Maths-
General
Easy

Question

Statement-I : If fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction then A=7 over 4
Statement-II : If fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction then p equals 2 comma q equals 3

Which of the above statements is true

  1. Only I is true
  2. Only II is true
  3. both I & II are true
  4. Neither I nor II true

Hint:

In this question we will solve both the questions to know whether they are true or not. In both statements we will first solve the fractions in RHS. Then we will find the coefficients of x squared, x and constant term after that we will compare both sides to get the value of A,B,C and p and q.

The correct answer is: Only I is true


    Statement-I : If fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction then A=7 over 4
    fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator A left parenthesis x plus 1 right parenthesis squared plus B left parenthesis x minus 1 right parenthesis left parenthesis x plus 1 right parenthesis plus C left parenthesis x minus 1 right parenthesis over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator A left parenthesis x squared plus 2 x plus 1 right parenthesis plus B left parenthesis x squared minus 1 right parenthesis plus C x minus C over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator A x squared plus A 2 x plus A plus B x squared minus B plus C x minus C over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator x squared left parenthesis A plus B right parenthesis plus x left parenthesis 2 A plus C right parenthesis plus 1 left parenthesis A minus B minus C right parenthesis over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction
N o w comma space o n space c o m p a r i n g space b o t h space s i d e s space comma space
A plus B equals 0 comma space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 A plus C equals 3 comma space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space A minus B minus C equals 4
rightwards double arrow A equals negative B space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow negative 2 B plus C equals 3 space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow negative B minus B minus C equals 4
p u t t i n g space B equals fraction numerator negative 7 over denominator 4 end fraction space space space space space space space space space space space space space space space space rightwards double arrow space minus 2 B equals 3 minus C space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow negative 2 B minus C equals 4
rightwards double arrow A equals 7 over 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space p u t t i n g C equals fraction numerator negative 1 over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space 3 minus C minus C equals 4
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow negative 2 B equals 3 plus 1 half space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space 3 minus 2 C equals 4
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow B equals fraction numerator negative 7 over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space C equals fraction numerator negative 1 over denominator 2 end fraction space space space space space space space space
    statement 1 is true.
    Statement-II : If fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction then p equals 2 comma q equals 3  
    fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction
space space space space space space space space space space space space space space space space equals fraction numerator left parenthesis 2 x minus 3 right parenthesis plus 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction
space space space space space space space space space space space space space space space space equals fraction numerator 2 x over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction
N o w comma space o n space c o m p a r i n g space b o t h space s i d e s space comma space
p equals 2 space a n d space q equals 0
    statement 2 is false.

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    Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
    Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

    Which of the above statements is true

    Statement-I: If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
    equals fraction numerator A left parenthesis x plus 1 right parenthesis squared plus B left parenthesis x plus 1 right parenthesis plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A left parenthesis x squared plus 2 x plus 1 right parenthesis plus B left parenthesis x plus 1 right parenthesis plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A x squared plus A 2 x plus A plus B x plus B plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A x squared plus x open parentheses 2 A plus B close parentheses plus A plus B plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
    Now, A=1         2A+B=3              A+B+C=1
    rightwards double arrow 2 plus B equals 3
rightwards double arrow B equals 1         rightwards double arrow 1 plus 1 plus C equals 1
rightwards double arrow C equals 1 minus 2
rightwards double arrow C equals negative 1
    A+B+C=1+1-1=-1
    Statement 1 is not true.
    Statement-II :If
    fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0
space space space space space space space space space space space space space space space space space space space space equals fraction numerator A x squared plus B x plus C over denominator x cubed end fraction
    Now, A=1, B=2, C=3
    A+B-C=0
    1+2-3=0
    So, statement 2 is true.

