Ncert Solutions Maths class 11th

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New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

f ( x ) = x + x 0 1 f ( t ) d t 0 1 t 0 f ( t ) d t

 Let 1 + 0 1 f ( t ) d t = α  

0 1 t f ( t ) d t = β  

So, f(x) = ax - b

Now,  α = 0 1 f ( t ) d t + 1  

α = 0 1 ( a t β ) d t + 1  

β = 0 1 t f ( t ) d t  

β = 4 1 3 , α = 1 8 1 3  

f(x) = ax – b

= 1 8 x 4 1 3

option (D) satisfies

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

Let S = 1 + 5 6 + 1 2 6 2 + 2 2 6 3 + . . . .  

S 6 = 1 6 + 5 6 2 + . . . . _  

5 S 6 = 1 + 4 6 + 7 6 2 + 1 0 6 3 + . . . .  

5 S 3 6 = 1 6 + 4 6 2 + 7 6 3 + . . . . . _  

2 5 S 3 6 = 1 + 3 6 + 3 6 2 + 3 6 3 + . . . . . .  

2 5 S 3 6 = 1 + 3 / 6 1 1 / 6  

2 5 S 3 6 = 1 + 3 / 5 1  

S = 2 8 8 1 2 5

New answer posted

2 weeks ago

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A
alok kumar singh

Contributor-Level 10

| z | = 3 circle with radius = 3

arg ( z 1 z + 1 ) = π 4 , part of a circle (with radius 2 ). no common points

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A
alok kumar singh

Contributor-Level 10

x4 + x2 + 1 = 0

x4 + 2x2 + 1 – x2 = 0

( x 2 + 1 + x ) ( x 2 x + 1 ) = 0

x = ± ω , ? ω 2

Now, =  α 1 0 1 1 + α 2 0 2 2 α 3 0 3 3

ω 1 0 1 1 + ω 2 0 2 2 ω 3 0 3 3

= 1 + 1 – 1 = 1

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

y + 2 x = 1 1 + 7 7  ………(i)

              2 y + x = 2 1 1 + 6 7        ………(ii)

              x + y = 1 1 + 1 3 3 7    ………(iii)

              Centre of the circle given by solving (i) & (ii)

              a s ( 8 7 3 , 1 1 + 5 7 3 )  

              Again 1 1 y 3 x = 5 7 7 3 is tangent to the circle.

              r = | 1 1 ( 1 1 + 5 7 3 ) 8 7 5 7 7 3 1 1 + 9 | = | 1 1 8 7 2 0 |  

              ( 5 h 8 k ) 2 + 5 r 2 = 8 1 6  

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

R ( 1 0 + α 3 , 8 3 )

| m A Q | = | m A P |

| 4 5 α | = | 3 2 α |

⇒a = -7 not possible α = 2 3 7 . 7 α + 3 β = 2 3 + 8 = 3 1

 

New question posted

2 weeks ago

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New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

kindly consider the following Image 

 

New answer posted

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R
Raj Pandey

Contributor-Level 9

t a n π 8 = 2 h 8 0 . . . . . . . . . ( i )

t a n θ = h 8 0 . . . . . . . . . . ( i i )

t a n π 8 = 2 t a n θ

t a n 2 θ = 3 2 2 4

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

x 2 + y 2 2 x 4 y = 0

Centre (1, 2) r = 5  

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒x + 2y = 0       ……… (i)

Equation of PQ is  x ( 1 + 5 ) + 2 y ( | x + 1 + 5 ) 2 ( y + 2 ) = 0  

Solving (i) & (ii),   Q ( 5 + 1 , 5 + 1 2 )  

              = 1 2 | 6 + 2 5 + 4 5 = 4 2 | = 6 5 + 1 0 4 = 3 5 + 5 2  

 

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