Maths-
General
Easy

Question

Which of the following relations is incorrect

  1. left parenthesis A plus B plus.... plus l right parenthesis to the power of ' end exponent equals A to the power of ' end exponent plus B to the power of ' end exponent plus.... plus l to the power of ' end exponent=A'+B'+....+l''/>    
  2. left parenthesis A B.... l right parenthesis to the power of ' end exponent equals A to the power of ' end exponent B to the power of ' end exponent.... l to the power of ' end exponent=A'B'....l''/>    
  3. left parenthesis k A right parenthesis to the power of ' end exponent equals k A to the power of ' end exponent=kA''/>    
  4. left parenthesis A right parenthesis to the power of ' end exponent equals A=A'/>    

The correct answer is: left parenthesis A B.... l right parenthesis to the power of ´ end exponent equals A to the power of ´ end exponent B to the power of ´ end exponent.... l to the power of ´ end exponent


    It is based on fundamental concept

    Related Questions to study

    General
    Maths-

    If A equals open square brackets table row 0 1 row 0 0 end table close square bracketsand A B equals O, then B =

    Since open square brackets table row 0 1 row 0 0 end table close square brackets open square brackets table row cell negative 1 end cell 0 row 0 0 end table close square brackets equals open square brackets table row 0 0 row 0 0 end table close square brackets equals O equals A B
    Þ B equals open square brackets table row cell negative 1 end cell 0 row 0 0 end table close square brackets

    If A equals open square brackets table row 0 1 row 0 0 end table close square bracketsand A B equals O, then B =

    Maths-General
    Since open square brackets table row 0 1 row 0 0 end table close square brackets open square brackets table row cell negative 1 end cell 0 row 0 0 end table close square brackets equals open square brackets table row 0 0 row 0 0 end table close square brackets equals O equals A B
    Þ B equals open square brackets table row cell negative 1 end cell 0 row 0 0 end table close square brackets
    General
    Maths-

    If A equals open square brackets table row cell 1 divided by 3 end cell 2 row 0 cell 2 x minus 3 end cell end table close square brackets comma B equals open square brackets table row 3 6 row 0 cell negative 1 end cell end table close square bracketsand A B equals I, then x =

    vertical line a d j A vertical line
    (As given)
    left right double arrow 3 minus 2 x equals 1 or x equals 1.

    If A equals open square brackets table row cell 1 divided by 3 end cell 2 row 0 cell 2 x minus 3 end cell end table close square brackets comma B equals open square brackets table row 3 6 row 0 cell negative 1 end cell end table close square bracketsand A B equals I, then x =

    Maths-General
    vertical line a d j A vertical line
    (As given)
    left right double arrow 3 minus 2 x equals 1 or x equals 1.
    General
    Maths-

    If A equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets, then A is

    capital delta equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets equals negative 1 not equal to 0, hence matrix is non-singular.

    If A equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets, then A is

    Maths-General
    capital delta equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets equals negative 1 not equal to 0, hence matrix is non-singular.
    parallel
    General
    Maths-

    The values of x comma y comma zin order of the system of equations 3 x plus y plus 2 z equals 3 comma 2 x minus 3 y minus z equals negative 3, x plus 2 y plus z equals 4 comma are

    Put the value left parenthesis x comma y comma z right parenthesis equals left parenthesis 1 , 2 comma negative 1 right parenthesis comma which satisfies the equation. Hence, (d) is correct.

    The values of x comma y comma zin order of the system of equations 3 x plus y plus 2 z equals 3 comma 2 x minus 3 y minus z equals negative 3, x plus 2 y plus z equals 4 comma are

    Maths-General
    Put the value left parenthesis x comma y comma z right parenthesis equals left parenthesis 1 , 2 comma negative 1 right parenthesis comma which satisfies the equation. Hence, (d) is correct.
    General
    Maths-

    If x is a positive integer, then capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar is equal to

    capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar
    = x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 1 cell left parenthesis x plus 1 right parenthesis end cell cell left parenthesis x plus 2 right parenthesis left parenthesis x plus 1 right parenthesis end cell row 1 cell left parenthesis x plus 2 right parenthesis end cell cell left parenthesis x plus 3 right parenthesis left parenthesis x plus 2 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    Applying R subscript 1 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript comma R subscript 2 end subscript rightwards arrow left parenthesis R subscript 3 end subscript minus R subscript 2 end subscript right parenthesis we get
    equals x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 0 1 cell 2 left parenthesis x plus 2 right parenthesis end cell row 0 1 cell 2 left parenthesis x plus 3 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    equals 2 x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial,(on simplification).
    Trick: Put x equals 1and then match the alternate.

