The differential equation representing the family of curves  ,

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 Multiple Choice QuestionsMultiple Choice Questions

11.

Let f : R → R be a continuous function defined
by f(x) = 1/ex + 2e-x
Statement - 1: f(c) = 1/3, for some c ∈ R.
Statement-2: 0 < f(x)≤ fraction numerator 1 over denominator 2 square root of 2 end fraction, for all x ∈ R.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is true, Statement-2 is false.

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12.

The equation of the tangent to the curvestraight y equals straight x space plus space 4 over straight x squared, that is parallel to the x-axis, is

  • y = 0

  • y = 1

  • y = 3

  • y = 3

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13.

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

  • second order and second degree

  • first order and second degree

  • first order and first degree

  • first order and first degree

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14.

The differential equation representing the family of curves straight y squared space equals space left parenthesis straight x plus square root of straight c right parenthesis , where c > 0, is a parameter, is of order and degree as follows:

  • order 1, degree 2

  • order 1, degree 1

  • order 1, degree 3

  • order 1, degree 3


C.

order 1, degree 3

straight y squared space equals 2 straight c space left parenthesis straight x space plus square root of straight c right parenthesis space..... space left parenthesis straight i right parenthesis
2 yy apostrophe space equals space 2 straight c.1 space or space yy apostrophe space equals space straight c space.... space left parenthesis ii right parenthesis
rightwards double arrow space straight y squared space equals space 2 yy apostrophe space left parenthesis straight x plus square root of yy apostrophe end root right parenthesis space left square bracket on space putting space value space of space cfrom space left parenthesis ii right parenthesis space and space left parenthesis straight i right parenthesis right square bracket
On space simplifying comma space we space get space
left parenthesis straight y minus 2 xy apostrophe right parenthesis squared space equals space 4 yy apostrophe cubed space left parenthesis iii right parenthesis
Hence space equation space left parenthesis iii right parenthesis space is space of space order space 1 space adn space degree space 3.
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15.

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

  • f(6) ≥ 8

  • f(6) < 8

  • f(6) < 5

  • f(6) < 5

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16.

If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2 , x, y ∈ R and f(0) = 0, then f(1) equals

  • -1

  • 0

  • 2

  • 2

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17.

If straight x dy over dx space equals space straight y space left parenthesis log space straight y space minus space log space straight x space plus 1 right parenthesis commathen the solution of the equation is

  • straight y space log space open parentheses straight x over straight y close parentheses space equals cx
  • space straight x space log space open parentheses straight y over straight x close parentheses space equals space cy
  • log space open parentheses straight y over straight x close parentheses space equals space cx
  • log space open parentheses straight y over straight x close parentheses space equals space cx
118 Views

18.

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3 /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

  • fraction numerator 1 over denominator 36 space straight pi end fraction space cm divided by min
  • fraction numerator 1 over denominator 18 space straight pi end fraction space cm divided by min
  • fraction numerator 1 over denominator 54 space straight pi end fraction space cm divided by min
  • fraction numerator 1 over denominator 54 space straight pi end fraction space cm divided by min
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19.

let straight f left parenthesis straight x right parenthesis space equals space fraction numerator 1 minus space tan space straight x over denominator 4 straight x minus straight pi end fraction space comma space straight x not equal to space straight pi over 4 space straight x space element of space open square brackets 0 comma space straight pi over 2 close square brackets . If f(x) is continuous in open square brackets 0 space comma space straight pi over 2 close square brackets comma then f (π/4) is

  • 1

  • -1/2

  • -1

  • -1

118 Views

20.

A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is

  • (x-1)2

  • (x-1)3

  • (x+1)3

  • (x+1)3

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