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If f colon R not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals x squared minus 2 x minus 3 text  then  end text f text  is  end text

Maths-General

  1. a function
  2. onto
  3. one one
  4. one one onto

    Answer:The correct answer is: a function

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    Related Questions to study

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    The distance travelled by a particle is given by S=3+2t +5t2The initial velocity of the particle is…..

    The distance travelled by a particle is given by S=3+2t +5t2The initial velocity of the particle is…..

    physicsGeneral
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    A freely falling particle covers a building of 45 m height in one second. Find the height of the point from where the particle was released.[g=10 ms-2]

    A freely falling particle covers a building of 45 m height in one second. Find the height of the point from where the particle was released.[g=10 ms-2]

    physicsGeneral
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    For the arrangement shown in figure the time interval after which the water jet ceases to cross the wall (area of cross section of tank is A and orifice is ‘a’)

    Velocity of efflux = square root of 2 g h end root
    Find ‘h’ to have range of ejected water greater or equal than x
    rightwards double arrow x equals square root of 2 g h end root. square root of fraction numerator 2 y over denominator g end fraction end root rightwards double arrow h equals fraction numerator x to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction
    time taken by liquid to drain out from H to h is equals fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open square brackets square root of H minus square root of h close square brackets

    For the arrangement shown in figure the time interval after which the water jet ceases to cross the wall (area of cross section of tank is A and orifice is ‘a’)

    physics-General
    Velocity of efflux = square root of 2 g h end root
    Find ‘h’ to have range of ejected water greater or equal than x
    rightwards double arrow x equals square root of 2 g h end root. square root of fraction numerator 2 y over denominator g end fraction end root rightwards double arrow h equals fraction numerator x to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction
    time taken by liquid to drain out from H to h is equals fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open square brackets square root of H minus square root of h close square brackets
    General
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    Figure shows a stream of fluid emerging from a tube in the base of an open fixed tank. The expression of ‘y’ (Maximum height traveled by jet of water) is

    y equals fraction numerator u subscript y end subscript superscript 2 end superscript over denominator 2 g end fraction
    u equals square root of 2 g h end root u subscript y end subscript equals square root of 2 g h end root sin invisible function application theta rightwards double arrow y equals fraction numerator open parentheses square root of 2 g h end root sin invisible function application theta close parentheses to the power of 2 end exponent over denominator 2 g end fraction equals h sin to the power of 2 end exponent invisible function application theta

    Figure shows a stream of fluid emerging from a tube in the base of an open fixed tank. The expression of ‘y’ (Maximum height traveled by jet of water) is

    physics-General
    y equals fraction numerator u subscript y end subscript superscript 2 end superscript over denominator 2 g end fraction
    u equals square root of 2 g h end root u subscript y end subscript equals square root of 2 g h end root sin invisible function application theta rightwards double arrow y equals fraction numerator open parentheses square root of 2 g h end root sin invisible function application theta close parentheses to the power of 2 end exponent over denominator 2 g end fraction equals h sin to the power of 2 end exponent invisible function application theta
    General
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    A thin movable plate is separated from two fixed plates P subscript 1 and P subscript 2 by two highly viscous liquids of coefficients of viscosity n subscript 1 and n subscript 2 as shown, where n subscript 2 end subscript equals 9 n subscript 1 end subscript. Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is ‘h’, then the distance ‘h subscript 1 end subscript’ of movable plate from upper plate such that movable plate can be moved with a finite velocity by applying the minimum possible force on movable plate is ( assume only linear velocity distribution in each liquid).

    Viscous force due to upper liquid equals n subscript 1 end subscript A open parentheses fraction numerator v minus o over denominator h subscript 1 end subscript end fraction close parentheses
    Viscous force due to lower liquid = n subscript 2 end subscript A open parentheses fraction numerator v minus o over denominator h minus h subscript 1 end subscript end fraction close parentheses
    If total force is minimum
    fraction numerator d over denominator d h subscript 1 end subscript end fraction open square brackets fraction numerator n subscript 1 end subscript over denominator h subscript 1 end subscript end fraction plus fraction numerator n subscript 2 end subscript over denominator h minus h subscript 1 end subscript end fraction close square brackets equals 0

    A thin movable plate is separated from two fixed plates P subscript 1 and P subscript 2 by two highly viscous liquids of coefficients of viscosity n subscript 1 and n subscript 2 as shown, where n subscript 2 end subscript equals 9 n subscript 1 end subscript. Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is ‘h’, then the distance ‘h subscript 1 end subscript’ of movable plate from upper plate such that movable plate can be moved with a finite velocity by applying the minimum possible force on movable plate is ( assume only linear velocity distribution in each liquid).

    physics-General
    Viscous force due to upper liquid equals n subscript 1 end subscript A open parentheses fraction numerator v minus o over denominator h subscript 1 end subscript end fraction close parentheses
    Viscous force due to lower liquid = n subscript 2 end subscript A open parentheses fraction numerator v minus o over denominator h minus h subscript 1 end subscript end fraction close parentheses
    If total force is minimum
    fraction numerator d over denominator d h subscript 1 end subscript end fraction open square brackets fraction numerator n subscript 1 end subscript over denominator h subscript 1 end subscript end fraction plus fraction numerator n subscript 2 end subscript over denominator h minus h subscript 1 end subscript end fraction close square brackets equals 0
    General
    physics-

    A vertical jet of water coming out of a nozzle with velocity 20 m/s supports a plate of mass M stationary at a height h = 15m, as shown in the figure. If the rate of water flow is 1 litre per second, the mass of the plate is (Assume the collision to be inelastic).

