Maths-
General
Easy

Question

The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

  1. r open parentheses C o s cubed space theta plus S i n cubed space theta close parentheses equals 3 a C o s space theta S i n space theta    
  2. r open parentheses C o s cubed space theta plus S i n cubed space theta close parentheses equals a C o s space 2 theta    
  3. r open parentheses C o s cubed space theta minus S i n cubed space theta close parentheses equals 3 a C o s space theta S i n space theta    
  4. r open parentheses C o s cubed space theta minus S i n cubed space theta close parentheses equals a C o s space 2 theta    

The correct answer is: r open parentheses C o s cubed space theta plus S i n cubed space theta close parentheses equals 3 a C o s space theta S i n space theta

Related Questions to study

General
maths-

If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

Let y = p + 1 and z = q + 2.
Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
rightwards double arrow x + p + q = 12
therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.

If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

maths-General
Let y = p + 1 and z = q + 2.
Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
rightwards double arrow x + p + q = 12
therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
= No. of non-negative integral solutions of x + p + q= 12
= Coeff. of x12 in (x0 + x1 + x2 + ……)3
= Coeff. of x12 in (1 – x)–3
= Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
= 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.
General
Maths-

If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

Maths-General
General
Maths-

Let p, q element of {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is

A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given  p, q element of {1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q24≥ ⇒ q≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q≥ 4
2,3,4
 
For p=2,q28
3,4
 
For p=3,q212
4
 
For p=4,q216
4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.

Let p, q element of {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is

Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given  p, q element of {1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
Now we know that for real roots, the discriminant is always greater than or equal to 0, so we have:
D=b2-4ac, applying this, we get:
q24≥ ⇒ q≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q≥ 4
2,3,4
 
For p=2,q28
3,4
 
For p=3,q212
4
 
For p=4,q216
4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
parallel
General
Maths-

ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as 00.
αβ = c/a as 00.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: α∣ β as α β 0.
So therefore: α∣ β

ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given a > 0, b < 0 and c < 0, the equation is ax2 + bx + c = 0.
Let the roots be α and β, where β>α, then:
α + β = -b/a >0 as 00.
αβ = c/a as 00.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: α∣ β as α β 0.
So therefore: α∣ β
General
Maths-

The cartesian equation of r squared c o s space 2 theta equals a squared is

A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r squared c o s space 2 theta equals a squared.
Now we know that:
cos space 2 theta equals cos squared theta minus sin squared theta.
So applying this, we get:
r squared left parenthesis cos squared theta minus sin squared theta right parenthesis equals a squared
Now lets substitute x=rcosθ and y=rsinθ, we get:
r squared cos squared theta minus r squared sin squared theta equals a squared
x squared minus y squared equals a squared


 

The cartesian equation of r squared c o s space 2 theta equals a squared is

Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r squared c o s space 2 theta equals a squared.
Now we know that:
cos space 2 theta equals cos squared theta minus sin squared theta.
So applying this, we get:
r squared left parenthesis cos squared theta minus sin squared theta right parenthesis equals a squared
Now lets substitute x=rcosθ and y=rsinθ, we get:
r squared cos squared theta minus r squared sin squared theta equals a squared
x squared minus y squared equals a squared


 

General
Maths-

The castesian equation of r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p is

A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p.
Now we know that:
cos space left parenthesis space theta space minus space alpha space right parenthesis equals cos space theta space cos space alpha space plus space sin space theta space sin space alpha.
So applying this, we get:
r left parenthesis cos space theta space cos space alpha space plus space sin space theta space sin space alpha right parenthesis equals p
Now lets substitute x=rcosθ and y=rsinθ, we get:
r cos space theta space cos space alpha space plus space r sin space theta space sin space alpha right parenthesis equals p
x space cos space alpha space plus space y space sin space alpha space equals space p


 

The castesian equation of r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p is

