Maths-
General
Easy
Question
The polar equation of
axy is
The correct answer is: 
Related Questions to study
maths-
If x, y, z are integers and x
0, y
1, z
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
0, p
0, q
0 and x + y + z = 15
x + p + q = 12
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 =
=
= 91.
Then x
= 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 =
If x, y, z are integers and x
0, y
1, z
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
0, p
0, q
0 and x + y + z = 15
x + p + q = 12
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 =
=
= 91.
Then x
= 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 =
Maths-
If
then the equation whose roots are 
If
then the equation whose roots are 
Maths-General
Maths-
Let p, q
{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
{1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
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:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q2 ≥ 4
q = 2,3,4
For p=2,q2≥8
q = 3,4
For p=3,q2≥12
q = 4
For p=4,q2≥16
q = 4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
Let p, q
{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
{1, 2, 3, 4}, the equation given is : px2 + qx + 1 = 0
ax² + bx + c = 0
where a, b, and c are constant terms and x is the unknown variable.
Now we have given p, q
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:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
q2−4p ≥ 0 ⇒ q2 ≥ 4p
Now the set includes 4 terms, putting each, we get:
For p=1,q2 ≥ 4
q = 2,3,4
For p=2,q2≥8
q = 3,4
For p=3,q2≥12
q = 4
For p=4,q2≥16
q = 4
So here we can see that total 7 seven solutions are possible so 7 equations can be formed.
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 a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
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 a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣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 a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
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 a > 0, b < 0.
αβ = c/a as a > 0, c < 0.
Now that the roots are of opposite signs, so β > 0 and α < 0.
So: ∣α∣ < β as α + β > 0.
So therefore: 0 < ∣α∣ < β
Maths-
The cartesian equation of
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
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

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
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
The cartesian equation of
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
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

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
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
Maths-
The castesian equation of
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
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

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
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
The castesian equation of
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
.
Now we know that:
.
So applying this, we get:

Now lets substitute x=rcosθ and y=rsinθ, we get:

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
Now we know that:
So applying this, we get:
Now lets substitute x=rcosθ and y=rsinθ, we get:
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
maths-
The equation of the directrix of the conic
is
The equation of the directrix of the conic
is
maths-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
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
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
maths-
(Area of
GPL) to (Area of
ALD) is equal to
(Area of
GPL) to (Area of
ALD) is equal to
maths-General
physics-
A small source of sound moves on a circle as shown in the figure and an observer is standing on
Let
and
be the frequencies heard when the source is at
and
respectively. Then

At point
source is moving away from observer so apparent frequency
(actual frequency) At point
source is coming towards observer so apparent frequency
and point
source is moving perpendicular to observer so 
Hence
Hence
A small source of sound moves on a circle as shown in the figure and an observer is standing on
Let
and
be the frequencies heard when the source is at
and
respectively. Then

physics-General
At point
source is moving away from observer so apparent frequency
(actual frequency) At point
source is coming towards observer so apparent frequency
and point
source is moving perpendicular to observer so 
Hence
Hence
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
physics-
Which of the following curves represents correctly the oscillation given by 

Given equation 
At
This is case with curve marked
At
This is case with curve marked
Which of the following curves represents correctly the oscillation given by 

physics-General
Given equation 
At
This is case with curve marked
At
This is case with curve marked