If ϕ(x)=1x∫π4x 42sin⁡t−3ϕ′(t)dt,x>0, then ϕ′ π4 is equal to

x⋅ϕ(x)=∫π4x 42sin⁡t−3θ1(t)dt,x>0,ϕ1π4=?,ϕπ4=0Diff both sidesx⋅ϕ1(x)+ϕ(x)⋅12x=42sin⁡x−3ϕ1(x)⋅1Put x=π4π4⋅ϕ1π4+0=42⋅12−3ϕ1π4ϕ'π4⋅3+π2=4ϕ1π4=86+π

If ϕ(x)=1x∫π4x 42sin⁡t−3ϕ′(t)dt,x>0, then ϕ′ π4 is equal to

x⋅ϕ(x)=∫π4x 42sin⁡t−3θ1(t)dt,x>0,ϕ1π4=?,ϕπ4=0Diff both sidesx⋅ϕ1(x)+ϕ(x)⋅12x=42sin⁡x−3ϕ1(x)⋅1Put x=π4π4⋅ϕ1π4+0=42⋅12−3ϕ1π4ϕ'π4⋅3+π2=4ϕ1π4=86+π

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If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) dt, x > 0, then ϕ^{'} (\frac{π}{4})$ is equal to

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$\frac{4}{6 + \sqrt{π}}$

$\frac{8}{6 + \sqrt{π}}$

$\frac{8}{\sqrt{π}}$

$\frac{4}{6 - \sqrt{π}}$

answer is B.

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Detailed Solution

$\sqrt{x} \cdot ϕ (x) = \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 θ^{1} (t)) dt, x > 0, ϕ^{1} (\frac{π}{4}) = ?, ϕ (\frac{π}{4}) = 0$
Diff both sides
$\sqrt{x} \cdot ϕ^{1} (x) + ϕ (x) \cdot \frac{1}{2 \sqrt{x}} = 4 \sqrt{2} \sin x - 3 ϕ^{1} (x) \cdot 1$
Put $x = \frac{π}{4}$
$\begin{array}{l} \sqrt{\frac{π}{4}} \cdot ϕ^{1} (\frac{π}{4}) + 0 = 4 \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 3 ϕ^{1} (\frac{π}{4}) \\ ϕ' (\frac{π}{4}) \cdot (3 + \frac{\sqrt{π}}{2}) = 4 \\ ϕ^{1} (\frac{π}{4}) = \frac{8}{6 + \sqrt{π}} \end{array}$