    If x is a positive integer, then capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar is equal to

    Maths-General
    capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar
    = x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 1 cell left parenthesis x plus 1 right parenthesis end cell cell left parenthesis x plus 2 right parenthesis left parenthesis x plus 1 right parenthesis end cell row 1 cell left parenthesis x plus 2 right parenthesis end cell cell left parenthesis x plus 3 right parenthesis left parenthesis x plus 2 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    Applying R subscript 1 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript comma R subscript 2 end subscript rightwards arrow left parenthesis R subscript 3 end subscript minus R subscript 2 end subscript right parenthesis we get
    equals x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 0 1 cell 2 left parenthesis x plus 2 right parenthesis end cell row 0 1 cell 2 left parenthesis x plus 3 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    equals 2 x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial,(on simplification).
    Trick: Put x equals 1and then match the alternate.
    General
    Maths-

    The system of equations lambda x plus y plus z equals 0 comma negative x plus lambda y plus z equals 0 comma negative x minus y plus lambda z equals 0, will have a non zero solution if real values of lambdaare given by

    Accordingly, open vertical bar table row lambda 1 1 row cell negative 1 end cell lambda 1 row cell negative 1 end cell cell negative 1 end cell lambda end table close vertical bar equals 0 rightwards double arrow lambda to the power of 3 end exponent plus 3 lambda equals 0
    Therefore lambda equals 0, since lambda equals i square root of 3does not exist.

    The system of equations lambda x plus y plus z equals 0 comma negative x plus lambda y plus z equals 0 comma negative x minus y plus lambda z equals 0, will have a non zero solution if real values of lambdaare given by

    Maths-General
    Accordingly, open vertical bar table row lambda 1 1 row cell negative 1 end cell lambda 1 row cell negative 1 end cell cell negative 1 end cell lambda end table close vertical bar equals 0 rightwards double arrow lambda to the power of 3 end exponent plus 3 lambda equals 0
    Therefore lambda equals 0, since lambda equals i square root of 3does not exist.
    parallel
    General
    Maths-

    Set of equations a plus b minus 2 c equals 0 , 2 a minus 3 b plus c equals 0 and a minus 5 b plus 4 c equals alphais consistent for alphaequal to

    a plus b minus 2 c equals 0
    2 a minus 3 b plus c equals 0
    a minus 5 b plus 4 c equals alpha
    System is consistent, if D equals open vertical bar table row 1 1 cell negative 2 end cell row 2 cell negative 3 end cell 1 row 1 cell negative 5 end cell 4 end table close vertical bar=0
    and D subscript 1 end subscript equals open vertical bar table row 0 1 cell negative 2 end cell row 0 cell negative 3 end cell 1 row alpha cell negative 5 end cell 4 end table close vertical bar= 0 and D subscript 2 end subscript also zero.
    Hence, value of alphais zero.

    Set of equations a plus b minus 2 c equals 0 , 2 a minus 3 b plus c equals 0 and a minus 5 b plus 4 c equals alphais consistent for alphaequal to

    Maths-General
    a plus b minus 2 c equals 0
    2 a minus 3 b plus c equals 0
    a minus 5 b plus 4 c equals alpha
    System is consistent, if D equals open vertical bar table row 1 1 cell negative 2 end cell row 2 cell negative 3 end cell 1 row 1 cell negative 5 end cell 4 end table close vertical bar=0
    and D subscript 1 end subscript equals open vertical bar table row 0 1 cell negative 2 end cell row 0 cell negative 3 end cell 1 row alpha cell negative 5 end cell 4 end table close vertical bar= 0 and D subscript 2 end subscript also zero.
    Hence, value of alphais zero.
    General
    Maths-

