Maths-

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

Maths-General

- 15
- 9
- 7
- 8

#### Answer:The correct answer is: 7

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#### ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

#### ax^{2} + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

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

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#### (Area of GPL) to (Area of ALD) is equal to

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#### 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

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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

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#### 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

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#### Which of the following curves represents correctly the oscillation given by

Given equation

At

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At

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#### Which of the following curves represents correctly the oscillation given by

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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

For each a

Also there is exactly one choice, viz., a

Therefore, a

P

_{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 subsetsP

_{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 -

maths-General

Let A = {a

For each a

Also there is exactly one choice, viz., a

Therefore, a

P

_{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 subsetsP

_{1}, P_{2}, ….. , P_{m}is (2^{m}– 1)^{n}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 –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

= (3k

& |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 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 –k y – x k x – k y x + k ….(3)

Number of points having integral coordinates

= (2k + 1)

= (3k

& |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 angle between the lines and is

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