Maths NCERT Exemplar Solutions Class 11th Chapter Seven

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R
Raj Pandey

Contributor-Level 9

A = {1, 2, 3, ….50}

R1 = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R2 = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

R 1 R 2  contains only 9 elements.

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R
Raj Pandey

Contributor-Level 9

kindly consider the following Image 

 

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R
Raj Pandey

Contributor-Level 9

t a n π 8 = 2 h 8 0 . . . . . . . . . ( i )

t a n θ = h 8 0 . . . . . . . . . . ( i i )

t a n π 8 = 2 t a n θ

t a n 2 θ = 3 2 2 4

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New answer posted

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R
Raj Pandey

Contributor-Level 9

P : ( x + 3 y z 6 ) = λ ( 6 x + 5 y z 7 ) = 0

passes ( 2 , 3 , 1 2 )

( 2 + 9 1 2 6 ) = λ ( 1 2 + 1 5 1 2 7 ) = 0

λ = 1

| 1 3 a | 2 d 2 = ( 1 3 ) 2 ( 9 3 ) ( 1 3 ) 2 = 9 3

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R
Raj Pandey

Contributor-Level 9

( 5 x + 8 y + 1 3 z 2 9 ) + λ ( 8 x 7 y + z 2 0 ) = 0  

P1 passing (2, 1, 3)

(10 + 8 + 39 – 29) + λ ( 1 6 7 + 3 2 0 ) = 0  

2 8 8 λ = 0 λ = 7 / 2  

 2X – Y + Z – 6 = 0      ….(i)

For P2 passes (0, 1, 2)

( 8 + 2 6 2 9 ) + λ ( 7 + 2 2 0 ) = 0

5 2 5 λ = 0 λ = 1 5  

P 2 : x + y + 2 z 5 = 0 . . . . . . . . ( i i )  

Acute angle between the planes

  c o s θ = 2 1 + 2 4 + 1 + 1 1 + 1 + 4 = 1 2  

θ = π 3  

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R
Raj Pandey

Contributor-Level 9

Let vector along L is x

  x = | i j k 1 2 1 3 5 2 |  

 = i ^ j ^ k ^  

Area of  Δ P Q R = 1 2 | P Q * P R |  

  = 1 2 | i ^ j ^ k ^ 1 4 1 5 3 4 3 1 1 3 |  

= 4 3 3 8  

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R
Raj Pandey

Contributor-Level 9

x 2 + y 2 2 x 4 y = 0

Centre (1, 2) r = 5  

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒x + 2y = 0       ……… (i)

Equation of PQ is  x ( 1 + 5 ) + 2 y ( | x + 1 + 5 ) 2 ( y + 2 ) = 0  

Solving (i) & (ii),   Q ( 5 + 1 , 5 + 1 2 )  

              = 1 2 | 6 + 2 5 + 4 5 = 4 2 | = 6 5 + 1 0 4 = 3 5 + 5 2  

 

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R
Raj Pandey

Contributor-Level 9

a 2 ( e 2 1 ) = b 2

e = 5 2 b 2 = 3 a 2 2

Length of latus rectum 2 b 2 a = 6 2  

3 a = 6 2 a = 2 2  

b = 2 3  

y = 2x + c is tangent to hyperbola

c 2 = a 2 m 2 b 2 = 2 0  

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R
Raj Pandey

Contributor-Level 9

f ( x ) = x 7 + 5 x 3 + 3 x + 1

f ' ( x ) = 7 x 6 + 1 5 x 4 + 3 > 0 x R

f ( x ) is increasing

 for x - , f (x)

x = 0, f (x) = 1

f ( x ) = 0 has only one real root.

 

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