Maths-
General
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Question

Let f colon open square brackets 1 half comma 1 close square brackets not stretchy rightwards arrow left square bracket negative 1 comma 1 right square bracket is given byf left parenthesis x right parenthesis equals 4 x cubed minus 3 x text  then  end text f to the power of negative 1 end exponent left parenthesis x right parenthesis text  is given by  end text

  1. cos invisible function application open square brackets 1 third cos to the power of negative 1 end exponent invisible function application x close square brackets
  2. 3 cos invisible function application open parentheses sin to the power of negative 1 end exponent invisible function application x close parentheses
  3. 3 sin to the power of negative 1 end exponent invisible function application left parenthesis cos invisible function application x right parenthesis
  4. sin invisible function application open parentheses 1 third cos to the power of negative 1 end exponent invisible function application x close parentheses

Hint:

In this question, we have to find inverse of function f(x) f colon open square brackets 1 half comma 1 close square brackets not stretchy rightwards arrow left square bracket negative 1 comma 1 right square bracket. Firstly, we will assume x to be cos theta and y to be cos 3theta. Then, we will find f(x) for x = cos theta and later find the value of theta 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: cos invisible function application open square brackets 1 third cos to the power of negative 1 end exponent invisible function application x close square brackets


    f left parenthesis x right parenthesis equals 4 x cubed minus 3 x
w h e r e comma space f open square brackets 1 half comma 1 close square brackets rightwards arrow open square brackets negative 1 comma 1 close square brackets
L e t space x equals space cos space theta space a n d space y equals cos space 3 theta
f left parenthesis cos space theta right parenthesis equals 4 cos cubed theta minus 3 space cos space theta
space space space space space space space space space space space space space equals space cos space 3 space theta
a s space 3 theta space equals cos to the power of negative 1 end exponent y
space space space space space space space space space space theta equals 1 third cos to the power of negative 1 end exponent y
f to the power of negative 1 end exponent left parenthesis cos space 3 theta right parenthesis equals cos space theta
a s space y space equals space f left parenthesis x right parenthesis rightwards double arrow f to the power of negative 1 end exponent left parenthesis y right parenthesis equals x
f to the power of negative 1 end exponent left parenthesis cos space 3 theta right parenthesis space equals space cos space theta
f to the power of negative 1 end exponent left parenthesis y right parenthesis space equals space cos space open parentheses 1 third cos to the power of negative 1 end exponent y close parentheses
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals cos open parentheses 1 third cos to the power of negative 1 end exponent x close parentheses

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    Related Questions to study

    General
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    If f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction text  then  end text f to the power of negative 1 end exponent left parenthesis x right parenthesis equals

    f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction equals y
rightwards double arrow fraction numerator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y
rightwards double arrow fraction numerator left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y
rightwards double arrow 10 to the power of 2 x end exponent minus 1 space equals y left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis
rightwards double arrow 10 to the power of 2 x end exponent minus 1 equals y.10 to the power of 2 x end exponent plus y
rightwards double arrow 10 to the power of 2 x end exponent minus y.10 to the power of 2 x end exponent equals y plus 1
rightwards double arrow 10 to the power of 2 x end exponent left parenthesis 1 minus y right parenthesis equals 1 plus y
rightwards double arrow 10 to the power of 2 x end exponent equals fraction numerator 1 plus y over denominator 1 minus y end fraction
P u t t i n g space log space o n space b o t h space t h e space s i d e s comma
rightwards double arrow log space 10 to the power of 2 x end exponent space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow 2 x space log space 10 space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow 2 x space equals fraction numerator log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses over denominator log space 10 end fraction
rightwards double arrow x space equals 1 half log subscript 10 open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals space 1 half log subscript 10 open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses

    If f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction text  then  end text f to the power of negative 1 end exponent left parenthesis x right parenthesis equals