    The number of solutions of the equations x plus 4 y minus z equals 0 comma 3 x minus 4 y minus z equals 0 comma x minus 3 y plus z equals 0 is

    The given system of homogeneous equations has capital delta equals open vertical bar table row 1 4 cell negative 1 end cell row 3 cell negative 4 end cell cell negative 1 end cell row 1 cell negative 3 end cell 1 end table close vertical bar equals 1 left parenthesis negative 4 minus 3 right parenthesis minus 4 left parenthesis 3 plus 1 right parenthesis minus 1 left parenthesis negative 9 plus 4 right parenthesis
    equals negative 7 minus 16 plus 5 not equal to 0.
    There exists only one trivial solution.

    The number of solutions of the equations x plus 4 y minus z equals 0 comma 3 x minus 4 y minus z equals 0 comma x minus 3 y plus z equals 0 is

    Maths-General
    The given system of homogeneous equations has capital delta equals open vertical bar table row 1 4 cell negative 1 end cell row 3 cell negative 4 end cell cell negative 1 end cell row 1 cell negative 3 end cell 1 end table close vertical bar equals 1 left parenthesis negative 4 minus 3 right parenthesis minus 4 left parenthesis 3 plus 1 right parenthesis minus 1 left parenthesis negative 9 plus 4 right parenthesis
    equals negative 7 minus 16 plus 5 not equal to 0.
    There exists only one trivial solution.
    General
    Maths-

    open vertical bar table row 0 a cell negative b end cell row cell negative a end cell 0 c row b cell negative c end cell 0 end table close vertical bar equals

    open vertical bar table row 0 a cell negative b end cell row cell negative a end cell 0 c row b cell negative c end cell 0 end table close vertical bar = 0 (Since value of determinant of skew-symmetric matrix of odd orders is 0)

    open vertical bar table row 0 a cell negative b end cell row cell negative a end cell 0 c row b cell negative c end cell 0 end table close vertical bar equals

    Maths-General
    open vertical bar table row 0 a cell negative b end cell row cell negative a end cell 0 c row b cell negative c end cell 0 end table close vertical bar = 0 (Since value of determinant of skew-symmetric matrix of odd orders is 0)
    parallel
    General
    Maths-

    If open vertical bar table row cell x plus 1 end cell 3 5 row 2 cell x plus 2 end cell 5 row 2 3 cell x plus 4 end cell end table close vertical bar equals 0, then x =

    By C subscript 1 end subscript rightwards arrow C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript,
    we have left parenthesis 9 plus x right parenthesis open vertical bar table row 1 3 5 row 1 cell x plus 2 end cell 5 row 1 3 cell x plus 4 end cell end table close vertical bar = 0
    rightwards double arrow left parenthesis x plus 9 right parenthesis open vertical bar table row 0 cell 1 minus x end cell 0 row 0 cell negative left parenthesis 1 minus x right parenthesis end cell cell 1 minus x end cell row 1 3 cell x plus 4 end cell end table close vertical bar equals 0
    rightwards double arrow left parenthesis x plus 9 right parenthesis left parenthesis 1 minus x right parenthesis to the power of 2 end exponent open vertical bar table row 0 1 0 row 0 cell negative 1 end cell 1 row 1 3 cell x plus 4 end cell end table close vertical bar equals 0
    rightwards double arrow x equals 1 , 1 comma negative 9, (Since the determinant = 1)

    If open vertical bar table row cell x plus 1 end cell 3 5 row 2 cell x plus 2 end cell 5 row 2 3 cell x plus 4 end cell end table close vertical bar equals 0, then x =

