Let f : R → R be a positive increasing function with  from M

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 Multiple Choice QuestionsMultiple Choice Questions

141.

If: R →R is a function defined by straight f space left parenthesis straight x right parenthesis space equals space left square bracket straight x right square bracket space cos space open parentheses fraction numerator 2 straight x minus 1 over denominator 2 end fraction close parentheses straight pi where [x] denotes the greatest integer function, then f is

  • continuous for every real x

  • discontinous only at x = 0

  • discontinuous only at non-zero integral values of x

  • discontinuous only at non-zero integral values of x

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142.

The value of p and q for which the function f(x) =open curly brackets table attributes columnalign left end attributes row cell table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator sin space left parenthesis straight p space plus 1 right parenthesis space straight x space plus space sin space straight x over denominator straight x end fraction comma end cell cell straight x less than 0 end cell row cell straight q comma end cell cell straight x space equals space 0 space is space continuous space for space all space straight x space in space straight R comma space are colon end cell end table
fraction numerator square root of straight x space plus straight x squared end root minus square root of straight x over denominator straight x to the power of 3 divided by 2 end exponent end fraction space comma space straight x greater than space 0 space end cell row space end table close

  • p = 1/2. q = -3/2

  • p = 5/2, q = 1/2

  • p = - 3/2, q = 1/2

  • p = - 3/2, q = 1/2

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143.

The domain of the function f(x) = fraction numerator 1 over denominator square root of vertical line straight x vertical line minus straight x end root end fraction

  • (-∞,∞)

  • (0,∞)

  • (-∞,0)

  • (-∞,0)

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144.

let f : (-1, 1) → R be a differentiable function
with f(0) = -1 and f'(0) = 1.
Let g(x) = [f(2f(x) + 2)]2. Then g'(0) =

  • 4

  • -4

  • 0

  • 0

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145.

Let f : R → R be a positive increasing function with limit as infinity space rightwards arrow 0 of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space equals 1 space then space limit as infinity space rightwards arrow 0 of space fraction numerator begin display style straight f left parenthesis 2 straight x right parenthesis end style over denominator begin display style straight f left parenthesis straight x right parenthesis end style end fraction space space is space equal space to

  • 1

  • 2/3

  • 3/2

  • 3/2


A.

1

Since f(x) is a positive increasing function.
⇒ 0< f(x)<f(2x)<f(3x)
⇒ 0<1<fraction numerator straight f left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction less than space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction
limit as straight x space rightwards arrow 0 of space less or equal than space stack lim space with straight x rightwards arrow infinity below space fraction numerator straight f space left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space less or equal than space limit as straight x rightwards arrow infinity of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction
by space sandwich space theorem comma space limit as straight x space rightwards arrow infinity of space fraction numerator straight f left parenthesis 2 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction space equals space 1

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146.

If for  x ∈ (0, 1/4)  the derivatives tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction close parentheses space is space square root of straight x. end root space straight g left parenthesis straight x right parenthesis comma then g(x) is equals to 

  • fraction numerator 3 over denominator 1 plus 9 straight x cubed end fraction
  • fraction numerator 9 over denominator 1 plus 9 straight x cubed end fraction
  • fraction numerator 3 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction
  • fraction numerator 3 straight x square root of straight x over denominator 1 minus 9 straight x cubed end fraction
211 Views

147. integral subscript 0 superscript straight pi[cot x]dx, where [.] denotes the greatest integer function, is equal to
  • π/2

  • 1

  • -1

  • -1

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148.

For real x, let f(x) = x3+ 5x + 1, then

  • f is one–one but not onto R

  • f is onto R but not one–one

  • f is one–one and onto R

  • f is one–one and onto R

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149.

Let f(x) = (x + 1)2– 1, x ≥ – 1
Statement – 1: The set {x : f(x) = f–1(x)} = {0, –1}.
Statement – 2: f is a bijection.

  • 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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150.

Let f(x) = x|x| and g(x) = sinx

Statement 1 : gof is differentiable at x = 0 and its derivative is continuous atthat point
Statement 2: gof is twice differentiable at x = 0

  • 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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