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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

The correct answer is: Only I is true

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If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

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If alpha comma beta be that roots 4 x squared minus 16 x plus lambda equals 0 where lambda element of R,  such that 1 less than alpha less than 2 and 2 less than beta less than 3 then the number of integral solutions of λ is

If alpha comma beta be that roots 4 x squared minus 16 x plus lambda equals 0 where lambda element of R,  such that 1 less than alpha less than 2 and 2 less than beta less than 3 then the number of integral solutions of λ is

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If α,β then the equationx squared minus 3 x plus 1 equals 0 with roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction will be

If α,β then the equationx squared minus 3 x plus 1 equals 0 with roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction will be

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The conic with length of latus rectum 6 and eccentricity is

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For the circle r equals 6 C o s space theta centre and radius are

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The polar equation of the circle of radius 5 and touching the initial line at the pole is

The polar equation of the circle of radius 5 and touching the initial line at the pole is

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The circle with centre at and radius 2 is

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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
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

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General
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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

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
General
maths-

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
= (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.

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.
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.