If f is a real-valued differentiable function satisfying |f(x) �

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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

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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


B.

0

straight f apostrophe left parenthesis straight x right parenthesis space equals space limit as straight h rightwards arrow 0 of space fraction numerator straight f left parenthesis straight x plus straight h right parenthesis minus straight f left parenthesis straight x right parenthesis over denominator straight h end fraction
vertical line straight f apostrophe space left parenthesis straight x right parenthesis vertical line space equals space limit as straight h rightwards arrow 0 of open vertical bar fraction numerator straight f left parenthesis straight x plus straight h right parenthesis minus straight f left parenthesis straight x right parenthesis over denominator straight h end fraction space close vertical bar less or equal than space limit as straight h rightwards arrow 0 of open vertical bar fraction numerator left parenthesis straight h right parenthesis squared over denominator straight h end fraction close vertical bar
⇒ |f′(x)| ≤ 0
⇒ f′(x) = 0
⇒ f(x) = constant As
f(0) = 0 ⇒ f(1) = 0
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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
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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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