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New answer posted

3 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Minimum distance Andhra Pradesh = OP – OA

= 3 2 2 = 2 2

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  l i m x π 4 π 4 . f ( s e c 2 x ) 2 s e c 2 x t a n x 2 x = 2f (2)

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

As ( A B ) = A B

( ( p v r ) ( q v r ) )

= ( p v r ) ( r q )

= ( p r ) ( r r ) ( q )

= ( p r )

New answer posted

3 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Required area

A = 2 ( ( s i n x + c o s x ) ( c o s x s i n x ) ) d x

= 2 0 π / 4 s i n x d x

= 2 2 ( 2 1 )    

New answer posted

3 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Total ways = 6!

Ways satisfying g (3) = 2g (1) is 3

Number of onto function 3 * 4!

Probability = 1 1 0

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

b 2 x 2 + a 2 y 2 = a 2 b 2

( b 2 x 2 + a 2 ( a b x 2 ) ) = a 2 b 2

x 2 = b a 2 ( b a ) b 2 a 2 , y 2 = a b 2 a + b

points of intersection

( a b a + b , b a a + b )

t a n θ = | m 1 m 2 1 + m 1 . m 2 |

= | a b a b |

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Δ = 0 a = 3 , 4  

 For a = 3, no solution

For a = 4, no solution

n (S1) = 2, n (S2) = 0,

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Equation of intersection of line

x 0 1 = y = z 0 1

              

Let r = i ^ k ^

Direction ratio of P Q

( λ + 1 ) ( 1 ) + ( 2 ) ( 0 ) + ( 2 λ ) ( 1 ) = 0

λ = 1 2 Q ( 1 2 , 0 , 1 2 )

P Q = 3 4 2

             

             

               

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

f ( x ) = l o g 5 ( 3 + 2 ( c o s x s i n x ) )

2 c o s x s i n x 2

0 f ( x ) 2

New answer posted

3 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

| α β 1 5 6 1 3 2 1 | = 2 4

4 α 2 β 8 = ± 2 4

4 α 2 β = 2 4 + 8 , 4 α 2 β = 2 4 + 8

2 ( 2 α β ) = 3 2 2 α β = 8

Distance of origin

D = α 2 + ( 2 α + 8 ) 2

α = 1 6 5

D = ( 1 6 5 ) 2 + ( 8 5 ) 2

if 2 - = 16

D = 1 6 5

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