Maths-
General
Easy

Question

If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

  1. (5, 1)    
  2. (6, 2)    
  3. (7, 3)    
  4. (9, 6)    

Hint:

U sin g space f o r m u l a comma space e x p a n d space t h e space a b o v e space g i v e n space e q u a t i o n s
P presuperscript n subscript r space equals space space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction
a n d space s i m p l i f y

The correct answer is: (6, 2)


    G i v e n space colon space P presuperscript left parenthesis m plus n right parenthesis end presuperscript subscript 2 equals space 56 space a n d space space P presuperscript m minus n end presuperscript subscript 2 space equals space 12

U sin g space f o r m u l a comma space e x p a n d space t h e space a b o v e space g i v e n space e q u a t i o n s
P presuperscript n subscript r space equals space space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction

P presuperscript left parenthesis m plus n right parenthesis end presuperscript subscript 2 equals space 56
rightwards double arrow fraction numerator left parenthesis m plus n right parenthesis factorial over denominator left parenthesis m plus n minus 2 right parenthesis factorial end fraction equals space 56
rightwards double arrow fraction numerator left parenthesis m plus n right parenthesis space left parenthesis m plus n minus 1 right parenthesis space left parenthesis m plus n minus 2 right parenthesis factorial over denominator left parenthesis m plus n minus 2 right parenthesis factorial end fraction equals space 56
rightwards double arrow left parenthesis m plus n right parenthesis space left parenthesis m plus n minus 1 right parenthesis space equals space 56 space

L e t space m space plus space n space equals space t

rightwards double arrow left parenthesis t right parenthesis space left parenthesis t minus 1 right parenthesis space equals space 56 space
rightwards double arrow t squared space minus space t space minus 56 space equals space 0

O n space f a c t o r i z i n g space b y space s p l i t t i n g space o f space m i d d l e space t e r m s
rightwards double arrow left parenthesis t minus 8 right parenthesis left parenthesis t space plus space 7 right parenthesis space equals space 0
rightwards double arrow space t space equals space 8 comma space minus 7 space left parenthesis n e g a t i v e space v a l u e space c a n n o t space b e space t a k e n right parenthesis
rightwards double arrow space t space equals space 8
rightwards double arrow space m space plus space n space equals space 8 space........ space left parenthesis 1 right parenthesis
    Similarly
    P presuperscript left parenthesis m minus n right parenthesis end presuperscript subscript 2 equals space 12
rightwards double arrow fraction numerator left parenthesis m minus n right parenthesis factorial over denominator left parenthesis m minus n minus 2 right parenthesis factorial end fraction equals space 56
rightwards double arrow fraction numerator left parenthesis m minus n right parenthesis space left parenthesis m minus n minus 1 right parenthesis space left parenthesis m minus n minus 2 right parenthesis factorial over denominator left parenthesis m minus n minus 2 right parenthesis factorial end fraction equals space 12
rightwards double arrow left parenthesis m minus n right parenthesis space left parenthesis m minus n minus 1 right parenthesis space equals space 12

L e t space m minus space n space equals space a

rightwards double arrow left parenthesis a right parenthesis space left parenthesis a minus 1 right parenthesis space equals space 12 space
rightwards double arrow a squared space minus space a space minus 12 space equals space 0

O n space f a c t o r i z i n g space b y space s p l i t t i n g space o f space m i d d l e space t e r m s
rightwards double arrow left parenthesis a minus 4 right parenthesis left parenthesis a space plus space 3 right parenthesis space equals space 0
rightwards double arrow space a space equals space 4 comma space minus 3 space left parenthesis n e g a t i v e space v a l u e space c a n n o t space b e space t a k e n right parenthesis
rightwards double arrow space a space equals space 4
rightwards double arrow space m space minus space n space equals space 4 space........ space left parenthesis 2 right parenthesis


    Adding (1) and (2), we get
    2 m space equals space 12
rightwards double arrow space m space equals space 6

    Substituting m= 6 in (2), we get
    6 minus n space equals space 4
rightwards double arrow n equals 2
     Thus, If P presuperscript left parenthesis m plus n right parenthesis end presuperscript subscript 2 equals space 56 and P presuperscript m minus n end presuperscript subscript 2 then (m, n) equals- (6,2)

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    General
    physics-

    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    physics-General
    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis
    General
    physics-

    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
    But stack a with rightwards arrow on top subscript 1 end subscript is in downward direction and in the upward direction i e, stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript
    therefore Acceleration of centre of mass
    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g

    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    physics-General
    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
    But stack a with rightwards arrow on top subscript 1 end subscript is in downward direction and in the upward direction i e, stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript
    therefore Acceleration of centre of mass
    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g
    General
    maths-

    If x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis. to straight infinity terms, vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis straight infinity terms, vertical line a vertical line less than 1, then x colon y

