Definite Integration Question 4
Question 4 - 24 January - Shift 2
Let $f$ be a differentiable function defined on $[0, \frac{\pi}{2}]$ such that $f(x)>0$ and
$f(x)+\int_0^{x} f(t) \sqrt{1-(\log _{e} f(t))^{2}} dt=e, \forall x \in[0, \frac{\pi}{2}]$.
Then $(6 \log _{e} f(\frac{\pi}{6}))^{2}$ is equal to
Show Answer
Answer: 27
Solution:
Formula: Leibnitz Theorem, Integration by substitution, Standard formulas for Indefinite Integration
$f(x)+\int_0^{x} f(t) \sqrt{1-(\log _{e} f(t))^{2}} d t=e$
$\Rightarrow f(0)=e$
$f^{\prime}(x)+f(x) \sqrt{1-(\ln f(x))^{2}}=0$
$f(x)=y$
$\frac{d y}{d x}=-y \sqrt{1-(\ln y)^{2}}$
$\int \frac{d y}{y \sqrt{1-(\ln y)^{2}}}=-\int dx$
Put $\ln y=t$
$\int \frac{dt}{\sqrt{1-t^{2}}}=-x+C$
$\sin ^{-1} t=-x+C \Rightarrow \sin ^{-1}(\ln y)=-x+C$
$\sin ^{-1}(\ln f(x))=-x+C$
$f(0)=e$
$\Rightarrow \frac{\pi}{2}=C$
$\Rightarrow \sin ^{-1}(\ln f(x))=-x+\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(\ln f(\frac{\pi}{6}))=\frac{-\pi}{6}+\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}(\ln f(\frac{\pi}{6}))=\frac{\pi}{3}$
$\Rightarrow \ln \mathrm{f}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$, we need $ 6 \times \ln \mathrm{f}\left(\frac{\pi}{6}\right) \Rightarrow \left(6 \times \frac{\sqrt{3}}{2}\right)^2=27$.
