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If ∫ ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) d x = π 4 ( π + α ) − 2

Then the value of 'α' is equal to

Option 1 -

1

Option 2 -

2

Option 3 -

3

Option 4 -

4

0 2 Views | Posted 3 weeks ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 weeks ago
    Correct Option - 3


    Detailed Solution:

    ∫ − π 2 π 2 ( x 2 c o s x 1 + π 2 + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) d x = π 4 ( π + α ) − 2

    ∫ 0 π 2 { ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) + ( x 2 c o s x 1 + π − x + 1 + s i n 2 x 1 + e − ( s i n x ) 2 0 2 3 ) } d x

    = π 4 ( π + α ) − 2

    ∫ 0 π 2 ( x 2 c o s x + 1 + s i n 2 x ) d x = π 4 ( π + α ) − 2

    ∫ 0 π 2 x 2 c o s x d x + ∫ 0 π 2 ( 1 + s i n 2 x ) d x = π 4 ( π + α ) − 2 .(1)

    Let I 1 = ∫ 0 π 2 ( 1 + s i n 2 x ) d x

    I 1 = ∫ 0 π 2 1 ⋅ d x + ∫ 0 π 2 ( 1 − c o s 2 x 2 ) d x

    I 1 = π 2 + 1 2 [ π 2 + 0 ]

    I 1 = 3 π 4

    Let I 2 = ∫ 0 π 2 x 2 c o s x d x

    I 2 = [ x 2 ( s i n x ) − ∫ 2 x ∫ c o s x d x ] 0 π 2

    I 2 = [ x 2 ( s i n x ) − 2 ∫ x s i n x ] 0 π 2

    I 2 = [ x 2 s i n x − 2 ( x ( − c o s x ) + ∫ c o s x ) ] 0 π 2

    I 2 = [ x 2 s i n x − 2 ( − x c o s x + s i n x ) ] 0 π 2

    I 2 = ( π 2 4 − 2 )

    Put l1 and l2 in (1)

    π 2 4 − 2 + 3 π 4

    π 2 4 + 3 π 4 − 2

    π 4 ( π + 3 ) − 2

    α = 3

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If | a → | = 1 , | b → | = 4 a → ⋅ b → = 2 and c → = 2 ( a → × b → ) − 3 b → Then the angle between  b → and c →  is
A
alok kumar singh

Given | a → | = 1 , | b → | = 4 , a → ⋅ b → = 2

c → = 2 ( a → × b → ) − 3 b →  

Dot product with  a → on both sides

c → ⋅ a → = − 6 ... (1)

Dot product with  b →  on both sides

b → ⋅ c → = − 4 8 ... (2)

c → ⋅ c → = 4 | a → × b → | 2 + 9 | b → | 2

| c → | 2 = 4 [ | a → | 2 ⋅ | b → | 2 − ( a → ⋅ b → ) 2 ] + 9 | b → | 2

| c → | 2 = 4 [ ( 1 ) ( 4 ) 2 − ( 4 ) ] + 9 ( 1 6 )

| c → | 2 = 4 [ 1 2 ] + 1 4 4

| c → | 2 = 4 8 + 1 4 4

| c → | 2 = 1 9 2

c o s θ = b → ⋅ c → | b → | | c → |

c o s θ = − 4 8 1 9 2 ⋅ 4

c o s θ = − 4 8 8 3 ⋅ 4

c o s θ = − 3 2 3

c o s θ = − 3 2 θ = c o s − 1 ( − 3 2 )

 

If the foot of perpendicular from (1, 2, 3) to the line x + 1 2 = y − 2 5 = z − 4 1  is (a, b, g) then find a + b + g
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alok kumar singh

(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

λ = 1 1 0  

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

= 8 λ + 5 = 8 1 0 + 5 = 5 . 8

If (α, β, γ) is mirror image of the point (2, 3, 4) with respect to the line x − 1 2 = y − 2 3 = z − 3 4 Then 2α + 3β + 4γ is
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alok kumar singh

Take x − 1 2 = y − 2 3 = z − 3 4 = λ  

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

  A B →  = (α − 2)  i ^ + (β − 3) j ^ + (γ − 4) k ^  

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

Let λ be an integer. IF the shortest distance between the lines x − λ = 2 y − 1 = − 2 z and x = y + 2 λ = z − λ     i s 7 2 2 , then the value of | λ | i s . . . . . . . . . . . . . . . . . . .  
V
Vishal Baghel

L 1 = x − λ 1 = y − 1 2 1 2 = z − 1 2

∴ S D = | − 2 λ + 3 ( 2 λ + 1 2 ) + λ | 1 4 = | 5 λ + 3 2 | 1 4

5 λ + 3 2 = − 7 2 ⇒ 5 λ = − 5 ⇒ λ = − 1

∴ | λ | = 1

The equation of the line through the point (0, 1, 2) perpendicular to the line

x − 1 2 = y + 1 3 = z − 1 − 2     i s :  
V
Vishal Baghel

x − 1 2 = y + 1 3 = z − 1 − 2 = r

For ‘B’, x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

{ ( 2 r + 1 ) − 0 } 2 + { ( 3 r − 1 ) − 1 } 3 + { ( − 2 r + 1 ) − 2 }

r = 2 1 7 ⇒ B ( 2 1 7 , − 1 1 7 , 1 3 1 7 )     

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

x − 3 = y − 1 4 = z − 2 3

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