    If x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis. to straight infinity terms, vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis straight infinity terms, vertical line a vertical line less than 1, then x colon y

    maths-General
    General
    Maths-

    The coefficient of x to the power of negative n end exponent in left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent is

    The coefficient of x to the power of negative n end exponent in left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent is

    Maths-General
    General
    maths-

    open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p

    open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p

    maths-General
    General
    chemistry-

    Compounds (A) and (B) are – 

    Compounds (A) and (B) are – 

    chemistry-General
    General
    Maths-

    2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

    2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

    Maths-General
    General
    maths-

    A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

    A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

    maths-General
    General
    Maths-

    If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

    If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

    Maths-General
    General
    Maths-

    If the quadratic equation a x squared plus 2 c x plus b equals 0 and a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis have a common root then a plus 4 b plus 4 c is equal to

    If the quadratic equation a x squared plus 2 c x plus b equals 0 and a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis have a common root then a plus 4 b plus 4 c is equal to

    Maths-General
    General
    physics-

    Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14m s to the power of negative 1 end exponent to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

    At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
    Hence,
    v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent

    Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14m s to the power of negative 1 end exponent to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

    physics-General
    At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
    Hence,
    v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent
    General
    Maths-

    In a Δabc if b+c=3a then cot invisible function application straight B over 2 times cot invisible function application straight C over 2 has the value equal to –

    In a Δabc if b+c=3a then cot invisible function application straight B over 2 times cot invisible function application straight C over 2 has the value equal to –

    Maths-General
    General
    Maths-

    In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

    In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

    Maths-General
    General
    Maths-

    In a triangle ABC, a: b: c = 4: 5: 6. Then 3A + B equals to :

    In a triangle ABC, a: b: c = 4: 5: 6. Then 3A + B equals to :

    Maths-General
    General
    physics-

    A bullet of mass m is fired with a velocity of 50 m s to the power of negative 1 end exponent at an angle theta with the horizontal. At the highest point of its trajectory, it collides had on with a bob of massless string of length l equals 10 divided by 3m and gets embedded in the bob. After the collision, the string moves to an angle of 120degree. What is the angle theta ?

    Velocity of bullet at highest point of its trajectory = 50 cos invisible function application theta in horizontal direction.
    As bullet of mass m collides with pendulum bob of mass 3m and two stick together, their common velocity
    v to the power of ´ end exponent equals fraction numerator m subscript 1 end subscript 50 cos invisible function application theta over denominator m plus 3 n end fraction equals fraction numerator 25 over denominator 2 end fraction cos invisible function application theta m s to the power of negative 1 end exponent
    As now under this velocity v ´ pendulum bob goes up to an angel 120degree, hence
    fraction numerator v to the power of ´ 2 end exponent over denominator 2 g end fraction equals h equals l open parentheses 1 minus cos invisible function application 120 degree close parentheses equals fraction numerator 10 over denominator 3 end fraction open square brackets 1 minus open parentheses f minus fraction numerator 1 over denominator 2 end fraction close parentheses close square brackets equals 5
    rightwards double arrow blank v to the power of ´ 2 end exponent equals 2 cross times 10 cross times 5 equals 100 or v to the power of ´ end exponent equals 10
    Comparing two answer of v ´, we get
    fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses

    A bullet of mass m is fired with a velocity of 50 m s to the power of negative 1 end exponent at an angle theta with the horizontal. At the highest point of its trajectory, it collides had on with a bob of massless string of length l equals 10 divided by 3m and gets embedded in the bob. After the collision, the string moves to an angle of 120degree. What is the angle theta ?

    physics-General
    Velocity of bullet at highest point of its trajectory = 50 cos invisible function application theta in horizontal direction.
    As bullet of mass m collides with pendulum bob of mass 3m and two stick together, their common velocity
    v to the power of ´ end exponent equals fraction numerator m subscript 1 end subscript 50 cos invisible function application theta over denominator m plus 3 n end fraction equals fraction numerator 25 over denominator 2 end fraction cos invisible function application theta m s to the power of negative 1 end exponent
    As now under this velocity v ´ pendulum bob goes up to an angel 120degree, hence
    fraction numerator v to the power of ´ 2 end exponent over denominator 2 g end fraction equals h equals l open parentheses 1 minus cos invisible function application 120 degree close parentheses equals fraction numerator 10 over denominator 3 end fraction open square brackets 1 minus open parentheses f minus fraction numerator 1 over denominator 2 end fraction close parentheses close square brackets equals 5
    rightwards double arrow blank v to the power of ´ 2 end exponent equals 2 cross times 10 cross times 5 equals 100 or v to the power of ´ end exponent equals 10
    Comparing two answer of v ´, we get
    fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses