential equation whose solution is aex + be2x + ce3x + d = 0, is

  • 1

  • 2

  • 3

  • 4


24.

The maximum value of the objective function Z = 3x + 2y for linear constraints x + y  7, 2x + 3y  16, x2  0, y2  0 is

  • 16

  • 21

  • 25

  • 28


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25.

The position vectors of vertices of a ABC are 4i^ - 2j^, i^ + 4j^ - 3k^ and - i^ + 5j^+ k^ respectively, then ABC is equal to

  • π6

  • π4

  • π3

  • π2


26.

A four digit number is to be formed using the digits 1, 2, 3 4, 5 6, 7 (no digit is being repeated in any number). Then, the probability that it is > 4000, is

  • 3/2

  • 1/2

  • 4/7

  • 3/7


27.

The equation of the plane which passes through (2, - 3, 1) and is normal to the line joining the points (3, 4, - 1) and (2, - 1, 5), is given by

  • x + 5y - 6z + 19 = 0

  • x - 5y + 6z - 19 = 0

  • x + 5y + 6z + 19 = 0

  • x - 5y - 6z - 19 = 0


28.

The angle between the lines x2 - xy - 6y2 - 7x + 31y - 18 = 0 is

  • π4

  • π6

  • π2

  • π3


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29.

The equation of the lines passing through the origin and having slopes 3 and - 13, is

  • 3y2 + 8xy - 3x2 = 0

  • 3x2 + 8xy + 3y2 = 0

  • 3y2 - 8xy - 3x2 = 0

  • 3x2 + 8xy - 3y2 = 0


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30.

The point where the line x - 12 = y - 2- 3 = z + 34 meets the plane 2x + 4y - z = 1, is

  • (3, - 1, 1)

  • (3, 1, 1)

  • (1, 1, 3)

  • (1, 3, 1)


A.

(3, - 1, 1)

Let point be (a, b, c), then

2a + 4b - c = 1           ...(i)

and a = 2k + 1, b = - 3k + 2 and c = 4k - 3

          [where k is constant]
On substituting these values in Eq. (i), we get

2(2k + 1) + 4 (- 3k + 2) - (4k - 3)= 1

                                            k = 1

Hence, required point is (3, - 1, 1).


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