    Maths-General
    f left parenthesis x right parenthesis equals fraction numerator 10 to the power of x minus 10 to the power of negative x end exponent over denominator 10 to the power of x plus 10 to the power of negative x end exponent end fraction equals y
rightwards double arrow fraction numerator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator 10 to the power of x left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y
rightwards double arrow fraction numerator left parenthesis 10 to the power of 2 x end exponent minus 1 right parenthesis over denominator left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis end fraction equals y
rightwards double arrow 10 to the power of 2 x end exponent minus 1 space equals y left parenthesis 10 to the power of 2 x end exponent plus 1 right parenthesis
rightwards double arrow 10 to the power of 2 x end exponent minus 1 equals y.10 to the power of 2 x end exponent plus y
rightwards double arrow 10 to the power of 2 x end exponent minus y.10 to the power of 2 x end exponent equals y plus 1
rightwards double arrow 10 to the power of 2 x end exponent left parenthesis 1 minus y right parenthesis equals 1 plus y
rightwards double arrow 10 to the power of 2 x end exponent equals fraction numerator 1 plus y over denominator 1 minus y end fraction
P u t t i n g space log space o n space b o t h space t h e space s i d e s comma
rightwards double arrow log space 10 to the power of 2 x end exponent space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow 2 x space log space 10 space equals log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow 2 x space equals fraction numerator log space open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses over denominator log space 10 end fraction
rightwards double arrow x space equals 1 half log subscript 10 open parentheses fraction numerator 1 plus y over denominator 1 minus y end fraction close parentheses
rightwards double arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals space 1 half log subscript 10 open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses
    General
    Maths-

    If f left parenthesis x right parenthesis equals fraction numerator e to the power of x plus e to the power of negative x end exponent over denominator 2 end fraction ,then the inverse function of f left parenthesis x right parenthesis text  is  end text

    If f left parenthesis x right parenthesis equals fraction numerator e to the power of x plus e to the power of negative x end exponent over denominator 2 end fraction ,then the inverse function of f left parenthesis x right parenthesis text  is  end text

    Maths-General
    General
    maths-

    If f left parenthesis x right parenthesis equals x squared plus x comma x equals 10 comma delta x equals 0.1 observe the following
    text  l)  end text delta f equals 2.11
    text  II)  end text d f equals 2.1
    text  III) Relative error in  end text x text  is  end text 1
    The true statements are

    If f left parenthesis x right parenthesis equals x squared plus x comma x equals 10 comma delta x equals 0.1 observe the following
    text  l)  end text delta f equals 2.11
    text  II)  end text d f equals 2.1
    text  III) Relative error in  end text x text  is  end text 1
    The true statements are

    maths-General
    General
    Maths-

    Statement I :f colon A not stretchy rightwards arrow B text  is one - one and  end text g colon B not stretchy rightwards arrow C text  is a one-one function, then gof  end text colon A not stretchy rightwards arrow C is one - one
    Statement II :text  If  end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A are two functions such that g o f equals I subscript A text  and  end text fog equals I subscript B comma text  then  end text f equalsg to the power of negative 1 end exponent
    Staatement III : f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text  then  end text f equals g

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

    Statement I :f colon A not stretchy rightwards arrow B text  is one - one and  end text g colon B not stretchy rightwards arrow C text  is a one-one function, then gof  end text colon A not stretchy rightwards arrow C is one - one
    Statement II :text  If  end text f colon A not stretchy rightwards arrow B comma g colon B not stretchy rightwards arrow A are two functions such that g o f equals I subscript A text  and  end text fog equals I subscript B comma text  then  end text f equalsg to the power of negative 1 end exponent
    Staatement III : f left parenthesis x right parenthesis equals sec squared invisible function application x minus tan squared invisible function application x g left parenthesis x right parenthesis equals cosec squared invisible function application x minus cot squared invisible function application x comma text  then  end text f equals g

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

    Maths-General
    General
    physics-

    A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 cross times 1 0 to the power of negative 5 end exponentm. The tube is immersed vertically into a liquid of surface tension 5.06 cross times 1 0 to the power of negative 2 end exponent N/m. To what length has the capillary to be immersed so that the liquid level inside and outside the capillary becomes the same?

