If a curve y=f(x) passes through the point (1, −1) and satisfi

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 Multiple Choice QuestionsMultiple Choice Questions

1.

If z = yxsinxy + cos1 + yx  , then xzx is equal to

  • yzy

  • - yxy

  • 2yzy

  • 2yzx


2. limit as straight n rightwards arrow infinity of space open parentheses fraction numerator left parenthesis straight n plus 1 right parenthesis left parenthesis straight n plus 2 right parenthesis....3 straight n over denominator straight n to the power of 2 straight n end exponent end fraction close parentheses to the power of 1 divided by straight n end exponent is equal to
  • 18/e4

  • 27/e2

  • 9/e2

  • 9/e2

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3.

The area (in sq. units) of the regionopen curly brackets left parenthesis straight x comma straight y right parenthesis colon straight y squared space greater or equal than space 2 straight x space and space straight x squared space plus straight y squared space less or equal than 4 straight x comma space straight x space greater or equal than 0 comma space straight y greater or equal than 0 close curly brackets is

  • straight pi minus 4 over 3
  • straight pi minus 8 over 3
  • straight pi minus fraction numerator 4 square root of 2 over denominator 3 end fraction
  • straight pi minus fraction numerator 4 square root of 2 over denominator 3 end fraction
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4.

If a curve y=f(x) passes through the point (1, −1) and satisfies the differential equation, y(1+xy) dx=x dy, then f(-1/2) is equal to

  • -2/5

  • -4/5

  • 2/5

  • 2/5


D.

2/5

Given differential equation is

 y( 1+ xy) dx = xdy

⇒ ydx + xy2 dx = xdy
fraction numerator xdy minus ydx over denominator straight y squared end fraction space equals space xdx
rightwards double arrow space minus fraction numerator left parenthesis ydx minus xdy right parenthesis over denominator straight y squared end fraction space equals space xdx
On space integrating space both space sides comma space we space get
fraction numerator negative straight x over denominator straight y end fraction space equals space straight x squared over 2 plus straight C
therefore comma space it space passes space through space left parenthesis 1 comma negative 1 right parenthesis
1 space equals 1 half space plus straight C
rightwards double arrow straight C space equals space 1 half
Now comma space from space eq space left parenthesis straight i right parenthesis comma
minus straight x over straight y space equals space straight x squared over 2 space plus space 1 half
rightwards double arrow space straight x squared space plus space 1 space equals space minus fraction numerator 2 straight x over denominator straight y end fraction
straight y equals space minus space fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction
therefore comma space straight f open parentheses negative fraction numerator begin display style 1 end style over denominator 2 end fraction close parentheses space equals space 4 over 5


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5.

If f and ga re differentiable  functions in (0,1) satisfying f(0) =2= g(1), g(0) = 0 and f(1) = 6, then for some c ε] 0,1[

  • 2f'(c) = g'(c)

  • 2f'(c) = 3g'(c)

  • f'(c) = g'(c)

  • f'(c) = g'(c)

149 Views

6.

The population p(t) at time t of a certain mouse species satisfies the differential equation fraction numerator dp space left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 0.5 space left parenthesis straight t right parenthesis space minus 450. if p (0) = 850, then the  time at which the population becomes zero is

  • 2 log 18

  • log 9

  • 1 half space log space 18
  • 1 half space log space 18
493 Views

7.

Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f′(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5).

  • Statement 1 is false, statement 2 is true

  • 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 true; statement 2 is not a correct explanation for statement 1

154 Views

8. fraction numerator straight d squared straight x over denominator dy squared end fraction equal to
  • open parentheses fraction numerator straight d squared straight x over denominator dy squared end fraction close parentheses to the power of negative 1 end exponent
  • negative open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses to the power of negative 1 end exponent open parentheses dy over dx close parentheses to the power of negative 3 end exponent
  • open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses open parentheses dy over dx close parentheses to the power of negative 2 end exponent
  • open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses open parentheses dy over dx close parentheses to the power of negative 2 end exponent
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9.

The shortest distance between line y - x = 1 and curve x = y2 is

  • √3/4

  • 3√2 /8

  • 8/3√2

  • 8/3√2

349 Views

10. limit as straight x space rightwards arrow 2 of space open parentheses fraction numerator square root of 1 minus cos space open curly brackets 2 left parenthesis straight x minus 2 right parenthesis close curly brackets end root over denominator straight x minus 2 end fraction close parentheses
  • does not exist

  • equal square root of 2

  • equal negative square root of 2

  • equal negative square root of 2

153 Views

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