Maths
General
Easy
Question
The polar equation of the circle with pole as centre and radius 3 is
 r=3
 2r=3


The correct answer is: r=3
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maths
(Area of GPL) to (Area of ALD) is equal to
(Area of GPL) to (Area of ALD) is equal to
mathsGeneral
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
physicsGeneral
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
mathsGeneral
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
physicsGeneral
Given equation
At
This is case with curve marked
At
This is case with curve marked
maths
A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is 
Let A = {a_{1}, a_{2},…..a_{n}}.
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
A is a set containing n elements. A subset P_{1} is chosen, and A is reconstructed by replacing the elements of P_{1}. The same process is repeated for subsets P_{1}, P_{2}, … , P_{m}, with m > 1. The Number of ways of choosing P_{1}, P_{2}, …, P_{m} so that P_{1} P_{2} … P_{m}= A is 
mathsGeneral
Let A = {a_{1}, a_{2},…..a_{n}}.
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
For each a_{i} (1 i n), either a_{i} P_{j} or a_{i} P_{j} (1 j m) . Thus, there are 2^{m} choices in which a_{i} (1 j n) may belong to the P_{j} s.
Also there is exactly one choice, viz., a_{i} P_{j} for j = 1, 2, …, m, for which a_{i} P_{1} P_{2} ... P_{m}.
Therefore, a_{i} P_{1} P_{2} …. P_{m} in (2^{m} – 1) ways . Since there are n elements in the set A, the number of ways of constructing subsets
P_{1}, P_{2}, ….. , P_{m} is (2^{m} – 1)^{n}
maths
The number of points in the Cartesian plane with integral coordinates satisfying the inequalities x k, y k, x – y k ; is
x k –k x k ….(1)
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
The number of points in the Cartesian plane with integral coordinates satisfying the inequalities x k, y k, x – y k ; is
mathsGeneral
x k –k x k ….(1)
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
& y k –k y k ….(2)
& x – y k y – x k –k y – x k x – k y x + k ….(3)
Number of points having integral coordinates
= (2k + 1)^{2} – 2
= (3k^{2} + 3k + 1).
Maths
The angle between the lines and is
The intersection of two perpendicular lines results in the formation of the cartesian plane, a twodimensional coordinate plane. The Xaxis is the horizontal line, and the Yaxis is the vertical line. The Cartesian coordinate point (x, y) indicates that the distance from the origin is x in the horizontal direction and y in the vertical direction.
Now the given lines are:
and
The cartesian form will be:
Now the given lines are:
and
The cartesian form will be:
$2x+5y=3$
$2y−5x=−4$
Slopes of these lines are $−5/2 $ and $2/5 $
Here, we can say that the product of slopes is $−1$
Hence, these lines are perpendicular so the angles between them is $90 degrees.$
The angle between the lines and is
MathsGeneral
The intersection of two perpendicular lines results in the formation of the cartesian plane, a twodimensional coordinate plane. The Xaxis is the horizontal line, and the Yaxis is the vertical line. The Cartesian coordinate point (x, y) indicates that the distance from the origin is x in the horizontal direction and y in the vertical direction.
Now the given lines are:
and
The cartesian form will be:
Now the given lines are:
and
The cartesian form will be:
$2x+5y=3$
$2y−5x=−4$
Slopes of these lines are $−5/2 $ and $2/5 $
Here, we can say that the product of slopes is $−1$
Hence, these lines are perpendicular so the angles between them is $90 degrees.$
maths
The polar equation of the straight line passing through and perpendicular to the initial line is
The polar equation of the straight line passing through and perpendicular to the initial line is
mathsGeneral
maths
The polar equation of the straight line passing through and parallel to the initial line is
The polar equation of the straight line passing through and parallel to the initial line is
mathsGeneral
maths
The equation of the line passing through pole and is
The equation of the line passing through pole and is
mathsGeneral
Maths
The polar equation of is
Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as:
Now lets xonsider: x=rcosθ and y=rsinθ
x^{2}+y^{2}=r^{2}
Now putting the values, we get:
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as:
Now lets xonsider: x=rcosθ and y=rsinθ
x^{2}+y^{2}=r^{2}
Now putting the values, we get:
The polar equation of is
MathsGeneral
Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as:
Now lets xonsider: x=rcosθ and y=rsinθ
x^{2}+y^{2}=r^{2}
Now putting the values, we get:
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as:
Now lets xonsider: x=rcosθ and y=rsinθ
x^{2}+y^{2}=r^{2}
Now putting the values, we get:
Maths
The cartesian equation of is
Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as
The cartesian equation of is
MathsGeneral
Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.
(r, +2n) or (r, +(2n+1)) are possible formulas, where n is an integer.
If measured in the opposite direction, the value of is positive.
When calculated counterclockwise, the value of is negative.
If laid off at the terminal side of, the value of r is positive.
If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.
The equation is given as
physics
Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be
or
On waxing the number of beats decreases hence
On waxing the number of beats decreases hence
Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be
physicsGeneral
or
On waxing the number of beats decreases hence
On waxing the number of beats decreases hence
Maths
If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are
a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are
MathsGeneral
a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
maths
Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is 
For 1 i 4, let x_{i} ( 3) be the number of blanks between i^{th} and (i + 1)^{th} letters. Then,
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is 
mathsGeneral
For 1 i 4, let x_{i} ( 3) be the number of blanks between i^{th} and (i + 1)^{th} letters. Then,
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence