The triangle formed by the tangent to the curve f (x) = x2 +

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 Multiple Choice QuestionsMultiple Choice Questions

401.

If the direction cosines of two lines are connected by the equations l + m + n = 0, l2 + m2 - n2 = 0, then the angle between the lines is

  • π4

  • π6

  • π2

  • π3


402.

The equation of the plane which contains the origin and the line of intersection of the planes r · a = d1 and r · b = d2, is

  • r . (d1a + d2b) = 0

  • r . (d2a - d1b) = 0

  • r . (d2a + d1b) = 0

  • r . (d1a - d2b) = 0


403.

If from a point P(a, b, c) perpendiculars PA and PB are drawn to YZ and ZX - planes, then the equation of the plane OAB is

  • bcx + cay + abz = 0

  • bcx + cay - abz = 0

  • - bcx + cay + abz = 0

  • bcx - cay + abz = 0


404.

If (2, 7, 3) is one end of a diameter of the sphere x2 + y+ z- 6x - 12y - 2z + 20 = 0, then the coordinates of the other end of the diameter are

  • (- 2, 5, - 1)

  • (4, 5, 1)

  • (2, - 5, 1)

  • (4, 5. - 1)


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405.

If a line segment OP makes angles of π4 and π3 with X-axis and Y-axis, respectively. Then, the direction cosines are

  • 12, 32, 12

  • 12, 12, 12

  • 1, 3, 1

  • 1, 13, 1


406.

If a plane passing through the point (2, 2, 1) and is perpendicular to the planes 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0. Then, the equation of the plane is

  • 2x - y - z - 1 = 0

  • 2x + 3y + z - 1 = 0

  • 2x + y + z + 3 = 0

  • x - y + z - 1 = 0


407.

If the points (1, 2, 3) and (2, - 1, 0) lie on the opposite sides of the plane 2x + 3y - 2z = k, then

  • k < 1

  • k > 2

  • k < 1 or k > 2

  • 1 < k < 2


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408.

The triangle formed by the tangent to the curve f (x) = x2 + bx - b at the point (1, 1) and the coordinate axes lies in the first quadrant. If its area is 2, then the value of b is

  • - 1

  • 3

  • - 3

  • 1


C.

- 3

Given curve is y = f(x) = x2 + bx - b.

On differentiating w.r.t. x, we get

                           f'(x) = 2x + b

The equation of tangent at point (1, 1) is

                               y - 1 = dydx1, 1x - 1                          y - 1 = b + 2x - 1               2 +bx - y = 1 +b x1 + b2 + b - y1 +b = 1

So,                     OA = 1 + b2 + band                     OB = - 1 + bNow, area of AOB = 12 × 1 + b . - 1 + b2 + b = 2     42 + b + 1 + b2 = 0 8 +4b + 1 + b2 +2b = 0                  b2 + 6b + 9 = 0                         b + 32 = 0                                     b = - 3


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409.

If a plane meets the coordinate axes at A, B and C such that the centroid of the triangle is (1, 2, 4), then the equation of the plane is

  • x + 2y + 4z = 12

  • 4x + 2y + z = 12

  • x + 2y + 4z = 3

  • 4x + 2y + z = 3


410.

The volume of the tetrahedron included between the plane 3x + 4y - 5z - 60 = 0 and the coordinate planes is

  • 60

  • 600

  • 720

  • 400


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