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New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

a → = i ^ + 2 j ^ + k ^                

b → = 2 i ^ + 4 j ^ − 5 k ^ c → = − λ i ^ + 2 j ^ + 3 k ^              

b → + c → = ( 2 − λ ) i ^ + 6 j ^ − 2 k ^

( 1 2 − λ ) 2 = ( 2 − λ ) 2 + 4 0

λ = 5

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

The 1st such digit is 11 * 19 = 209

Sum = [209 + 220 + 231 + .+ 495] - [231 + 319 + 341 + 418 + 451] = 7744

New answer posted

a year ago

0 Follower 30 Views

A
alok kumar singh

Contributor-Level 10

| 2 A | = 2 3 | A |

replace A by adj 2A

  | 2 a d j     2 A | = 2 3 | a d j     2 A |           

= 2 3 | 2 A | 2 = 2 3 ( 2 3 | A | ) 2

= 2 9 | A | 2              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 29 |adj 2A|4

= 29 (|2A|2)4

->|A2| = 4

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

a + b = 1   α + γ = 1 0 3

α β = 2 λ α γ = 9 λ

β γ = 2 9 , β − γ = 1 − 1 0 3 = − 7 3

⇒ β = 2 3 γ = 3

α = 1 3 , λ = 1 9

∴ β γ λ = 1 8                                  

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

f ( x ) = 2 x 3 − 3 x 2 − 1 2 x  

  f ' ( x ) = 6 x 2 − 6 x − 1 2             

= 6 (x – 2) (x + 1)

a = -1, b = 2

A = ∫ − 1 0 ( 2 x 3 − 3 x 2 − 1 2 x ) d x − ∫ 0 2 ( 2 x 3 − 3 x 2 − 1 2 x ) d x = 5 7 2

->4A = 114

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

 = coefficient of x4 in (1 + x)21 + coefficient of x4 in (1 + x)21

= 2 1 C 4 + 2 1 C 4 = 2 . 2 1 C 4              

A 3 = coefficient of x3 in (1 + x)21 + coefficient of x3 in (1 + x)21

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Equation of required plane

( x + y + 4 z − 1 6 ) + λ ( − x + y + z − 6 ) = 0 it passes (1, 2, 3)

⇒ − 1 + λ ( − 2 ) = 0

⇒ λ = − 1 2

∴ Equation of plane

( 1 − λ ) x + ( 1 + λ ) y + ( 4 + λ ) z − 1 6 − 6 λ = 0              

3 2 x + 1 2 y + 7 2 z − 1 3 = 0              

⇒ 3 x + y + 7 z = 2 6              

  (4, 2, 2) not satisfying the plane

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

  2 − 4 = − 1 2

y – 4 = 2 (x – 3)

y = 2x – 2

x2 + (2x – 2)2 = 25

  5 x 2 − 8 x − 2 1 = 0

z ( − 7 5 , − 2 4 5 )              

               

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

x = 2t A ( 2 t , t 2 3 )

y = t 2 3  S (0, 3)

  3 y = ( x 2 ) 2 B ( 0 , λ )

3 k = t 2 3 + 3 + λ = 2 t 2 3 + 3 − 1 2 t 2 9 − t 2

l i m t → 1 3 k = 2 3 + 3 − 1 2 8 = 3 − 5 6 = 1 3 6

 

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

y 2 d x + ( x 2 − x y + y 2 ) d y = 0

d x d y + x 2 − x y + y 2 y 2 = 0 ⇒ d x d y + ( x y ) 2 − ( x y ) + 1 = 0

Put x = vy ⇒ v + y d v d y + v 2 − v + 1 = 0

⇒ y d v d y + v 2 + 1 = 0

⇒ π 6 + l n | y | = π 4

l n | y | = π 1 2

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