Class 11th

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New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

  f ( x ) = m i n { 1 , 1 + x s i n x } , 0 x 2 π

f ( x ) = { 1 , 0 x < π 1 + x s i n x , π x 2 π

Now at x = π,

f ( x ) is not differentiable at x = π

(m, n) = (1, 0)

New answer posted

11 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  l i m x 0 c o s ( s i n x ) c o s x x 4 = l i m x 0 2 s i n ( x + s i n x ) . s i n ( x s i n x 2 ) x 4

= l i m x 0 2 . ( ( x + s i n x 2 ) ( x s i n x 2 ) x 4 )

= l i m x 0 1 2 . ( 2 x 2 3 ! + x 4 5 ! . . . . . . . . ) ( 1 3 ! x 2 5 ! 1 )

= 1 6

New answer posted

11 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

A = n = 1 ( 1 ) n ( 3 + ( 1 ) n ) n

B = n = 1 ( 1 ) n ( 3 + ( 1 ) n ) n

A = 1 1 1 5 , B = 9 1 5

A B = 1 1 9

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

(xa)n+ (yb)n=2

na (xa)n1+nb (yb)n1dydx=0

dydx=ba (bxay)n1

dydx (a, b)=ba

So line always touches the given curve.

New answer posted

11 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

20212 mod (7)

(2021)2023 (2)2023mod (7) …… (i)

Now,   (2)31mod (7)

(2)2023 (2)mod (7)5mod (7) ……. (ii)

(i) & (ii)

(2021)20235mod (7)

 Remainder = 5

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

System of equation can be written as

(32158921a) (xyz)= (b31)

(321152427633a) (xyz)= (b93)

R32R1, R25R1

for no solution

3a + 9 = 0 but 329b20

a=3b13

New answer posted

11 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

|adj (24A)|=|adj (3adj (2A))|

|24A|2=|3adj (2A)|2

246|A|2=36. (23)4|A|4

|A|2=24636.212=218.3636.212=26

New answer posted

11 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

pH of acidic buffer is given by,

pH = pKa + log10 [ S a l t ] [ a c i d ]

or, p = p k a + l o g 1 0 [ C H 3 C H 2 C O O ] [ C H 3 C H 2 C O O H ]

or, [ H + ] = k a [ C H 3 C H 2 C O O H ] [ C H 3 C H 2 C O O ]

[ C H 3 C H 2 C O O ] [ C H 3 C H 2 C O O H ] = k a [ H + ] = 1 . 3 * 1 0 5 1 0 4 = 0 . 1 3

 

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Geometry of SF4 is trigonal bipyramidal, in which there is one lonepair which occupy equatorial position as,

There are two lone pair – bond pair repulsions at 90°

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

Values of principal quantum number, n = 1, 2, 3, .

Values of azimuthal quantum number,   l = 0 , 1 , 2 . . . . . . . . . . . , n 1

Number of values of magnetic quantum number are 2 l + 1 .  

Values of spin quantum number are ± 1 2 .

Number of orbitals for particular value of l a r e 2 l + 1 , s o f o r l = 5 ,  no. of orbitals = 11.

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