उत्तर: (3)

समाधान:

सूत्र: लेब्निज के प्रमेय , निश्चित समाकलन के गुणधर्म

$\phi(x)=\dfrac{1}{\sqrt{x}} \int _{\dfrac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$\sqrt{x} \ \phi(x)=\int _{\dfrac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t , \quad \phi(\dfrac{\pi}{4})=0$

लेब्निज प्रमेय के अनुसार, हम प्राप्त करते हैं

$ \dfrac{1}{2\sqrt{x}}\phi(x) +\sqrt{x}\phi^{\prime}(x)=[(4 \sqrt{2} \sin x-3 \phi^{\prime}(x)) \cdot 1-0]$

$\Rightarrow \ \sqrt{x} \phi ^{\prime}(x) + 3 \phi ^{\prime} (x) = 4 \sqrt{2} \sin x - \dfrac{1}{2 \sqrt{x}} \phi (x)$

$ x = \dfrac{\pi }{4} $ पर

$\Rightarrow \ \phi ^{\prime }(\dfrac{\pi}{4}) \left( \dfrac{\pi}{4}+3\right) = 4 \sqrt{2} \sin \dfrac{\pi}{4}- \dfrac{1}{2 \sqrt{\dfrac{\pi}{4}}} \phi (\dfrac{\pi}{4})$

$\Rightarrow \ \phi ^{\prime}(\dfrac{\pi}{4}) \left(\dfrac{\sqrt{\pi}}{2} + 3\right) = 4- \dfrac{1}{2} \cdot 0$

$\Rightarrow \ \phi ^{\prime} (\dfrac{\pi}{4}) = \dfrac{4}{\dfrac{\sqrt{\pi}}{2}+3}$

$ \Rightarrow \phi^{\prime}(\dfrac{\pi}{4})=\dfrac{8}{\sqrt{\pi}+6}$