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Question

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=

2
1
-1
-2

hint Hint:

Taking LCM on RHS and simplifying , then take any value of x and find a,
Equate the coefficients and find the required value.

The correct answer is: -1

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

Take LCM on RHS
$rightwards double arrow fraction numerator 2 x plus 1 over denominator left parenthesis x minus 1 right parenthesis open parentheses x squared plus 2 close parentheses end fraction equals fraction numerator A open parentheses x squared plus 2 close parentheses space plus left parenthesis B x space plus c right parenthesis space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis open parentheses x squared plus 2 close parentheses end fraction C a n c e l l i n g space d e n o m i n a t o r s space o n space b o t h space s i d e space rightwards double arrow 2 x plus 1 space equals space A open parentheses x squared plus 2 close parentheses space plus left parenthesis B x space plus c right parenthesis space left parenthesis x minus 1 right parenthesis W h e n space x space equals space 1 rightwards double arrow 2 plus 1 space equals space A open parentheses 1 plus 2 close parentheses space plus space 0 rightwards double arrow space A space equals space 1 E q u a t i n g space c o e f f i c i e n t s space o f space x squared space a n d space x space b y space c o m p a r i n g space L H S thin space a n d space R H S comma space w e space g e t rightwards double arrow A space plus space B space equals space 0 space rightwards double arrow space B space equals space minus A rightwards double arrow T h u s comma space B space equals space minus 1$

Related Questions to study

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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Reason : f(x) = sgn x is always a positive function.

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

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

parallel

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parallel

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

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parallel

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Statement II : $left parenthesis x minus 1 right parenthesis to the power of 2 end exponent equals fraction numerator 4 y minus 3 over denominator 1 minus y end fraction$ .

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parallel

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Statement II : sin x and {x} are both periodic with period $2 pi$ and 1 respectively.

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Statement II : sin x and {x} are both periodic with period $2 pi$ and 1 respectively.

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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

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

parallel

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