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General
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Question

If fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction then ascending order of A, B, C.

  1. A,B,C
  2. B,C,A
  3. C,A,B
  4. B,A,C

Hint:

In this question using the equation we will find the value of A, B and C. After finding the values we will arrange the values in ascending order to find the required sequence.

The correct answer is: B,C,A


    space space space fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction
rightwards double arrow A left parenthesis 1 minus x squared right parenthesis plus B left parenthesis 1 plus 2 x right parenthesis left parenthesis 1 minus x right parenthesis plus C left parenthesis 1 plus 2 x right parenthesis left parenthesis 1 plus x right parenthesis equals 1
rightwards double arrow A minus A x squared plus B plus B x minus 2 x squared B plus C plus 3 x C plus 2 x squared C equals 1
rightwards double arrow x squared left parenthesis 2 C minus 2 B minus A right parenthesis plus x open parentheses B plus 3 C close parentheses plus open parentheses A plus B plus C close parentheses equals 1
rightwards double arrow A plus B plus C equals 1 space o r space 2 C minus 2 B minus A equals 0 space o r space B plus 3 C equals 0

space space space space space B plus 3 C equals 0
rightwards double arrow B equals negative 3 C

space space space space A plus B plus C equals 1
rightwards double arrow A minus 3 C plus C equals 1
rightwards double arrow A minus 2 C equals 1
rightwards double arrow A equals 1 plus 2 C

space space space space space 2 C minus 2 B minus A equals 0
rightwards double arrow 2 C minus 2 left parenthesis negative 3 C right parenthesis minus left parenthesis 1 plus 2 C right parenthesis equals 0
rightwards double arrow 2 C plus 6 C minus 1 minus 2 C equals 0
rightwards double arrow 6 C equals 1
rightwards double arrow C equals 1 over 6

A equals 1 plus 2 cross times 1 over 6 equals 1 plus 1 third equals 4 over 3

B equals negative 3 C equals negative 3 cross times 1 over 6 equals fraction numerator negative 1 over denominator 2 end fraction

S o comma space v a l u e s space i n space a s s c e n d i n g space o r d e r space i s space B comma space C comma space A.

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    Related Questions to study

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    The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

    Complete step by step solution:
    We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
    Total number of Digits  = 7

    We are given that the digit two occurs exactly twice in each number.
    Thus, the digit two occurs twice in the seven digit number.
    Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
    Total number of ways that the digit two occurs exactly twice in each number =C presuperscript 7 subscript 2

    Now, the remaining five digits can be written using two digits 1 and 3 in 2 to the power of 5ways.
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     Total number of seven digit number = fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5
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    Total number of seven digit number = fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5
    Total number of seven digit number  = fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5
    Simplifying the expression, we get
    Total number of seven digit number  = 7 cross times 6 cross times 2 to the power of 4

    Multiplying the terms, we get
    Total number of seven digit number  672
    Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

     

    The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

    Maths-General
    Complete step by step solution:
    We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
    Total number of Digits  = 7

    We are given that the digit two occurs exactly twice in each number.
    Thus, the digit two occurs twice in the seven digit number.
    Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
    Total number of ways that the digit two occurs exactly twice in each number =C presuperscript 7 subscript 2

    Now, the remaining five digits can be written using two digits 1 and 3 in 2 to the power of 5ways.
    We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
    Total number of seven digit number = C presuperscript 7 subscript 2 space cross times space 2 to the power of 5
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    C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript , we get
     Total number of seven digit number = fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5
    We know that the factorial can be written by the formula  n! = ncross times(n-1)! , so we get
    Total number of seven digit number = fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5
    Total number of seven digit number  = fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5
    Simplifying the expression, we get
    Total number of seven digit number  = 7 cross times 6 cross times 2 to the power of 4

    Multiplying the terms, we get
    Total number of seven digit number  672
    Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

     
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