    P subscript i n end subscript equals P subscript 0 end subscript But P subscript 0 end subscript L equals P subscript i n end subscript open parentheses L minus x close parentheses

    A glass capillary sealed at the upper end is of length 0.11 m and internal diameter 2 cross times 1 0 to the power of negative 5 end exponentm. The tube is immersed vertically into a liquid of surface tension 5.06 cross times 1 0 to the power of negative 2 end exponent 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
    P subscript i n end subscript equals P subscript 0 end subscript But P subscript 0 end subscript L equals P subscript i n end subscript open parentheses L minus x close parentheses
    General
    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: 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

    physics-General
    Application of Bernoulli’s theorem
    General
    physics-

    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 0.6 g c m to the power of negative 3 end exponent. The mass of block is

    Weight of block
    = Weight of displaced oil + Weight of displaced water
    Þ m g equals V subscript 1 end subscript rho subscript 0 end subscript g plus V subscript 2 end subscript rho subscript W end subscript g
    Þ m equals left parenthesis 10 cross times 10 cross times 6 right parenthesis cross times 0.6 plus left parenthesis 10 cross times 10 cross times 4 right parenthesis cross times 1= 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 0.6 g c m to the power of negative 3 end exponent. The mass of block is

    physics-General
    Weight of block
    = Weight of displaced oil + Weight of displaced water
    Þ m g equals V subscript 1 end subscript rho subscript 0 end subscript g plus V subscript 2 end subscript rho subscript W end subscript g
    Þ m equals left parenthesis 10 cross times 10 cross times 6 right parenthesis cross times 0.6 plus left parenthesis 10 cross times 10 cross times 4 right parenthesis cross times 1= 760 gm
    General
    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
    General
    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
    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 log subscript e invisible function application x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 2 log subscript e invisible function application 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 log subscript e invisible function application x comma x element of left parenthesis 0 comma 1 right parenthesis comma f left parenthesis x right parenthesis equals 2 log subscript e invisible function application x comma x element of left square bracket 1 comma straight infinity right parenthesis text  is  end text

    maths-General
    General
    physics-

    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
    General
    physics-

    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 rho = density of water) by


    If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
    Now, pi r to the power of 2 end exponent h subscript 1 end subscript equals pi left parenthesis n r right parenthesis to the power of 2 end exponent h subscript 2 end subscriptÞ h subscript 1 end subscript equals n to the power of 2 end exponent h subscript 2 end subscript
    Now, pressure at point A = pressure at point B
    h rho g equals left parenthesis h subscript 1 end subscript plus h subscript 2 end subscript right parenthesis rho ´ g
    Þ h = left parenthesis n to the power of 2 end exponent h subscript 2 end subscript plus h subscript 2 end subscript right parenthesis s g open parentheses text As end text   s equals fraction numerator rho ´ over denominator rho end fraction close parentheses Þ h subscript 2 end subscript equals fraction numerator h over denominator left parenthesis n to the power of 2 end exponent plus 1 right parenthesis s end fraction

    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 rho = density of water) by

    physics-General

    If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
    Now, pi r to the power of 2 end exponent h subscript 1 end subscript equals pi left parenthesis n r right parenthesis to the power of 2 end exponent h subscript 2 end subscriptÞ h subscript 1 end subscript equals n to the power of 2 end exponent h subscript 2 end subscript
    Now, pressure at point A = pressure at point B
    h rho g equals left parenthesis h subscript 1 end subscript plus h subscript 2 end subscript right parenthesis rho ´ g
    Þ h = left parenthesis n to the power of 2 end exponent h subscript 2 end subscript plus h subscript 2 end subscript right parenthesis s g open parentheses text As end text   s equals fraction numerator rho ´ over denominator rho end fraction close parentheses Þ h subscript 2 end subscript equals fraction numerator h over denominator left parenthesis n to the power of 2 end exponent plus 1 right parenthesis s end fraction
    General
    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 rho. 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) = F equals F subscript B end subscript minus F subscript A end subscript equals fraction numerator d p subscript B end subscript over denominator d t end fraction minus fraction numerator d p subscript A end subscript over denominator d t end fraction
    equals a v subscript B end subscript rho cross times v subscript B end subscript minus a v subscript A end subscript rho cross times v subscript A end subscript
    \ F equals a rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses…(i)
    According to Bernoulli's theorem
    p subscript A end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript A end subscript superscript 2 end superscript plus rho g h equals p subscript B end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript B end subscript superscript 2 end superscript plus 0
    Þ fraction numerator 1 over denominator 2 end fraction rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses equals rho g h Þ v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript equals 2 g h
    From equation (i), F equals 2 a rho g h.

