Class 11th

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New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

2a² + 3b² = 35
a=0, b≠I
a=±1, b²=11 ⇒ b≠I
a=±2, b=9 ⇒ b=3 or -3
a=±3, b²=17/3 ⇒ b≠I
a=±4, b²=1 ⇒ b=±1

New answer posted

3 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

v m a x = A ω = A k m

Momentum conservation   m v max   + 0 = m v A + 2 m v B

v A + 2 v B = v m a x

v 2 - v 1 = u 1 - u 2 v B - v A = v m a x

By (1) & (2) v A = - A ω 3

v A = A new   ω

A ω 3 = A new   ω A new   = A 3 ; f = 1 3

New answer posted

3 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

log x = t
(t² - 1/t)¹²
T? = ¹²C? (t²)¹²? (-1/t)? = ¹²C? t²? ³? (-1)?
For constant term r = 8 & coefficient ¹²C?
= (12*11*10*9) / 24 = 45*11 = 495

New answer posted

3 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

tan? ¹ (x) = t
t² - 4t + 3 > 0
t ∈ (-∞, 1) U (3, ∞)
tan? ¹ (x) ∈ (-π/2, 1)
x ∈ (-∞, tan1)
the largest integral x = 1

New answer posted

3 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

z - 2² = z + 2²
⇒ x=0
Hence minimum '0'

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(Var)new = k² (Var)old
= 4 * 5 = 20

New answer posted

3 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  x 2 + y 2 2 x 6 y + 6 = 0 centre (1, 3)

r = 1 + 9 6 = 2 C M = 1 + 4 = 5           

r = 5 + 4 = 3

 

           

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

~ (p v (~ p v q)
= p ^ ( (~p v q) = ~p ^ (p ^ ~q)

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Sum of elements A ( B C ) = 2 7 4 * 4 0 0  

In set B numbers of the form 9k + 2 are {101, 109, .992}

  s u m = 1 0 0 2 ( 1 0 1 + 9 9 2 ) = 1 0 0 * 1 0 9 3 2 . . . . . . ( i )         

Another possible number is 9k + 5 forms are {104, .995}

  s u m = 1 0 0 2 ( 1 0 4 + 9 9 5 ) = 1 0 0 2 * 1 0 9 9 . . . . . . . . . ( i i )

T o t a l = 1 0 0 2 * [ 1 0 9 3 + 1 0 9 9 ] = 1 0 0 * 1 0 9 6 = 2 7 4 * 4 * 1 0 0 = 2 7 4 * 4 0 0           

possible value of l = 5

New answer posted

3 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  4 s i n x + 1 1 s i n x = a x ( 0 , π 2 )

Let sin x = t, t  (0, 1)

g ( t ) = 4 t + 1 1 t         

g ' ( t ) = 0 t = 2 3           

g ' ' ( 2 3 ) > 0           

g ( t ) m i n i m u m = 4 2 / 3 + 1 1 2 / 3 = 9          

Minimum value of a for which solution exist = 9

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