General
Easy
Maths-

Range of function f(x) = fraction numerator x to the power of 2 end exponent plus 2 x plus 3 over denominator x end fraction, x element of times R is given by -

Maths-General

  1. pi    
  2. pi divided by 2    
  3. 3 pi divided by 2    
  4. 2 pi    

    Answer:The correct answer is: 3 pi divided by 2]
    (C) left parenthesis negative infinity comma 2 minus 2 square root of 3 right square bracket union left square bracket 2 plus 2 square root of 3 comma infinity right parenthesis

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    Related Questions to study

    General
    physics-

    A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 blank k gand 0.72 blank k g. Taking g equals 10 blank m divided by s to the power of 2 end exponent, find the work done (in joules) by the string on the block of mass 0.36 blank k g during the first second after the system is released from rest

    In the given condition tension in the string

    T equals fraction numerator 2 m subscript 1 end subscript m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction g equals fraction numerator 2 cross times 0.36 cross times 0.72 over denominator 1.08 end fraction cross times 10
    T equals 4.8 blank N
    And acceleration of each block
    a equals open parentheses fraction numerator m subscript 2 end subscript minus m subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g equals open parentheses fraction numerator 0.72 minus 0.36 over denominator 0.72 plus 0.36 end fraction close parentheses g equals fraction numerator 10 over denominator 3 end fraction blank m divided by s to the power of 2 end exponent
    Let ‘S’ is the distance covered by block of mass 0.36 blank k g in first sec
    S equals u t plus fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent rightwards double arrow S equals 0 plus fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 10 over denominator 3 end fraction close parentheses cross times 1 to the power of 2 end exponent equals fraction numerator 10 over denominator 6 end fraction blank m e t e r
    therefore Work done by the string W equals T S equals 4.8 cross times fraction numerator 10 over denominator 6 end fraction
    rightwards double arrow W equals 8 blank J o u l e

    A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 blank k gand 0.72 blank k g. Taking g equals 10 blank m divided by s to the power of 2 end exponent, find the work done (in joules) by the string on the block of mass 0.36 blank k g during the first second after the system is released from rest

    physics-General
    In the given condition tension in the string

    T equals fraction numerator 2 m subscript 1 end subscript m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction g equals fraction numerator 2 cross times 0.36 cross times 0.72 over denominator 1.08 end fraction cross times 10
    T equals 4.8 blank N
    And acceleration of each block
    a equals open parentheses fraction numerator m subscript 2 end subscript minus m subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g equals open parentheses fraction numerator 0.72 minus 0.36 over denominator 0.72 plus 0.36 end fraction close parentheses g equals fraction numerator 10 over denominator 3 end fraction blank m divided by s to the power of 2 end exponent
    Let ‘S’ is the distance covered by block of mass 0.36 blank k g in first sec
    S equals u t plus fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent rightwards double arrow S equals 0 plus fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 10 over denominator 3 end fraction close parentheses cross times 1 to the power of 2 end exponent equals fraction numerator 10 over denominator 6 end fraction blank m e t e r
    therefore Work done by the string W equals T S equals 4.8 cross times fraction numerator 10 over denominator 6 end fraction
    rightwards double arrow W equals 8 blank J o u l e
    General
    physics-

    An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel theta with the vertical. Work done by the force F is

    W equals increment K or W subscript T end subscript plus W subscript g end subscript plus W subscript F end subscript equals 0
    (Since, change in kinetic energy is zero)

    Here, W subscript T end subscript equals work done by tension = 0
    W subscript g end subscript equals work done by fore of gravity
    equals negative m g h
    equals negative m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    therefore blank W subscript F end subscript equals negative W subscript g end subscript equals m g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angel theta with the vertical. Work done by the force F is

    physics-General
    W equals increment K or W subscript T end subscript plus W subscript g end subscript plus W subscript F end subscript equals 0
    (Since, change in kinetic energy is zero)

    Here, W subscript T end subscript equals work done by tension = 0
    W subscript g end subscript equals work done by fore of gravity
    equals negative m g h
    equals negative m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    therefore blank W subscript F end subscript equals negative W subscript g end subscript equals m g L left parenthesis 1 minus cos invisible function application theta right parenthesis
    General
    physics-

    A block of mass m equals 25kg sliding on a smooth horizontal surface with a velocity v equals 3 m s to the power of negative 1 end exponent meets the spring of spring constant k equals 100 N m to the power of negative 1 end exponent fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are

    When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator m v to the power of 2 end exponent over denominator k end fraction end root
    equals square root of fraction numerator 25 cross times 3 to the power of 2 end exponent over denominator 100 end fraction end root square root of fraction numerator 9 over denominator 4 end fraction end root equals fraction numerator 3 over denominator 2 end fraction equals 1.5 blank m
    When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
    H e n c e v equals negative 3 m s to the power of negative 1 end exponent

    A block of mass m equals 25kg sliding on a smooth horizontal surface with a velocity v equals 3 m s to the power of negative 1 end exponent meets the spring of spring constant k equals 100 N m to the power of negative 1 end exponent fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are

    physics-General
    When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator m v to the power of 2 end exponent over denominator k end fraction end root
    equals square root of fraction numerator 25 cross times 3 to the power of 2 end exponent over denominator 100 end fraction end root square root of fraction numerator 9 over denominator 4 end fraction end root equals fraction numerator 3 over denominator 2 end fraction equals 1.5 blank m
    When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
    H e n c e v equals negative 3 m s to the power of negative 1 end exponent
    General
    physics-

    The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x equals 1 mto x equals 5m will be

    Work done=area enclosed by F minus xgraph
    =area of ABNM + area of CDEN - area of EFGH + area of HIJ

    =1 cross times 10 plus 1 cross times 5 minus 1 cross times 5 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 10
    equals 10 plus 5 minus 5 plus 5 equals 15 blank J

    The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x equals 1 mto x equals 5m will be

    physics-General
    Work done=area enclosed by F minus xgraph
    =area of ABNM + area of CDEN - area of EFGH + area of HIJ

    =1 cross times 10 plus 1 cross times 5 minus 1 cross times 5 plus fraction numerator 1 over denominator 2 end fraction cross times 1 cross times 10
    equals 10 plus 5 minus 5 plus 5 equals 15 blank J
    General
    physics-

    Three objects A comma B and C are kept in a straight line on a frictionless horizontal surface. These have masses m comma blank 2 m and m respectively. The object A moves towards B with a speed 9 blank m divided by s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m divided by s) of the object C


    v subscript 2 end subscript equals fraction numerator 2 m subscript 1 end subscript v subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 2 cross times m cross times 9 over denominator m plus 2 m end fraction equals 6 blank m divided by s
    i. e. After elastic collision B strikes to C with velocity of 6 blank m divided by s. Now collision between B and C is perfectly inelastic

    By the law of conservation of momentum
    2 m cross times 6 plus 0 equals 3 m cross times v subscript s y s end subscript
    rightwards double arrow v subscript s y s end subscript equals 4 blank m divided by s

    Three objects A comma B and C are kept in a straight line on a frictionless horizontal surface. These have masses m comma blank 2 m and m respectively. The object A moves towards B with a speed 9 blank m divided by s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m divided by s) of the object C

    physics-General

    v subscript 2 end subscript equals fraction numerator 2 m subscript 1 end subscript v subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 2 cross times m cross times 9 over denominator m plus 2 m end fraction equals 6 blank m divided by s
    i. e. After elastic collision B strikes to C with velocity of 6 blank m divided by s. Now collision between B and C is perfectly inelastic

    By the law of conservation of momentum
    2 m cross times 6 plus 0 equals 3 m cross times v subscript s y s end subscript
    rightwards double arrow v subscript s y s end subscript equals 4 blank m divided by s
    General
    physics-

    The relation between the displacement X of an object produced by the application of the variable force F is represented by a graph shown in the figure. If the object undergoes a displacement from X equals 0.5 blank m to X equals 2.5 blank m the work done will be approximately equal to

    Work done = Area under curve and displacement axis
    = Area of trapezium
    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses s u m blank o f blank t w o blank p a r a l l e l blank l i n e s close parentheses cross times d i s t a n c e blank b e t w e e n blank t h e m
    equals fraction numerator 1 over denominator 2 end fraction open parentheses 10 plus 4 close parentheses cross times open parentheses 2.5 minus 0.5 close parentheses equals fraction numerator 1 over denominator 2 end fraction 14 cross times 2 equals 14 blank J
    As the area actually is not trapezium so work done will be more than 14 blank J i. e. approximately 16 blank J

    The relation between the displacement X of an object produced by the application of the variable force F is represented by a graph shown in the figure. If the object undergoes a displacement from X equals 0.5 blank m to X equals 2.5 blank m the work done will be approximately equal to