    Maths-General
    By C subscript 1 end subscript rightwards arrow C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript,
    we have left parenthesis 9 plus x right parenthesis open vertical bar table row 1 3 5 row 1 cell x plus 2 end cell 5 row 1 3 cell x plus 4 end cell end table close vertical bar = 0
    rightwards double arrow left parenthesis x plus 9 right parenthesis open vertical bar table row 0 cell 1 minus x end cell 0 row 0 cell negative left parenthesis 1 minus x right parenthesis end cell cell 1 minus x end cell row 1 3 cell x plus 4 end cell end table close vertical bar equals 0
    rightwards double arrow left parenthesis x plus 9 right parenthesis left parenthesis 1 minus x right parenthesis to the power of 2 end exponent open vertical bar table row 0 1 0 row 0 cell negative 1 end cell 1 row 1 3 cell x plus 4 end cell end table close vertical bar equals 0
    rightwards double arrow x equals 1 , 1 comma negative 9, (Since the determinant = 1)
    General
    Maths-

    open vertical bar table row cell a minus b end cell cell b minus c end cell cell c minus a end cell row cell x minus y end cell cell y minus z end cell cell z minus x end cell row cell p minus q end cell cell q minus r end cell cell r minus p end cell end table close vertical bar equals

    open vertical bar table row cell a minus b end cell cell b minus c end cell cell c minus a end cell row cell x minus y end cell cell y minus z end cell cell z minus x end cell row cell p minus q end cell cell q minus r end cell cell r minus p end cell end table close vertical bar equals open vertical bar table row 0 cell b minus c end cell cell c minus a end cell row 0 cell y minus z end cell cell z minus x end cell row 0 cell q minus r end cell cell r minus p end cell end table close vertical bar = 0

    open vertical bar table row cell a minus b end cell cell b minus c end cell cell c minus a end cell row cell x minus y end cell cell y minus z end cell cell z minus x end cell row cell p minus q end cell cell q minus r end cell cell r minus p end cell end table close vertical bar equals

    Maths-General
    open vertical bar table row cell a minus b end cell cell b minus c end cell cell c minus a end cell row cell x minus y end cell cell y minus z end cell cell z minus x end cell row cell p minus q end cell cell q minus r end cell cell r minus p end cell end table close vertical bar equals open vertical bar table row 0 cell b minus c end cell cell c minus a end cell row 0 cell y minus z end cell cell z minus x end cell row 0 cell q minus r end cell cell r minus p end cell end table close vertical bar = 0
    General
    Maths-

    If x comma y comma zare three consecutive positive integers, then fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 2 x z plus 1 end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 2 x z plus 1 end fraction close parentheses to the power of 3 end exponent plus.... equals

    Since x comma y comma zare three consecutive positive integers, therefore 2 y equals x plus z.
    rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x plus z right parenthesis to the power of 2 end exponent rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x minus z right parenthesis to the power of 2 end exponent plus 4 x z
    rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis negative 2 right parenthesis to the power of 2 end exponent plus 4 x z comma left parenthesis because z minus x equals negative 2 right parenthesis
    rightwards double arrow y to the power of 2 end exponent equals 1 plus x z ....(i)
    Now fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 1 plus 2 x z end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus....
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x plus log subscript e end subscript invisible function application z close plus 2 open open curly brackets open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus.... close curly brackets close square brackets
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus fraction numerator 1 over denominator 1 plus 2 x z end fraction over denominator 1 minus fraction numerator 1 over denominator 1 plus 2 x z end fraction end fraction close parentheses close square brackets
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus x z over denominator x z end fraction close parentheses close square brackets
    equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application left parenthesis 1 plus x z right parenthesis equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application y to the power of 2 end exponent equals log subscript e end subscript invisible function application y.

    If x comma y comma zare three consecutive positive integers, then fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 2 x z plus 1 end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 2 x z plus 1 end fraction close parentheses to the power of 3 end exponent plus.... equals

