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New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

take z = x + iy

z 2 + z ¯ = 0               

⇒ x 2 − y 2 + x + i     2 x y − y i = 0               

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

if y = 0 Þ x = 0, -1

i f     x = 1 2 ⇒ y = ± 3 2               

Σ ( R e ( z ) + l m ( z ) ) = ( 0 − 1 + 1 2 + 1 2 ) + ( 0 + 0 + 3 2 − 3 2 ) = 0               

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

( x + 1 ) 2 λ + 5 = ( y + 1 ) 2 λ + 5 4 = 1  

length of latus rectum = 2 b 2 a = 2 ( λ + 5 4 ) 5 + λ = 4  

λ + 5 = 8 ⇒ λ = 5 9                

Major axis = 2 λ + 5 = 1 6  

New answer posted

a year ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Let A = [ a b c d e f 9 h i ]  

Now ATA

trace will be    a 2 + b 2 + c 2 + d 2 + e 2 + f 2 + 9 2 + h 2 = 6

total ways = 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

| f ( x ) | ≤ 8 0 0 ⇒ 2 n 2 − n − 1 ≤ 8 0 0  

⇒ 2 n 2 − n − 8 0 1 ≤ 0                

∑ x ∈ S f ( x ) = ∑ ( 2 x 2 − x − 1 )                

= 2 ( 1 9 2 + 1 8 2 + . . . . . . . . 1 2 + 0 2 + 1 2 + . . . . . + 2 0 2 )                

= 10620

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

y5 – 9xy + 2x = 0

differentiate 5y4 – 9x d y d x - 9y + 2 = 0

d y d x = 9 y − 2 5 y 4 − 9 x        

 For horizontal tangent d y d x = 0 ⇒ y = 2 9  which does not satisfy the equation so no horizontal

For vertical tangent -> 5 y 4 − 9 x = 0  

->m = 0, N = 2

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

s i n ( 2 x 2 ) . 1 0 9 e ( t a n x 2 ) d y + 4 x y     d x = 4 2 . x . ( s i n x 2 c o s π 4 − c o s x 2 s i n π 4 ) d x  

⇒ l n ( t a n x 2 ) d y + 4 x s i n ( 2 x 2 ) y d x = 4 x ( s i n x 2 − c o s x 2 ) s i n ( 2 x 2 ) d x

Integrate

⇒ y . l n ( t a n x 2 ) = 2 . l n ( s i n x 2 + c o s x 2 − 1 s i n x 2 + c o s x 2 + 1 ) + C

x = π 6 , y = 1  calculate C.

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

e H = 1 + 6 4 4 9 = 1 1 3 7

⇒ e H . e E = 1 2
⇒ 1 1 3 4 9 . ( 6 4 − a 2 ) 6 4 = 1 4 ⇒ a 2 − 6 4 = 3 2 2 1 1 3
l = 2 a 2 b = 2 ( 6 4 + 3 2 2 1 1 3 ) . 1 8
1 1 3 l = 1 5 5 2

 

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is    x ( 4 a + 2 λ ) + y ( − 1 − 5 λ ) + z ( 5 − λ ) = 7 a + 3 λ

This plane contains 4, -1, 0

->9a + 1 + 10l = 0          …… (i)

Plane contains the line x − 4 1 = y + 1 − 2 = z 1  

-> 4 a + 1 1 λ + 7 = 0 ……. (ii)

From (i) & (ii) a = 1,   λ  =-1

Equation of plane π ≡ x + 2 y + 3 z − 2 = 0  

⇒ 7 P + 3 − 2 P + 4 − 1 2 P + 9 − 2 = 0 ⇒ P = 2

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

x ¯ = ∑ i = 1 1 0 x i 1 0 = 1 5 ; ∑ i = 1 1 0 x i 2 1 0 − ( x ¯ ) 2 = 1 5

⇒ Σ x i = 1 5 0 ; Σ x i 2 = 2 4 0 0

Actual mean x ¯ = Σ x i + 1 5 − 2 5 1 0 = 1 4 0 1 0 = 1 4

Actual variance =  Σ x i 2 + 1 5 2 − 2 5 2 1 0 − ( 1 4 ) 2

= 2 4 0 0 − 4 0 0 1 0 − 1 9 6

σ 2 = 4 σ = 2

 

New answer posted

a year ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

use s i n − 1 x = c o s − 1 1 − x 2  

t a n − 1 1 − x 2 = c o t − 1 1 1 − x 2               

s i n − 1 x 1 − x 2 = c o s − 1 1 − 2 x 2 1 − x 2               

Sum of roots ->b = -1 + 2   ( k 2 − 1 ) k 2 − 2

Product of roots -> -5 = -2  ( k 2 − 1 k 2 − 2 )  

b = 4, k2 = 1 3  

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