Maths-

General

Easy

### Question

#### Let is given by

### Hint:

In this question, we have to find inverse of function f(x) . Firstly, we will assume x to be cos and y to be cos 3. Then, we will find f(x) for x = cos and later find the value of from that equation. Now, finding inverse of f(y). Further, replacing y with x to get the required inverse of f(x).

#### The correct answer is:

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### Related Questions to study

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#### If

#### If

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#### If ,then the inverse function of

#### If ,then the inverse function of

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#### If observe the following

The true statements are

#### If observe the following

The true statements are

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#### Statement I : is one - one

Statement II : are two functions such that

Staatement III :

Which of the above statement/s is/are true.

#### Statement I : is one - one

Statement II : are two functions such that

Staatement III :

Which of the above statement/s is/are true.

Maths-General

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#### A glass capillary sealed at the upper end is of length 0.11 m and internal diameter m. The tube is immersed vertically into a liquid of surface tension N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

But

#### A glass capillary sealed at the upper end is of length 0.11 m and internal diameter m. The tube is immersed vertically into a liquid of surface tension N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

physics-General

But

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#### Water flows through a frictionless duct with a cross-section varying as shown in fig. . Pressure p at points along the axis is represented by: A resume water to be non-viscour

Application of Bernoulli’s theorem

#### Water flows through a frictionless duct with a cross-section varying as shown in fig. . Pressure p at points along the axis is represented by: A resume water to be non-viscour

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Application of Bernoulli’s theorem

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#### A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is . The mass of block is

Weight of block

= Weight of displaced oil + Weight of displaced water

Þ

Þ = 760 gm

= Weight of displaced oil + Weight of displaced water

Þ

Þ = 760 gm

#### A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is . The mass of block is

physics-General

Weight of block

= Weight of displaced oil + Weight of displaced water

Þ

Þ = 760 gm

= Weight of displaced oil + Weight of displaced water

Þ

Þ = 760 gm

physics-

#### Water flows through a frictionless duct with a cross-section varying as shown in fig. Pressure p at points along the axis is represented by

When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli’s theorem, the pressure P decreased at that place

#### Water flows through a frictionless duct with a cross-section varying as shown in fig. Pressure p at points along the axis is represented by

physics-General

When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli’s theorem, the pressure P decreased at that place

physics-

#### The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like

When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero

#### The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like

physics-General

When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero

maths-

maths-General

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#### A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

#### A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

physics-General

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#### Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and = density of water) by

If the level in narrow tube goes down by h

_{1}then in wider tube goes up to h

_{2},

Now, Þ

Now, pressure at point A = pressure at point B

Þ h = Þ

#### Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and = density of water) by

physics-General

If the level in narrow tube goes down by h

_{1}then in wider tube goes up to h

_{2},

Now, Þ

Now, pressure at point A = pressure at point B

Þ h = Þ

physics-

#### There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density . The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

Net force (reaction) =

\ …(i)

According to Bernoulli's theorem

Þ Þ

From equation (i),

#### There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density . The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

physics-General

Net force (reaction) =

\ …(i)

According to Bernoulli's theorem

Þ Þ

From equation (i),

physics-

#### A cylinder containing water up to a height of 25 cm has a hole of cross-section in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out

Let A = The area of cross section of the hole

v = Initial velocity of efflux

d = Density of water,

Initial volume of water flowing out per second = Av

Initial mass of water flowing out per second = Avd

Rate of change of momentum = Adv

Initial downward force on the flowing out water = Adv

So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

v = Initial velocity of efflux

d = Density of water,

Initial volume of water flowing out per second = Av

Initial mass of water flowing out per second = Avd

Rate of change of momentum = Adv

^{2}Initial downward force on the flowing out water = Adv

^{2}So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

#### A cylinder containing water up to a height of 25 cm has a hole of cross-section in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out

physics-General

Let A = The area of cross section of the hole

v = Initial velocity of efflux

d = Density of water,

Initial volume of water flowing out per second = Av

Initial mass of water flowing out per second = Avd

Rate of change of momentum = Adv

Initial downward force on the flowing out water = Adv

So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

v = Initial velocity of efflux

d = Density of water,

Initial volume of water flowing out per second = Av

Initial mass of water flowing out per second = Avd

Rate of change of momentum = Adv

^{2}Initial downward force on the flowing out water = Adv

^{2}So equal amount of reaction acts upwards on the cylinder.

\ Initial upward reaction = [As ]

Initial decrease in weight

gm-wt.

physics-

#### Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s^{2})

Let A = cross-section of tank

a = cross-section hole

V = velocity with which level decreases

v = velocity of efflux

From equation of continuity

By using Bernoulli's theorem for energy per unit volume

Energy per unit volume at point A

= Energy per unit volume at point B

Þ

a = cross-section hole

V = velocity with which level decreases

v = velocity of efflux

From equation of continuity

By using Bernoulli's theorem for energy per unit volume

Energy per unit volume at point A

= Energy per unit volume at point B

Þ

#### Water is filled in a cylindrical container to a height of 3m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s^{2})

physics-General

Let A = cross-section of tank

a = cross-section hole

V = velocity with which level decreases

v = velocity of efflux

From equation of continuity

By using Bernoulli's theorem for energy per unit volume

Energy per unit volume at point A

= Energy per unit volume at point B

Þ

a = cross-section hole

V = velocity with which level decreases

v = velocity of efflux

From equation of continuity

By using Bernoulli's theorem for energy per unit volume

Energy per unit volume at point A

= Energy per unit volume at point B

Þ