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### Question

#### The polar equation of the straight line passing through and perpendicular to the initial line is

#### The correct answer is:

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### Related Questions to study

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#### 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

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#### The equation of the line passing through pole and is

#### The equation of the line passing through pole and is

maths-General

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#### The polar equation of is

#### The polar equation of is

maths-General

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#### The cartesian equation of is

#### The cartesian equation of is

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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

physics-General

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

#### If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are

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#### 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

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{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

#### 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 -

maths-General

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{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

maths-

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

maths-General

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

maths-

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

maths-General

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

maths-

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

maths-General

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

maths-

#### The number of points in the Cartesian plane with integral co-ordinates 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 ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

maths-General

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).maths-

#### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.#### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

maths-General

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.maths-

#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

maths-General

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.maths-

#### If the line is a normal to the hyperbola then

#### If the line is a normal to the hyperbola then

maths-General

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#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

maths-General