Maths-
General
Easy

Question

Domain of definition of the functionf left parenthesis x right parenthesis equals fraction numerator 3 over denominator 4 minus x to the power of 2 end exponent end fraction plus l o g subscript 10 end subscript invisible function application open parentheses x to the power of 3 end exponent minus x close parentheses, is

  1. left parenthesis negative 1 , 0 right parenthesis union left parenthesis 1 , 2 right parenthesis union left parenthesis 2 comma infinity right parenthesis    
  2. left parenthesis 1 , 2 right parenthesis    
  3. left parenthesis negative 1 , 0 right parenthesis union left parenthesis 1 , 2 right parenthesis    
  4. left parenthesis 1 , 2 right parenthesis union left parenthesis 2 comma infinity right parenthesis    

The correct answer is: left parenthesis negative 1 , 0 right parenthesis union left parenthesis 1 , 2 right parenthesis union left parenthesis 2 comma infinity right parenthesis

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Related Questions to study

General
physics-

If w subscript 1 end subscript comma w subscript 2 blank a n d blank W subscript 3 end subscript end subscript represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between w subscript 1 end subscript comma w subscript 2 end subscript a n d blank w subscript 3 end subscript

Gravitational field is a conservative force field. In a conservative force field work done is path independent.
therefore blank W subscript 1 end subscript equals W subscript 2 end subscript equals W subscript 3 end subscript

If w subscript 1 end subscript comma w subscript 2 blank a n d blank W subscript 3 end subscript end subscript represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively(as shown)in the gravitational field of a point mass m. Find the correct relation between w subscript 1 end subscript comma w subscript 2 end subscript a n d blank w subscript 3 end subscript

physics-General
Gravitational field is a conservative force field. In a conservative force field work done is path independent.
therefore blank W subscript 1 end subscript equals W subscript 2 end subscript equals W subscript 3 end subscript
General
physics-

Given below is a graph between a variable force left parenthesis F right parenthesis (along y-axis) and the displacement left parenthesis X right parenthesis (along x-axis) of a particle in one dimension. The work done by the force in the displacement interval between 0 blank m and 30 blank m is

Given below is a graph between a variable force left parenthesis F right parenthesis (along y-axis) and the displacement left parenthesis X right parenthesis (along x-axis) of a particle in one dimension. The work done by the force in the displacement interval between 0 blank m and 30 blank m is

physics-General
General
physics-

A particle of mass m moving with a velocity u makes an elastic one dimensional collision with a stationary particle of mass m establishing a contact with it for extremely small time T. Their force of contact increases from zero to F subscript 0 end subscript linearly in time T divided by 4, remains constant for a further time T divided by 2 and decreases linearly from F subscript 0 end subscript to zero in further time T divided by 4 as shown. The magnitude possessed by F subscript 0 end subscript is

Change in momentum = Impulse
= Area under force-time graph
therefore m v equals Area of trapezium
rightwards double arrow m v equals fraction numerator 1 over denominator 2 end fraction open parentheses T plus fraction numerator T over denominator 2 end fraction close parentheses F subscript 0 end subscript rightwards double arrow m v equals fraction numerator 3 T over denominator 4 end fraction F subscript 0 end subscript rightwards double arrow F subscript 0 end subscript equals fraction numerator 4 m u over denominator 3 T end fraction

A particle of mass m moving with a velocity u makes an elastic one dimensional collision with a stationary particle of mass m establishing a contact with it for extremely small time T. Their force of contact increases from zero to F subscript 0 end subscript linearly in time T divided by 4, remains constant for a further time T divided by 2 and decreases linearly from F subscript 0 end subscript to zero in further time T divided by 4 as shown. The magnitude possessed by F subscript 0 end subscript is

physics-General
Change in momentum = Impulse
= Area under force-time graph
therefore m v equals Area of trapezium
rightwards double arrow m v equals fraction numerator 1 over denominator 2 end fraction open parentheses T plus fraction numerator T over denominator 2 end fraction close parentheses F subscript 0 end subscript rightwards double arrow m v equals fraction numerator 3 T over denominator 4 end fraction F subscript 0 end subscript rightwards double arrow F subscript 0 end subscript equals fraction numerator 4 m u over denominator 3 T end fraction
General
Maths-

Let A be a set containing 10 distinct elements, then the total number of distinct functions from A to A is-

Let A be a set containing 10 distinct elements, then the total number of distinct functions from A to A is-

Maths-General
General
maths-

The number of bijective functions from set A to itself when a contains 106 elements-

The number of bijective functions from set A to itself when a contains 106 elements-

maths-General
General
physics-

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 blank m is

Work done W equals area under F minus S graph
= area of trapezium A B C D plus area of trapezium C E F D
equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 10 plus 15 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
equals 125 plus 75 equals 200 blank J

The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 blank m is

physics-General
Work done W equals area under F minus S graph
= area of trapezium A B C D plus area of trapezium C E F D
equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 10 plus 15 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
equals 125 plus 75 equals 200 blank J
General
physics-

A particle is acted upon by a force Fwhich varies with position xas shown in figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

Work done=area between the graph force displacement curve and displacement
W equals fraction numerator 1 over denominator 2 end fraction cross times 6 cross times 10 minus 5 cross times 4 plus 5 cross times 4 minus 5 cross times 2
W equals 20 blank J
According to work energy theorem
increment equals K subscript E end subscript equals W
K subscript E end subscript subscript f end subscript equals W plus increment K
=20+25
=45J

A particle is acted upon by a force Fwhich varies with position xas shown in figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

physics-General
Work done=area between the graph force displacement curve and displacement
W equals fraction numerator 1 over denominator 2 end fraction cross times 6 cross times 10 minus 5 cross times 4 plus 5 cross times 4 minus 5 cross times 2
W equals 20 blank J
According to work energy theorem
increment equals K subscript E end subscript equals W
K subscript E end subscript subscript f end subscript equals W plus increment K
=20+25
=45J
General
physics-

A vertical spring with force constant K is fixed on a table. A ball of mass mat a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

Gravitational potential energy of ball gets converted into elastic potential energy of the spring m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent
Net work done equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0

A vertical spring with force constant K is fixed on a table. A ball of mass mat a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

physics-General
Gravitational potential energy of ball gets converted into elastic potential energy of the spring m g open parentheses h plus d close parentheses equals fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent
Net work done equals m g open parentheses h plus d close parentheses minus fraction numerator 1 over denominator 2 end fraction K d to the power of 2 end exponent equals 0
General
physics-

Figure shows the F-xgraph. Where F is the force applied and x is the distance covered

By the body along a straight line path. Given that F is in n e w t o n and x blankin m e t r e, what is the work done?

Work done =area under curve and displacement axis
equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J

Figure shows the F-xgraph. Where F is the force applied and x is the distance covered

By the body along a straight line path. Given that F is in n e w t o n and x blankin m e t r e, what is the work done?

physics-General
Work done =area under curve and displacement axis
equals 1 cross times 10 minus 1 cross times 10 plus 1 cross times 10 equals 10 blank J
General
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 -

]
(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

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
]
(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
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