    Maths-General
    Since x comma y comma zare three consecutive positive integers, therefore 2 y equals x plus z.
    rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x plus z right parenthesis to the power of 2 end exponent rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis x minus z right parenthesis to the power of 2 end exponent plus 4 x z
    rightwards double arrow 4 y to the power of 2 end exponent equals left parenthesis negative 2 right parenthesis to the power of 2 end exponent plus 4 x z comma left parenthesis because z minus x equals negative 2 right parenthesis
    rightwards double arrow y to the power of 2 end exponent equals 1 plus x z ....(i)
    Now fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application x plus fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application z plus fraction numerator 1 over denominator 1 plus 2 x z end fraction plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus....
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x plus log subscript e end subscript invisible function application z close plus 2 open open curly brackets open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction open parentheses fraction numerator 1 over denominator 1 plus 2 x z end fraction close parentheses to the power of 3 end exponent plus.... close curly brackets close square brackets
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus fraction numerator 1 over denominator 1 plus 2 x z end fraction over denominator 1 minus fraction numerator 1 over denominator 1 plus 2 x z end fraction end fraction close parentheses close square brackets
    equals fraction numerator 1 over denominator 2 end fraction open square brackets log subscript e end subscript invisible function application x z plus log subscript e end subscript invisible function application open parentheses fraction numerator 1 plus x z over denominator x z end fraction close parentheses close square brackets
    equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application left parenthesis 1 plus x z right parenthesis equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application y to the power of 2 end exponent equals log subscript e end subscript invisible function application y.
    parallel
    General
    Maths-

    The value of log subscript e end subscript invisible function application open parentheses 1 plus a x to the power of 2 end exponent plus a to the power of 2 end exponent plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses is

    log subscript e end subscript invisible function application open square brackets 1 plus a x to the power of 2 end exponent plus a to the power of 2 end exponent plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close square brackets
    equals log subscript e end subscript invisible function application left parenthesis 1 plus a x to the power of 2 end exponent right parenthesis open parentheses 1 plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses
    equals log subscript e end subscript invisible function application left parenthesis 1 plus a x to the power of 2 end exponent right parenthesis plus log subscript e end subscript invisible function application open parentheses 1 plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses
    = open square brackets a x to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction a to the power of 2 end exponent x to the power of 4 end exponent plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent x to the power of 6 end exponent minus........... close square brackets plus open square brackets fraction numerator a over denominator x to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator 2 end fraction a to the power of 2 end exponent open parentheses fraction numerator 1 over denominator x to the power of 4 end exponent end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent open parentheses fraction numerator 1 over denominator x to the power of 6 end exponent end fraction close parentheses minus..... close square brackets
    equals a open parentheses x to the power of 2 end exponent plus fraction numerator 1 over denominator x to the power of 2 end exponent end fraction close parentheses minus fraction numerator 1 over denominator 2 end fraction. a to the power of 2 end exponent open parentheses x to the power of 4 end exponent plus fraction numerator 1 over denominator x to the power of 4 end exponent end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent open parentheses x to the power of 6 end exponent plus fraction numerator 1 over denominator x to the power of 6 end exponent end fraction close parentheses minus...

    The value of log subscript e end subscript invisible function application open parentheses 1 plus a x to the power of 2 end exponent plus a to the power of 2 end exponent plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses is

    Maths-General
    log subscript e end subscript invisible function application open square brackets 1 plus a x to the power of 2 end exponent plus a to the power of 2 end exponent plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close square brackets
    equals log subscript e end subscript invisible function application left parenthesis 1 plus a x to the power of 2 end exponent right parenthesis open parentheses 1 plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses
    equals log subscript e end subscript invisible function application left parenthesis 1 plus a x to the power of 2 end exponent right parenthesis plus log subscript e end subscript invisible function application open parentheses 1 plus fraction numerator a over denominator x to the power of 2 end exponent end fraction close parentheses
    = open square brackets a x to the power of 2 end exponent minus fraction numerator 1 over denominator 2 end fraction a to the power of 2 end exponent x to the power of 4 end exponent plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent x to the power of 6 end exponent minus........... close square brackets plus open square brackets fraction numerator a over denominator x to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator 2 end fraction a to the power of 2 end exponent open parentheses fraction numerator 1 over denominator x to the power of 4 end exponent end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent open parentheses fraction numerator 1 over denominator x to the power of 6 end exponent end fraction close parentheses minus..... close square brackets
    equals a open parentheses x to the power of 2 end exponent plus fraction numerator 1 over denominator x to the power of 2 end exponent end fraction close parentheses minus fraction numerator 1 over denominator 2 end fraction. a to the power of 2 end exponent open parentheses x to the power of 4 end exponent plus fraction numerator 1 over denominator x to the power of 4 end exponent end fraction close parentheses plus fraction numerator 1 over denominator 3 end fraction a to the power of 3 end exponent open parentheses x to the power of 6 end exponent plus fraction numerator 1 over denominator x to the power of 6 end exponent end fraction close parentheses minus...
    General
    Maths-