    Force by liquid = Mg

    But and

    A vertical jet of water coming out of a nozzle with velocity 20 m/s supports a plate of mass M stationary at a height h = 15m, as shown in the figure. If the rate of water flow is 1 litre per second, the mass of the plate is (Assume the collision to be inelastic).

    physics-General
    Force by liquid = Mg

    But and

    General
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    A conical flask of mass 10 kg and base area as 103 cm2 is floating in liquid of relative density 1.2, as shown in the figure. The force that liquid exerts on curved surface of conical flask will be (Given g = 10 m/s2)


    = Buoyant force = weight
    F = 20 N in down ward direction

    A conical flask of mass 10 kg and base area as 103 cm2 is floating in liquid of relative density 1.2, as shown in the figure. The force that liquid exerts on curved surface of conical flask will be (Given g = 10 m/s2)

    physics-General

    = Buoyant force = weight
    F = 20 N in down ward direction
    General
    physics-

    A sphere of radius ‘R’ floats in a liquid of density ‘s’ such that its diameter x-x is below distance ‘h’ from free surface as shown. The density of sphere is r. The sphere is depressed slightly and released. The frequency of small oscillation is

    negative pi left parenthesis R to the power of 2 end exponent minus h to the power of 2 end exponent right parenthesis times left parenthesis x right parenthesis times sigma times g equals fraction numerator 4 pi R to the power of 3 end exponent rho left parenthesis a right parenthesis over denominator 3 end fraction
    and a equals negative omega to the power of 2 end exponent left parenthesis x right parenthesis

    A sphere of radius ‘R’ floats in a liquid of density ‘s’ such that its diameter x-x is below distance ‘h’ from free surface as shown. The density of sphere is r. The sphere is depressed slightly and released. The frequency of small oscillation is

    physics-General
    negative pi left parenthesis R to the power of 2 end exponent minus h to the power of 2 end exponent right parenthesis times left parenthesis x right parenthesis times sigma times g equals fraction numerator 4 pi R to the power of 3 end exponent rho left parenthesis a right parenthesis over denominator 3 end fraction
    and a equals negative omega to the power of 2 end exponent left parenthesis x right parenthesis
    General
    maths-

    The function f colon R not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals x minus left square bracket x right square bracket comma straight for all x element of R text  is  end text

    The function f colon R not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals x minus left square bracket x right square bracket comma straight for all x element of R text  is  end text

    maths-General
    General
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    A particle is thrown in upward direction with Velocity V0 It passes through a point p of height h at time t1 and t2 so t1+t1

    A particle is thrown in upward direction with Velocity V0 It passes through a point p of height h at time t1 and t2 so t1+t1

    physicsGeneral
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    The displacement of a particle in x direction is given by x.=9-5t+4t2 Find the Velocity at time t=0

    The displacement of a particle in x direction is given by x.=9-5t+4t2 Find the Velocity at time t=0

    physicsGeneral
    General
    maths-

    f colon R to the power of plus not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals 2 to the power of x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 3 to the power of x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

    f colon R to the power of plus not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals 2 to the power of x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 3 to the power of x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

    maths-General
    General
    physics-

    Two hollow hemispheres of metal are fitted together to form a sphere of radius R. Air from the shall is pumped out and pressure remained left inside is fraction numerator 1 over denominator 4 end fractionth the outside atmospheric pressure P0. The force required to separate the hemisphers is

    Excess pressure exists outside

    Two hollow hemispheres of metal are fitted together to form a sphere of radius R. Air from the shall is pumped out and pressure remained left inside is fraction numerator 1 over denominator 4 end fractionth the outside atmospheric pressure P0. The force required to separate the hemisphers is

    physics-General
    Excess pressure exists outside
    General
    physics-

    An iron block and a wooden block are positioned in a vessel containing water as shown in the figure. The iron block (I) hangs from a massless string with a rigid support from the top while the wooden block floats being tied to the bottom through a massless string. If now the vessel starts accelerating upwards, the incorrect option is

    An iron block and a wooden block are positioned in a vessel containing water as shown in the figure. The iron block (I) hangs from a massless string with a rigid support from the top while the wooden block floats being tied to the bottom through a massless string. If now the vessel starts accelerating upwards, the incorrect option is

    physics-General
    General
    physics-

    A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water

    If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb is 4 times as that of left limb.
    therefore Level of water in left limb is (36 + 4x) cm.

    Now equating pressure at interface of Hg and water (at A' B')
    left parenthesis 36 plus 4 x right parenthesis cross times 1 cross times g equals 5 x cross times 13.6 cross times g
    By solving we get x = 0.56 cm.

    A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water

    physics-General
    If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb is 4 times as that of left limb.
    therefore Level of water in left limb is (36 + 4x) cm.

    Now equating pressure at interface of Hg and water (at A' B')
    left parenthesis 36 plus 4 x right parenthesis cross times 1 cross times g equals 5 x cross times 13.6 cross times g
    By solving we get x = 0.56 cm.