Definite Integration
Integral with φ — Differentiating Both Sides
nta_pyq_2023_jan
Grade 12
Question:
If $\phi(x)=\dfrac{1}{\sqrt{x}}\displaystyle\int_{\pi/4}^x(4\sqrt{2}\sin t-3\phi'(t))\,dt$, $x>0$, then $\phi'\!\left(\dfrac{\pi}{4}\right)$ is equal to:
\dfrac{8}{\sqrt{\pi}}
\dfrac{4}{6+\sqrt{\pi}}
\dfrac{8}{6+\sqrt{\pi}}
\dfrac{4}{6-\sqrt{\pi}}
Step-by-Step Solution
Key Concept: Differentiate: $\phi'(x)\sqrt{x}+\phi(x)/(2\sqrt{x})=4\sqrt{2}\sin x-3\phi'(x)$. At $x=\pi/4$: $\phi(\pi/4)=0$.
$\phi'(\pi/4)=\dfrac{8}{6+\sqrt{\pi}}$.
Correct Answer: 3