    The sum of the products of the elements of any row of a determinant A with the same row is always equal to

    We know that the row to row multiplication of a determinant is always equal to the value of the determinant i.e., | A|.

    The sum of the products of the elements of any row of a determinant A with the same row is always equal to

    Maths-General
    We know that the row to row multiplication of a determinant is always equal to the value of the determinant i.e., | A|.
    General
    Maths-

    open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell a b end cell cell c a end cell row cell a b end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b c end cell row cell c a end cell cell b c end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar is divisor of

    open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell a b end cell cell c a end cell row cell a b end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b c end cell row cell c a end cell cell b c end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Multiply C subscript 1 end subscript comma C subscript 2 end subscript comma C subscript 3 end subscript by a comma b comma crespectively and hence divide by abc
    \ capital delta equals fraction numerator 1 over denominator a b c end fraction open vertical bar table row cell a left parenthesis a to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell cell a b to the power of 2 end exponent end cell cell c to the power of 2 end exponent a end cell row cell a to the power of 2 end exponent b end cell cell b left parenthesis b to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell cell b c to the power of 2 end exponent end cell row cell c a to the power of 2 end exponent end cell cell b to the power of 2 end exponent c end cell cell c left parenthesis c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell end table close vertical bar
    Now take out a, b and c common from R subscript 1 end subscript comma R subscript 2 end subscript and R subscript 3 end subscript,
    \ capital delta equals open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row cell a to the power of 2 end exponent end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row cell a to the power of 2 end exponent end cell cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Now applying C subscript 1 end subscript rightwards arrow C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript
    rightwards double arrow capital delta equals left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis open vertical bar table row 1 cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row 1 cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row 1 cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Þ capital delta equals x to the power of 4 end exponent left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis
    Hence, it is divisible by x to the power of 2 end exponent

    open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell a b end cell cell c a end cell row cell a b end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b c end cell row cell c a end cell cell b c end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar is divisor of

    Maths-General
    open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell a b end cell cell c a end cell row cell a b end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b c end cell row cell c a end cell cell b c end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Multiply C subscript 1 end subscript comma C subscript 2 end subscript comma C subscript 3 end subscript by a comma b comma crespectively and hence divide by abc
    \ capital delta equals fraction numerator 1 over denominator a b c end fraction open vertical bar table row cell a left parenthesis a to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell cell a b to the power of 2 end exponent end cell cell c to the power of 2 end exponent a end cell row cell a to the power of 2 end exponent b end cell cell b left parenthesis b to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell cell b c to the power of 2 end exponent end cell row cell c a to the power of 2 end exponent end cell cell b to the power of 2 end exponent c end cell cell c left parenthesis c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis end cell end table close vertical bar
    Now take out a, b and c common from R subscript 1 end subscript comma R subscript 2 end subscript and R subscript 3 end subscript,
    \ capital delta equals open vertical bar table row cell a to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row cell a to the power of 2 end exponent end cell cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row cell a to the power of 2 end exponent end cell cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Now applying C subscript 1 end subscript rightwards arrow C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript
    rightwards double arrow capital delta equals left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis open vertical bar table row 1 cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row 1 cell b to the power of 2 end exponent plus x to the power of 2 end exponent end cell cell c to the power of 2 end exponent end cell row 1 cell b to the power of 2 end exponent end cell cell c to the power of 2 end exponent plus x to the power of 2 end exponent end cell end table close vertical bar
    Þ capital delta equals x to the power of 4 end exponent left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus c to the power of 2 end exponent plus x to the power of 2 end exponent right parenthesis
    Hence, it is divisible by x to the power of 2 end exponent
    parallel

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