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New answer posted

8 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

89 – k + k + 98 – k + x = 100

=> 8 7 k 8 9

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

f ( x ) = 1 x 1 + x , 0 < x < 1 (on simplification)

f ' ( x ) = 2 ( 1 + x ) 2

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

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Given equation

c o s x 1 + s i n x = 2 | s i n x | c o s x | c o s 2 x s i n 2 x |

| 1 2 s i n 2 x | = 2 | s i n x | ( 1 + s i n x )

Put sin x = t

then equation,

| 1 2 t 2 | = 2 | t | ( 1 + t ) , t ( 1 , 1 ) { ± 1 2 }

Case I : For 1 2

the equation has no solution

Case II : For 0 t < 1 2

Equation t = 5 1 4 x = 1 8 °

Case III : F o r 1 2 < t < 0

Equation t = 1 2 x = 3 0 °

Case IV : For T < 1 2

Equation t = 5 + 1 4  x = 54°

Sum of solutions = 18° - 30° - 54° = 66°

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Total number = 20 + 8 + 24 = 52

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

f ( 0 ) = f ( 0 ) = a

f ( 0 + ) = l i m x 0 + s i n 2 x ( 1 c o s 2 x 1 ) b x 3 = 4 b

ab = 4 10 – ab = 14

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

S = 2 1 + k = 1 2 1 ( z k + 1 z k ) 3 = ( z k + z ¯ k ) 3 = 8 ( R e ( z k ) ) 3

( A s | z | = 1 1 z = z ¯ )

S = 2 k = 1 2 1 c o s k π + 6 k = 1 2 1 c o s k π 3 + 2 1

= 2 ( 1 ) + 6 R e ( α + α 2 + α 3 ) + 2 1

Where α = e i ( 2 π / 6 )

= 2 ( 1 ) + 6 ( 1 2 1 2 1 ) + 2 1

= 8 + 2 1 = 1 3

New answer posted

8 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

x 2 + y 2 + ( x 1 ) 2 + y 2 + x 2 + ( y 1 ) 2 + ( x 1 ) 2 + ( y 1 ) 2 = 1 8

x 2 + y 2 x y 7 2 = 0

r 2 = 1 4 + 1 4 + 7 2 = 4

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

(x + y) = 4xy

d y d x = 4 x y + 4 y 2 1 1 4 x 2 4 x y ,

d 2 y d x 2 = ( 1 4 x 2 4 x y ) ( 4 y + 4 x y ' + 8 y y ' ) ( 4 x y + 4 y 2 1 ) ( 8 x 4 y 4 x y ' ) ( 1 4 x 2 4 x y ) 2

New answer posted

8 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

B ( 1 + 2 λ , 3 + λ , 4 + 2 λ )

x – 2y – z = 3 => λ = 6 B ( 1 1 , 3 , 8 )

N ( 1 + t , 3 2 t , 4 t )  

x 2 y z = 3 t = 2 N ( 3 , 1 , 2 )  

Line NB : x 3 7 = y + 1 1 = 3 2 5

d 2 = P M 2 = | P Q * R | 2 | R | 2

| R | 2 = 7 5

d 2 = 1 9 5 0 7 5 = 2 6  

New answer posted

8 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

L H S = r = 1 1 5 r ( r ! )

= ( r + 1 1 ) r ! = r = 1 1 5 ( r + 1 ) ! r !

= ( 2 ! 1 ! ) + ( 3 ! 2 ! ) + . . . . + ( 1 6 ! 1 5 ! )

= 1 6 ! 1 = 1 6 P 1 6 1

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