Integral Calculus
Hyperbola Eccentricity
MMTS_Full_Test_11
Grade 12
Question:
Let $e$ be the eccentricity of a hyperbola and $f(e)$ be the eccentricity of its conjugate hyperbola. Then $\int(f(e)+f(f(e)))de=g(e)$ and $g(\sqrt{2})=2$, then $g(e)=$
$\frac{1}{2}\sqrt{e^2-1}+\frac{e^2}{2}$
$\frac{1}{2}\sqrt{e^2+1}+\frac{e^2}{2}$
$\sqrt{e^2-1}+\frac{e^2}{2}$
$\sqrt{e^2+1}+\frac{e^2}{2}$
Step-by-Step Solution
Key Concept: $f(e)=\frac{e}{\sqrt{e^2-1}}$; $f(f(e))=e$; integrate $\frac{e}{\sqrt{e^2-1}}+e$
$f(e)+f(f(e))=\frac{e}{\sqrt{e^2-1}}+e$. $\int(\frac{e}{\sqrt{e^2-1}}+e)de=\sqrt{e^2-1}+\frac{e^2}{2}+C$. $g(\sqrt{2})=1+1+C=2\Rightarrow C=0$. $g(e)=\sqrt{e^2-1}+\frac{e^2}{2}$.
Correct Answer: 3