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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 2 days ago
Asked by Shiksha User

  • 1 Answer

  • 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 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

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

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

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= 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

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x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

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Now,

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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 . . . . . . . . . . . . . . . . . . .  
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L 1 = x − λ 1 = y − 1 2 1 2 = z − 1 2

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Equation of AB

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