If φ(x)=(1/√ x ) ∫ limits(π/4)x(4 √2 sin t-3 φ prime(t)) dt , x0, then φ prime((π/4)) is equal to :

Question

If φ(x)=(1/√ x ) ∫ limits(π/4)x(4 √2 sin t-3 φ prime(t)) dt , x>0, then φ prime((π/4)) is equal to : (A) (4/6+√π) (B) (8/6+√π) (C) (8/√π) (D) (4/6-√π)

Tardigrade Admin · Accepted Answer

φ prime( x )=(1/√ x )[(4 √2 sin x -3 φ prime( x )) ⋅ 1-0]-(1/2) x -3 / 2 ∫ limits(π/4) x (4 √2 sin t -3 φ prime( t )) dt φ prime((π/4))=(2/√π)[4-3 φ prime((π/4))]+0 (1+(6/√π)) φ prime((π/4))=(8/√π) φ prime((π/4))=(8/√π+6)

Q. If $ϕ (x) = \frac{1}{x} \frac{π}{4} \int x (42 sin t - 3 ϕ^{'} (t)) d t, x > 0$ , then $ϕ^{'} (\frac{π}{4})$ is equal to :

$\frac{4}{6 + π}$

$\frac{8}{6 + π}$

$\frac{8}{π}$

$\frac{4}{6 - π}$

Q. If ϕ(x)=x​1​4π​∫x​(42​sint−3ϕ′(t))dt,x>0, then ϕ′(4π​) is equal to :

6+π​4​

6+π​8​

π​8​

6−π​4​

ϕ′(x)=x​1​[(42​sinx−3ϕ′(x))⋅1−0]−21​x−3/2 4π​∫x​(42​sint−3ϕ′(t))dt ϕ′(4π​)=π​2​[4−3ϕ′(4π​)]+0 (1+π​6​)ϕ′(4π​)=π​8​ ϕ′(4π​)=π​+68​

Q. If $ϕ (x) = \frac{1}{x} \frac{π}{4} \int x (42 sin t - 3 ϕ^{'} (t)) d t, x > 0$ , then $ϕ^{'} (\frac{π}{4})$ is equal to :

$\frac{4}{6 + π}$

$\frac{8}{6 + π}$

$\frac{8}{π}$

$\frac{4}{6 - π}$