    Statement-I : If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
    Statement-II :If fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0

    Which of the above statements is true

    Maths-General
    Statement-I: If fraction numerator x squared plus 3 x plus 1 over denominator x squared plus 2 x plus 1 end fraction equals A plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction text  then  end text bold italic A plus bold italic B plus bold italic C equals bold 0
    equals fraction numerator A left parenthesis x plus 1 right parenthesis squared plus B left parenthesis x plus 1 right parenthesis plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A left parenthesis x squared plus 2 x plus 1 right parenthesis plus B left parenthesis x plus 1 right parenthesis plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A x squared plus A 2 x plus A plus B x plus B plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
equals fraction numerator A x squared plus x open parentheses 2 A plus B close parentheses plus A plus B plus C over denominator left parenthesis x plus 1 right parenthesis squared end fraction
    Now, A=1         2A+B=3              A+B+C=1
    rightwards double arrow 2 plus B equals 3
rightwards double arrow B equals 1         rightwards double arrow 1 plus 1 plus C equals 1
rightwards double arrow C equals 1 minus 2
rightwards double arrow C equals negative 1
    A+B+C=1+1-1=-1
    Statement 1 is not true.
    Statement-II :If
    fraction numerator x squared plus 2 x plus 3 over denominator x cubed end fraction equals A over x plus B over x squared plus C over x cubed text  then  end text A plus B minus C equals 0
space space space space space space space space space space space space space space space space space space space space equals fraction numerator A x squared plus B x plus C over denominator x cubed end fraction
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    If x, y are rational number such that x plus y plus left parenthesis x minus 2 y right parenthesis square root of 2 equals 2 x minus y plus left parenthesis x minus y minus 1 right parenthesis square root of 6 Then

    A s space t h e y space a r e space r a t i o n a l space n u m b e r comma space c o f f i c i e n t s space o f space square root of 2 comma square root of 6 s h o u l d space b e space 0.
t h u s comma space x minus 2 y equals 0 space space _ _ _ _ space left parenthesis i right parenthesis space a n d space x minus y minus 1 equals 0 space _ _ _ _ _ _ _ left parenthesis i i right parenthesis
S u b t r a c t i n g space left parenthesis i right parenthesis space a n d space left parenthesis i i right parenthesis comma space w e space g e t colon
rightwards double arrow negative y plus 1 equals 0
rightwards double arrow y equals 1
S u b s t i t u t i n g space v a l u e space o f space y space i n space r q u a t i o n space left parenthesis i right parenthesis comma
rightwards double arrow x minus 2 cross times 1 equals 0
rightwards double arrow x equals 2
S o comma space x equals 2 space a n d space y space equals 1.

    If x, y are rational number such that x plus y plus left parenthesis x minus 2 y right parenthesis square root of 2 equals 2 x minus y plus left parenthesis x minus y minus 1 right parenthesis square root of 6 Then

    Maths-General
    A s space t h e y space a r e space r a t i o n a l space n u m b e r comma space c o f f i c i e n t s space o f space square root of 2 comma square root of 6 s h o u l d space b e space 0.
t h u s comma space x minus 2 y equals 0 space space _ _ _ _ space left parenthesis i right parenthesis space a n d space x minus y minus 1 equals 0 space _ _ _ _ _ _ _ left parenthesis i i right parenthesis
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rightwards double arrow negative y plus 1 equals 0
rightwards double arrow y equals 1
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rightwards double arrow x minus 2 cross times 1 equals 0
rightwards double arrow x equals 2
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    A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

    Number of groups having 4 boys and 1 girl
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    and number of groups having 3 boys and 2 girls
    = (4C3) (gC2) = 2g(g – 1)
    Thus, the number of dolls distributed
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    = 4g2 – 3g
    We are given 4g2 – 3g = 85 rightwards double arrow g = 5.

    A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

    maths-General
    Number of groups having 4 boys and 1 girl
    = (4C4) (gC1) = g
    and number of groups having 3 boys and 2 girls
    = (4C3) (gC2) = 2g(g – 1)
    Thus, the number of dolls distributed
    = g(1) + (2)[2g (g – 1)]
    = 4g2 – 3g
    We are given 4g2 – 3g = 85 rightwards double arrow g = 5.
    General
    maths-

    There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

    Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
    x + 2x + y = 20 or 3x + y = 20
    rightwards double arrowy = 20 – 3x.
    As 0 less or equal than y 20, we get 0 less or equal than 20 – 3x less or equal than 20
    rightwards double arrow 0 less or equal than 3x less or equal than 20 or 0 less or equal thanless or equal than 6
    thereforeThe number of ways of selecting the balls is 7.

