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CBSE Class 12-science Answered

<div>If&nbsp;q&sup2;&minus;4pr=0&cedil;p&gt;0&cedil;then find the domain of the function f(x)=log(px&sup3;+(p+q)x&sup2;+(q+r)x+r).</div>
Asked by yeeshuraj3 | 22 Jul, 2016, 06:46: PM
Expert Answer
begin mathsize 16px style straight f left parenthesis straight x right parenthesis equals log left parenthesis px cubed plus left parenthesis straight p plus straight q right parenthesis squared straight x plus left parenthesis straight q plus straight r right parenthesis straight x plus straight r right parenthesis
Since space log subscript straight a straight x space is space defined space for space all space straight x greater than 0 comma
therefore comma space straight f left parenthesis straight x right parenthesis space is space defined space if space px cubed plus left parenthesis straight p plus straight q right parenthesis squared straight x plus left parenthesis straight q plus straight r right parenthesis straight x plus straight r space greater than space 0
Since space straight x plus 1 space is space straight a space factor space of space px cubed plus left parenthesis straight p plus straight q right parenthesis squared straight x plus left parenthesis straight q plus straight r right parenthesis straight x plus straight r.... left parenthesis By space Synthetic space division right parenthesis
rightwards double arrow px cubed plus left parenthesis straight p plus straight q right parenthesis squared straight x plus left parenthesis straight q plus straight r right parenthesis straight x plus straight r equals left parenthesis px squared plus qx plus straight r right parenthesis left parenthesis straight x plus 1 right parenthesis.... left parenthesis straight i right parenthesis
straight f left parenthesis straight x right parenthesis equals log left parenthesis px cubed plus left parenthesis straight p plus straight q right parenthesis squared straight x plus left parenthesis straight q plus straight r right parenthesis straight x plus straight r right parenthesis
rightwards double arrow straight f left parenthesis straight x right parenthesis equals log left square bracket left parenthesis px squared plus qx plus straight r right parenthesis left parenthesis straight x plus 1 right parenthesis right square bracket... left parenthesis from space left parenthesis straight i right parenthesis right parenthesis
rightwards double arrow straight f left parenthesis straight x right parenthesis equals log left parenthesis px squared plus qx plus straight r right parenthesis plus log left parenthesis straight x plus 1 right parenthesis
rightwards double arrow px squared plus qx plus straight r space greater than 0 space and space straight x plus 1 greater than 0
rightwards double arrow straight x equals fraction numerator negative straight q plus-or-minus square root of straight q squared minus 4 pr end root over denominator 2 straight p end fraction greater than 0 space and space straight x greater than negative 1
rightwards double arrow straight x equals fraction numerator negative straight q over denominator 2 straight p end fraction greater than 0 space and space straight x greater than negative 1........ left parenthesis Since space given space that space straight q squared minus 4 pr equals 0 space and space straight p greater than 0 right parenthesis
rightwards double arrow straight x space greater than fraction numerator negative straight q over denominator 2 straight p end fraction
Hence comma space the space domain space of space the space function space is space straight x space greater than fraction numerator negative straight q over denominator 2 straight p end fraction. end style
Answered by Rebecca Fernandes | 25 Jul, 2016, 10:33: AM
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