The equation of the tangent to the curve (1 + x2)y = 2 - x where

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Value of the determinant 1 +a1a111 + b1111 +c is :

  • 1 + abc + ab + bc + ca

  • abc

  • 4abc

  • abc1 + 1a + 1b + 1c


2.

If the value of the determinant x + 1112x + 2233x + 3 is equal to zero, then x is :

  • 0 and - 6

  • 0 and 6

  • 6

  • - 6


3.

The value of a for which the matrix A = a224 is singular :

  • a  1

  • a = 1

  • a = 0

  • a = - 1


4.

If A = 2- 1- 12 and I is the unit matrix of order two, then A2 is equal to :

  • 4A - 3I

  • 3A - 4I

  • A - I

  • A + I


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5.

If A and B are two square matrices of the same order, then (A - B)2 :

  • A2 - AB - BA + B2

  • A2 - 2AB + B2

  • A2 - 2BA + B2

  • A2 - B2


6.

If P = i0- i0- ii- ii0 and Q = - ii00i- i, then PQ is equal to

  • - 221- 11- 1

  • 2- 2- 11- 11

  • 2- 2- 11

  • 100010001


7.

If I is the unit matrix of order 10, then the determinant of I is equal to :

  • 10

  • 1

  • 110

  • 9


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8.

The equation of the tangent to the curve (1 + x2)y = 2 - x where it crosses the x-axis, is :

  • x + 5y = 2

  • x - 5y = 2

  • 5x - y = 2

  • 5x + y - 2 = 0


A.

x + 5y = 2

The given equation of curve is (1 + x2)y = 2 - x, it meets x-axis at (2, 0)

It can be rewritten as y = 2 - x1 + x2

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

           dydx = 1 + x2- 1 - 2 - x2x1 + x22                 = - 1 - x2 - 4x + 2x21 + x22 dydx2, 0 = - 1 - 4 - 8 + 825 = - 15  Required equation of tangent is     y - 0 = - 15x - 2 x + 5y = 2


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9.

The sides of an equilateral triangle are increasing at the rate of 2 cm/s. The rate at which the area increases, when the side is 10 cm, is:

  • 3 sq cm/s

  • 10 sq cm/s

  • 103 sq cm/s

  • 103 sq cm/s


10.

If f(x) = x2 - 9x - 3,    if x  32x + k,    otherwise is continuous at x = 3, then k is equal to :

  • 3

  • 0

  • - 6

  • 16


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