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Question

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=

  1. fraction numerator 1 plus square root of 1 minus x to the power of 2 end exponent end root over denominator x end fraction minus 2 t a n to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root plus C    
  2. fraction numerator 1 plus square root of 1 minus x to the power of 2 end exponent end root over denominator x end fraction minus 2 t a n to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 plus x over denominator 1 minus x end fraction end root plus C    
  3. fraction numerator 1 plus square root of 1 minus x to the power of 2 end exponent end root over denominator x end fraction plus 2 t a n to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 plus x over denominator 1 minus x end fraction end root plus C    
  4. fraction numerator 1 plus square root of 1 minus x to the power of 2 end exponent end root over denominator x end fraction plus 2 stack stack t a n to the power of negative 1 end exponent with _ below with _ below square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root plus C    

The correct answer is: fraction numerator 1 plus square root of 1 minus x to the power of 2 end exponent end root over denominator x end fraction plus 2 t a n to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 plus x over denominator 1 minus x end fraction end root plus C

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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

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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 -

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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

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

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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:

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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:

maths-General
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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

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General
maths-

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
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maths-

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
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
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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
maths-

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
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
parallel
General
maths-

Statement- 1 colon If f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line Where 2 less than x less than 3 is an identity function.
Statement- 2 colon f colon A rightwards arrowR defined by f left parenthesis x right parenthesis equals x is an identity function.

(I) f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line
equals 2 less than x less than 3
equals x identity function (correct)
(II) f left parenthesis x right parenthesis equals x is an identity function (correct)

Statement- 1 colon If f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line Where 2 less than x less than 3 is an identity function.
Statement- 2 colon f colon A rightwards arrowR defined by f left parenthesis x right parenthesis equals x is an identity function.

maths-General
(I) f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line
equals 2 less than x less than 3
equals x identity function (correct)
(II) f left parenthesis x right parenthesis equals x is an identity function (correct)
General
Maths-

Assertion (A) : Graph of open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text  and  end text x comma y element of R close curly brackets
Reason (R) : In the expression am/n, where a, m, n × J+, m represents the power to which a is be raised, whereas n determines the root to be taken; these two processes may be administered in either order with the same result.

Assertion (A) : Graph of open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text  and  end text x comma y element of R close curly brackets
Reason (R) : In the expression am/n, where a, m, n × J+, m represents the power to which a is be raised, whereas n determines the root to be taken; these two processes may be administered in either order with the same result.

Maths-General
General
maths-

Assertion: The period of f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket is 1/2.
Reason: The period of x – [x] is 1.

Assertion: The period of f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket is 1/2.
Reason: The period of x – [x] is 1.

maths-General
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