Ncert Solutions Maths class 11th

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New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

? f ( x ) = l n ( x + 1 + x 2 ) f ( x ) = f ( x ) .

Hence f(x) is an odd function.

N o w g ( t ) = π 2 π 2 c o s ( π 4 t + f ( x ) ) d x

Put t = 0, g(0) = π 2 π 2 c o s ( f ( x ) ) d x = 2 0 π 2 c o s ( f ( x ) ) d x . . . . . . . . . . . . . . . ( i )

where cos(f(x)) is an even function.

Now again put t = 1,

g ( 1 ) = 1 2 * g ( 0 ) , f r o m ( i )

g ( 0 ) = 2 g ( 1 )

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Centre of the smallest circle is A

Centre of the largest circle is B

r 1 = | C P + C B | = 3 2 + 3 and

r 2 = | C P C B | = 3 2 3    

r 1 r 2 = 3 2 + 3 3 2 3 = 3 + 2 2

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

Let mid point of PQ is R (h, k)

h = α + 4 α 2 + α + 1 2 2 a n d

   

k = 4 α 2 + 1 + 4 α 2 + α + 1 2 2   

Eliminate a from above these two, we get

2 (3x – y)2 + (x – 3y) + 2 = 0

New answer posted

7 months ago

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V
Vishal Baghel

Contributor-Level 10

t a n 1 ( 2 * 3 5 1 9 2 5 ) + s i n 1 5 1 3 t a n 1 1 5 8 + t a n 1 5 1 2 = t a n 1 1 5 8 + 5 1 2 1 1 5 8 5 1 2 = t a n 1 2 2 0 2 1

t a n ( t a n 1 2 2 0 2 1 ) = 2 2 0 2 1

New answer posted

7 months ago

0 Follower 33 Views

A
alok kumar singh

Contributor-Level 10

6 3 = 2 t t = 1

 ->R (-1,0)

    P R 2 + R Q 2 = 2 0 + 5 = 2 5          

 

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

7 months ago

0 Follower 24 Views

A
alok kumar singh

Contributor-Level 10

ar (ABC) = 4ar (DEF)

= 4 * 1 2 * | 2 ( 2 5 ) + 1 ( 5 3 ) + 7 ( 3 2 ) | = 2 | 6 + 2 + 7 | = 6

 

New question posted

7 months ago

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New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  ( p q ) = p q

 

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

For every a, there  must be a2 – 2. So, there will be infinitely many pairs (a, b)

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