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not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fractionis

Maths-General

  1. l o g invisible function application left parenthesis t a n invisible function application h x right parenthesis plus c    
  2. 1 half straight e to the power of 2 straight x end exponent plus straight c    
  3. x plus c    
  4. l o g invisible function application left parenthesis c o t blank h x right parenthesis plus c    

    Answer:The correct answer is: x plus c

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    integral open parentheses x minus blank to the power of 10 end exponent C subscript 1 end subscript x to the power of 2 end exponent plus blank to the power of 10 end exponent C subscript 2 end subscript x to the power of 3 end exponent minus blank to the power of 10 end exponent C subscript 3 end subscript x to the power of 4 end exponent horizontal ellipsis horizontal ellipsis plus blank to the power of 10 end exponent C subscript 10 end subscript x to the power of 11 end exponent close parentheses d xdx equals:

    integral open parentheses x minus blank to the power of 10 end exponent C subscript 1 end subscript x to the power of 2 end exponent plus blank to the power of 10 end exponent C subscript 2 end subscript x to the power of 3 end exponent minus blank to the power of 10 end exponent C subscript 3 end subscript x to the power of 4 end exponent horizontal ellipsis horizontal ellipsis plus blank to the power of 10 end exponent C subscript 10 end subscript x to the power of 11 end exponent close parentheses d xdx equals:

    maths-General
    General
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    If I = not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction, then I equals:

    If I = not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction, then I equals:

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    General
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    If f left parenthesis x right parenthesis equals t a n to the power of negative 1 end exponent invisible function application x plus l space l n invisible function application square root of 1 plus x end root minus l n square root of 1 minus x end root. The integral of 1 divided by 2 space f apostrophe left parenthesis x right parenthesis with respect to x to the power of 4 end exponent is -

    If f left parenthesis x right parenthesis equals t a n to the power of negative 1 end exponent invisible function application x plus l space l n invisible function application square root of 1 plus x end root minus l n square root of 1 minus x end root. The integral of 1 divided by 2 space f apostrophe left parenthesis x right parenthesis with respect to x to the power of 4 end exponent is -

    maths-General
    General
    maths-

    The indefinite integral of left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent is, for any arbitrary constant -


    The indefinite integral of left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent is, for any arbitrary constant -

    maths-General

    General
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    If f open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses = x + 2 then not stretchy integral f left parenthesis x right parenthesis d x is equal to

    If f open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses = x + 2 then not stretchy integral f left parenthesis x right parenthesis d x is equal to

    maths-General
    General
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    Let x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N, then integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x is equal to:

    Let x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N, then integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x is equal to:

    maths-General
    General
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    Evaluate not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fractiondx

    Evaluate not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fractiondx

    maths-General
    General
    maths-

    fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction=

    fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction=

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    General
    maths-

    Assertion : f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis can never become positive.
    Reason : f(x) = sgn x is always a positive function.

    Assertion : f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis can never become positive.
    Reason : f(x) = sgn x is always a positive function.

    maths-General
    General
    maths-

    Statement I : y = f(x) =fraction numerator x to the power of 2 end exponent minus 2 x plus 4 over denominator x to the power of 2 end exponent minus 2 x plus 5 end fraction, xelement ofR Range of f(x) is [3/4, 1)
    Statement II : left parenthesis x minus 1 right parenthesis to the power of 2 end exponent equals fraction numerator 4 y minus 3 over denominator 1 minus y end fraction.

    Statement I : y = f(x) =fraction numerator x to the power of 2 end exponent minus 2 x plus 4 over denominator x to the power of 2 end exponent minus 2 x plus 5 end fraction, xelement ofR Range of f(x) is [3/4, 1)
    Statement II : left parenthesis x minus 1 right parenthesis to the power of 2 end exponent equals fraction numerator 4 y minus 3 over denominator 1 minus y end fraction.

    maths-General
    General
    maths-

    Statement I : Function f(x) = sinx + {x} is periodic with period 2 pi
    Statement II : sin x and {x} are both periodic with period 2 pi and 1 respectively.

    Statement I : Function f(x) = sinx + {x} is periodic with period 2 pi
    Statement II : sin x and {x} are both periodic with period 2 pi and 1 respectively.

    maths-General
    General
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    If the remainders of the polynomial f(x) when divided by x minus 1 comma x minus 2 are 2,5 then the remainder of f left parenthesis x right parenthesis when divided by left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis is

    If the remainders of the polynomial f(x) when divided by x minus 1 comma x minus 2 are 2,5 then the remainder of f left parenthesis x right parenthesis when divided by left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis is

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    General
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    Statement 1 : f : R rightwards arrow R and f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponentis bijective.
    Statement 2 : f colon R rightwards arrow R comma space f left parenthesis x right parenthesis equals e to the power of x minus e to the power of negative x end exponentis bijective.

    Statement 1 : f : R rightwards arrow R and f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponentis bijective.
    Statement 2 : f colon R rightwards arrow R comma space f left parenthesis x right parenthesis equals e to the power of x minus e to the power of negative x end exponentis bijective.

    maths-General
    General
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    If fraction numerator 2 x plus 1 over denominator left parenthesis x minus 1 right parenthesis open parentheses x to the power of 2 end exponent plus 2 close parentheses end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B x plus c over denominator x to the power of 2 end exponent plus 2 end fraction then B=

    If fraction numerator 2 x plus 1 over denominator left parenthesis x minus 1 right parenthesis open parentheses x to the power of 2 end exponent plus 2 close parentheses end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B x plus c over denominator x to the power of 2 end exponent plus 2 end fraction then B=

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    General
    maths-

    Statement I : Graph of y = tan x is symmetrical about origin
    Statement II : Graph of y equals sec squared invisible function application x is symmetrical about y-axis

    y = tan x is odd function so must be symmetrical about origin & y equals s e c to the power of 2 end exponent invisible function application x is decretive of y = tan x so it must be even imply symmetrical about y-axis or vice-versa

    Statement I : Graph of y = tan x is symmetrical about origin
    Statement II : Graph of y equals sec squared invisible function application x is symmetrical about y-axis

    maths-General
    y = tan x is odd function so must be symmetrical about origin & y equals s e c to the power of 2 end exponent invisible function application x is decretive of y = tan x so it must be even imply symmetrical about y-axis or vice-versa