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New answer posted

3 months ago

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Vishal Baghel

Contributor-Level 10

AAT = ATA = l

A ( A T B A ) 2 0 2 1 A T              

B 2 0 2 1 = ( I + P ) 2 0 2 1 = l + 2 0 2 1 P = [ 1 0 2 0 2 1 i 1 ]

( B 2 0 2 1 ) 1 = [ 1 0 2 0 2 1 i 1 ]

where P = [ 0 0 i 0 ]

New answer posted

3 months ago

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Vishal Baghel

Contributor-Level 10

S 1 x 1 = 2 1 0 1 ( x 2 1 0 0 ) 2 1

For x = 2, S = 1 2 1 0 1 4 1 0 1 1

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Vishal Baghel

Contributor-Level 10

r p l a n e s

n 1 , n 2 = 0 λ = 3 4

Where required plane P is

( 1 + λ ) x + ( 2 λ ) y + ( 3 λ ) z + 1 6 λ = 0

2 x + y + 2 z 5 = 0

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Vishal Baghel

Contributor-Level 10

Δ = 0

sin 3 θ=  1 2

but 0 < θ < π 2

θ= 70°

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Vishal Baghel

Contributor-Level 10

Centre (0, 1)

Radius = 2

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Vishal Baghel

Contributor-Level 10

89 – k + k + 98 – k + x = 100

=> 8 7 k 8 9

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Vishal Baghel

Contributor-Level 10

f ( x ) = 1 x 1 + x , 0 < x < 1 (on simplification)

f ' ( x ) = 2 ( 1 + x ) 2

( 1 x ) 2 f ' ( x ) = 2 ( f ( x ) ) 2

New answer posted

3 months ago

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Vishal Baghel

Contributor-Level 10

Given equation

c o s x 1 + s i n x = 2 | s i n x | c o s x | c o s 2 x s i n 2 x |

| 1 2 s i n 2 x | = 2 | s i n x | ( 1 + s i n x )

Put sin x = t

then equation,

| 1 2 t 2 | = 2 | t | ( 1 + t ) , t ( 1 , 1 ) { ± 1 2 }

Case I : For 1 2

the equation has no solution

Case II : For 0 t < 1 2

Equation t = 5 1 4 x = 1 8 °

Case III : F o r 1 2 < t < 0

Equation t = 1 2 x = 3 0 °

Case IV : For T < 1 2

Equation t = 5 + 1 4  x = 54°

Sum of solutions = 18° - 30° - 54° = 66°

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Total number = 20 + 8 + 24 = 52

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