    There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density rho. 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) = F equals F subscript B end subscript minus F subscript A end subscript equals fraction numerator d p subscript B end subscript over denominator d t end fraction minus fraction numerator d p subscript A end subscript over denominator d t end fraction
    equals a v subscript B end subscript rho cross times v subscript B end subscript minus a v subscript A end subscript rho cross times v subscript A end subscript
    \ F equals a rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses…(i)
    According to Bernoulli's theorem
    p subscript A end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript A end subscript superscript 2 end superscript plus rho g h equals p subscript B end subscript plus fraction numerator 1 over denominator 2 end fraction rho v subscript B end subscript superscript 2 end superscript plus 0
    Þ fraction numerator 1 over denominator 2 end fraction rho open parentheses v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript close parentheses equals rho g h Þ v subscript B end subscript superscript 2 end superscript minus v subscript A end subscript superscript 2 end superscript equals 2 g h
    From equation (i), F equals 2 a rho g h.
    General
    physics-

    A cylinder containing water up to a height of 25 cm has a hole of cross-section fraction numerator 1 over denominator 4 end fraction c m to the power of 2 end exponent 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 = Adv2
    Initial downward force on the flowing out water = Adv2
    So equal amount of reaction acts upwards on the cylinder.
    \ Initial upward reaction =A d v to the power of 2 end exponent [As v equals square root of 2 g h end root]
    therefore Initial decrease in weight equals A d left parenthesis 2 g h right parenthesis
    equals 2 A d g h equals 2 cross times open parentheses fraction numerator 1 over denominator 4 end fraction close parentheses cross times 1 cross times 980 cross times 25 equals 12.5 gm-wt.

    A cylinder containing water up to a height of 25 cm has a hole of cross-section fraction numerator 1 over denominator 4 end fraction c m to the power of 2 end exponent 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 = Adv2
    Initial downward force on the flowing out water = Adv2
    So equal amount of reaction acts upwards on the cylinder.
    \ Initial upward reaction =A d v to the power of 2 end exponent [As v equals square root of 2 g h end root]
    therefore Initial decrease in weight equals A d left parenthesis 2 g h right parenthesis
    equals 2 A d g h equals 2 cross times open parentheses fraction numerator 1 over denominator 4 end fraction close parentheses cross times 1 cross times 980 cross times 25 equals 12.5 gm-wt.
    General
    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/s2)

    Let A = cross-section of tank
    a = cross-section hole
    V = velocity with which level decreases
    v = velocity of efflux

    From equation of continuity a v equals A V rightwards double arrow V equals fraction numerator a v over denominator A end fraction
    By using Bernoulli's theorem for energy per unit volume
    Energy per unit volume at point A
    = Energy per unit volume at point B
    P plus rho g h plus fraction numerator 1 over denominator 2 end fraction rho V to the power of 2 end exponent equals P plus 0 plus fraction numerator 1 over denominator 2 end fraction rho v to the power of 2 end exponent
    Þ v to the power of 2 end exponent equals fraction numerator 2 g h over denominator 1 minus open parentheses fraction numerator a over denominator A end fraction close parentheses to the power of 2 end exponent end fraction equals fraction numerator 2 cross times 10 cross times left parenthesis 3 minus 0.525 right parenthesis over denominator 1 minus left parenthesis 0.1 right parenthesis to the power of 2 end exponent end fraction equals 50 left parenthesis m divided by sec invisible function application right parenthesis to the power of 2 end exponent

    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/s2)

    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 a v equals A V rightwards double arrow V equals fraction numerator a v over denominator A end fraction
    By using Bernoulli's theorem for energy per unit volume
    Energy per unit volume at point A
    = Energy per unit volume at point B
    P plus rho g h plus fraction numerator 1 over denominator 2 end fraction rho V to the power of 2 end exponent equals P plus 0 plus fraction numerator 1 over denominator 2 end fraction rho v to the power of 2 end exponent
    Þ v to the power of 2 end exponent equals fraction numerator 2 g h over denominator 1 minus open parentheses fraction numerator a over denominator A end fraction close parentheses to the power of 2 end exponent end fraction equals fraction numerator 2 cross times 10 cross times left parenthesis 3 minus 0.525 right parenthesis over denominator 1 minus left parenthesis 0.1 right parenthesis to the power of 2 end exponent end fraction equals 50 left parenthesis m divided by sec invisible function application right parenthesis to the power of 2 end exponent