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Maths-

General

Easy

Question

The solution of $fraction numerator d y over denominator d x end fraction plus square root of open parentheses fraction numerator 1 minus y to the power of 2 end exponent over denominator 1 minus x to the power of 2 end exponent end fraction close parentheses end root equals 0$ is

$\tan^{- 1} x + \cot^{- 1} x = c$
$\sin^{- 1} x + \sin^{- 1} y = c$
$\sec^{- 1} x + cose c^{- 1} x = c$
None of these

The correct answer is: $sin to the power of negative 1 end exponent invisible function application x plus sin to the power of negative 1 end exponent invisible function application y equals c$

Given equation is $not stretchy integral fraction numerator d y over denominator square root of 1 minus y to the power of 2 end exponent end root end fraction plus not stretchy integral fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root end fraction equals 0$
Integrating we get, $sin to the power of negative 1 end exponent invisible function application y plus sin to the power of negative 1 end exponent invisible function application x equals c$ .

Related Questions to study

Solution of $y d x minus x d y equals x to the power of 2 end exponent y d x$ is

Solution of $y d x minus x d y equals x to the power of 2 end exponent y d x$ is

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals 2 x y$ is

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals 2 x y$ is

The solution of the differential equation $x left parenthesis e to the power of 2 y end exponent minus 1 right parenthesis d y plus left parenthesis x to the power of 2 end exponent minus 1 right parenthesis e to the power of y end exponent d x equals 0$ is

The solution of the differential equation $x left parenthesis e to the power of 2 y end exponent minus 1 right parenthesis d y plus left parenthesis x to the power of 2 end exponent minus 1 right parenthesis e to the power of y end exponent d x equals 0$ is

parallel

The solution of the equation $sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator d y over denominator d x end fraction close parentheses equals x plus y$ is

The solution of the equation $sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator d y over denominator d x end fraction close parentheses equals x plus y$ is

The solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator left parenthesis 1 plus x right parenthesis y over denominator left parenthesis y minus 1 right parenthesis x end fraction$ is

The solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator left parenthesis 1 plus x right parenthesis y over denominator left parenthesis y minus 1 right parenthesis x end fraction$ is

The solution of the differential equation $left parenthesis 1 plus cos invisible function application x right parenthesis d y equals left parenthesis 1 minus cos invisible function application x right parenthesis d x$ is

The solution of the differential equation $left parenthesis 1 plus cos invisible function application x right parenthesis d y equals left parenthesis 1 minus cos invisible function application x right parenthesis d x$ is

parallel

The solution of the differential equation $left parenthesis sin invisible function application x plus cos invisible function application x right parenthesis d y plus left parenthesis cos invisible function application x minus sin invisible function application x right parenthesis d x equals 0$ is

The solution of the differential equation $left parenthesis sin invisible function application x plus cos invisible function application x right parenthesis d y plus left parenthesis cos invisible function application x minus sin invisible function application x right parenthesis d x equals 0$ is

The solution of the differential equation $left parenthesis 1 plus x to the power of 2 end exponent right parenthesis fraction numerator d y over denominator d x end fraction equals x$ is

The solution of the differential equation $left parenthesis 1 plus x to the power of 2 end exponent right parenthesis fraction numerator d y over denominator d x end fraction equals x$ is

Let f : R → R such that for all ‘x’ and y in R $vertical line f left parenthesis x right parenthesis minus f left parenthesis y right parenthesis vertical line less or equal than vertical line x minus y vertical line to the power of 3 end exponent text then end text f left parenthesis x right parenthesis$

Let f : R → R such that for all ‘x’ and y in R $vertical line f left parenthesis x right parenthesis minus f left parenthesis y right parenthesis vertical line less or equal than vertical line x minus y vertical line to the power of 3 end exponent text then end text f left parenthesis x right parenthesis$

parallel

In Δ ABC if A = (0, 0, 4); AB = 4, BC = 3, CA = 5, I = (1, 0, 1) is the incentre and the internal bisector of $straight angle straight A$ intersects BC at D then Dx =

In Δ ABC if A = (0, 0, 4); AB = 4, BC = 3, CA = 5, I = (1, 0, 1) is the incentre and the internal bisector of $straight angle straight A$ intersects BC at D then Dx =

$stack P Q with rightwards arrow on top comma stack Q R with rightwards arrow on top comma stack R S with rightwards arrow on top comma stack S T with rightwards arrow on top comma stack T U with rightwards arrow on top text and end text stack U P with rightwards arrow on top$ represent the sides of a regular hexagon
Statement I: $stack P Q with rightwards arrow on top cross times open parentheses stack R S with rightwards arrow on top plus stack S T with rightwards arrow on top close parentheses not equal to 0$
Statement II: $stack P Q with rightwards arrow on top cross times stack R S with rightwards arrow on top equals 0$ and $stack P Q with rightwards arrow on top cross times stack S T with rightwards arrow on top not equal to stack 0 with rightwards arrow on top$

$stack P Q with rightwards arrow on top comma stack Q R with rightwards arrow on top comma stack R S with rightwards arrow on top comma stack S T with rightwards arrow on top comma stack T U with rightwards arrow on top text and end text stack U P with rightwards arrow on top$ represent the sides of a regular hexagon
Statement I: $stack P Q with rightwards arrow on top cross times open parentheses stack R S with rightwards arrow on top plus stack S T with rightwards arrow on top close parentheses not equal to 0$
Statement II: $stack P Q with rightwards arrow on top cross times stack R S with rightwards arrow on top equals 0$ and $stack P Q with rightwards arrow on top cross times stack S T with rightwards arrow on top not equal to stack 0 with rightwards arrow on top$

A square PQRS is folded along the diagonal PR so that the planes PRQ and PRS are perpendicular to one another then the shortest distance between PQ ' and RS interms of P (the length of side of the square) (Q → Q' after folding)

A square PQRS is folded along the diagonal PR so that the planes PRQ and PRS are perpendicular to one another then the shortest distance between PQ ' and RS interms of P (the length of side of the square) (Q → Q' after folding)

parallel

If the resultant of two forces is equal in magnitude to one of the components and perpendicular to it in direction, find the other component using the vector method

If the resultant of two forces is equal in magnitude to one of the components and perpendicular to it in direction, find the other component using the vector method

In the below experiment, the value of P is –

In the below experiment, the value of P is –

chemistry-General

VversusTcurvesatconstantpressureP1 and P2foranidealgasareshowninfig.Whichis correct-

VversusTcurvesatconstantpressureP1 and P2foranidealgasareshowninfig.Whichis correct-

chemistry-General

parallel

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