Maths-General
A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given the equation as r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p.
Now we know that:
cos space left parenthesis space theta space minus space alpha space right parenthesis equals cos space theta space cos space alpha space plus space sin space theta space sin space alpha.
So applying this, we get:
r left parenthesis cos space theta space cos space alpha space plus space sin space theta space sin space alpha right parenthesis equals p
Now lets substitute x=rcosθ and y=rsinθ, we get:
r cos space theta space cos space alpha space plus space r sin space theta space sin space alpha right parenthesis equals p
x space cos space alpha space plus space y space sin space alpha space equals space p


 

parallel
General
maths-

The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is

The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is

maths-General
General
maths-

The equation of the directrix of the conic r C o s to the power of 2 end exponent invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses equals 5 is

The equation of the directrix of the conic r C o s to the power of 2 end exponent invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses equals 5 is

maths-General
General
maths-

The equation of the circle touching the initial line at pole and radius 2 is

The equation of the circle touching the initial line at pole and radius 2 is

maths-General
parallel
General
maths-

The equation of the circle passing through pole and centre at (4,0) is

The equation of the circle passing through pole and centre at (4,0) is

maths-General
General
Maths-

The polar equation of the circle with pole as centre and radius 3 is

The polar equation of the circle with pole as centre and radius 3 is

Maths-General
General
maths-

(Area of incrementGPL) to (Area of incrementALD) is equal to

(Area of incrementGPL) to (Area of incrementALD) is equal to

maths-General
parallel
General
physics-

A small source of sound moves on a circle as shown in the figure and an observer is standing on O. Let n subscript 1 end subscript comma n subscript 2 end subscript and n subscript 3 end subscript be the frequencies heard when the source is at A comma blank B blankand C respectively. Then

At point A comma source is moving away from observer so apparent frequency n subscript 1 end subscript less than n (actual frequency) At point B source is coming towards observer so apparent frequency n subscript 2 end subscript greater than n and point C source is moving perpendicular to observer so n subscript 3 end subscript equals n
Hence n subscript 2 end subscript greater than n subscript 3 end subscript greater than n subscript 1 end subscript

A small source of sound moves on a circle as shown in the figure and an observer is standing on O. Let n subscript 1 end subscript comma n subscript 2 end subscript and n subscript 3 end subscript be the frequencies heard when the source is at A comma blank B blankand C respectively. Then

physics-General
At point A comma source is moving away from observer so apparent frequency n subscript 1 end subscript less than n (actual frequency) At point B source is coming towards observer so apparent frequency n subscript 2 end subscript greater than n and point C source is moving perpendicular to observer so n subscript 3 end subscript equals n
Hence n subscript 2 end subscript greater than n subscript 3 end subscript greater than n subscript 1 end subscript
General
Maths-

In a triangle ABC, if a : b : c = 7 : 8 : 9, then cos A : cos B equals to

In a triangle ABC, if a : b : c = 7 : 8 : 9, then cos A : cos B equals to

Maths-General
General
physics-

Which of the following curves represents correctly the oscillation given by y equals y subscript 0 end subscript sin invisible function application open parentheses omega t minus ϕ close parentheses comma blank w h e r e blank 0 less than ϕ less than 90

Given equation y equals y subscript 0 end subscript sin invisible function application left parenthesis omega t minus ϕ right parenthesis
At t equals 0 comma blank y equals negative y subscript 0 end subscript sin invisible function application ϕ
This is case with curve marked D

Which of the following curves represents correctly the oscillation given by y equals y subscript 0 end subscript sin invisible function application open parentheses omega t minus ϕ close parentheses comma blank w h e r e blank 0 less than ϕ less than 90

physics-General
Given equation y equals y subscript 0 end subscript sin invisible function application left parenthesis omega t minus ϕ right parenthesis
At t equals 0 comma blank y equals negative y subscript 0 end subscript sin invisible function application ϕ
This is case with curve marked D
parallel

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