    There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

    maths-General
    Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
    x + 2x + y = 20 or 3x + y = 20
    rightwards double arrowy = 20 – 3x.
    As 0 less or equal than y 20, we get 0 less or equal than 20 – 3x less or equal than 20
    rightwards double arrow 0 less or equal than 3x less or equal than 20 or 0 less or equal thanless or equal than 6
    thereforeThe number of ways of selecting the balls is 7.
    parallel
    General
    maths-

    If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -

    We have (a3 – a) + (b3 – b) + (c3 – c)
    = (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
    Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
    6 | {(a3 – a) + (b3 – b) + (c3 – c)}
    rightwards double arrow6 | (a3 + b3 + c3) as 6|(a + b + c).

    If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -

    maths-General
    We have (a3 – a) + (b3 – b) + (c3 – c)
    = (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
    Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
    6 | {(a3 – a) + (b3 – b) + (c3 – c)}
    rightwards double arrow6 | (a3 + b3 + c3) as 6|(a + b + c).
    General
    maths-

    For x element of R, let [x] denote the greatest integer less or equal than x, then value ofopen square brackets negative fraction numerator 1 over denominator 3 end fraction close square brackets+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 100 end fraction close square brackets+ open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 2 over denominator 100 end fraction close square brackets+…+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 99 over denominator 100 end fraction close square bracketsis -

    For 0 less or equal thanless or equal than 66, 0 less or equal than fraction numerator r over denominator 100 end fraction< fraction numerator 2 over denominator 3 end fraction
    rightwards double arrowfraction numerator 2 over denominator 3 end fraction < – fraction numerator r over denominator 100 end fraction less or equal than 0
    rightwards double arrowfraction numerator 1 over denominator 3 end fractionfraction numerator 2 over denominator 3 end fraction< –fraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –1 for 0 less or equal thanless or equal than 66
    Also, for 67 less or equal thanless or equal than 100, fraction numerator 67 over denominator 100 end fractionless or equal than fraction numerator r over denominator 100 end fraction less or equal than 1
    rightwards double arrow–1 less or equal thanfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 67 over denominator 100 end fraction
    rightwards double arrowfraction numerator 1 over denominator 3 end fraction – 1 less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator 67 over denominator 100 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –2 for 67 less or equal thanless or equal than 100
    Hence, not stretchy sum subscript r equals 0 end subscript superscript 100 end superscript open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= 67(–1) + 2(–34) = –135.

    For x element of R, let [x] denote the greatest integer less or equal than x, then value ofopen square brackets negative fraction numerator 1 over denominator 3 end fraction close square brackets+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 100 end fraction close square brackets+ open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 2 over denominator 100 end fraction close square brackets+…+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 99 over denominator 100 end fraction close square bracketsis -

    maths-General
    For 0 less or equal thanless or equal than 66, 0 less or equal than fraction numerator r over denominator 100 end fraction< fraction numerator 2 over denominator 3 end fraction
    rightwards double arrowfraction numerator 2 over denominator 3 end fraction < – fraction numerator r over denominator 100 end fraction less or equal than 0
    rightwards double arrowfraction numerator 1 over denominator 3 end fractionfraction numerator 2 over denominator 3 end fraction< –fraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –1 for 0 less or equal thanless or equal than 66
    Also, for 67 less or equal thanless or equal than 100, fraction numerator 67 over denominator 100 end fractionless or equal than fraction numerator r over denominator 100 end fraction less or equal than 1
    rightwards double arrow–1 less or equal thanfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 67 over denominator 100 end fraction
    rightwards double arrowfraction numerator 1 over denominator 3 end fraction – 1 less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator 67 over denominator 100 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –2 for 67 less or equal thanless or equal than 100
    Hence, not stretchy sum subscript r equals 0 end subscript superscript 100 end superscript open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= 67(–1) + 2(–34) = –135.
    General
    maths-

    The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

    Let t = z + 1
    Equation reduces to x + y + t = 25
    less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
    Required number of ways
    = Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
    = Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
    = Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
    = Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
    = Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
    = 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
    = 14C29C28C2 + 3C2
    = 91 – 36 – 28 + 3 = 30.

    The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

    maths-General
    Let t = z + 1
    Equation reduces to x + y + t = 25
    less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
    Required number of ways
    = Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
    = Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
    = Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
    = Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
    = Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
    = 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
    = 14C29C28C2 + 3C2
    = 91 – 36 – 28 + 3 = 30.
    parallel

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