Question

Let $e$ be the eccentricity of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,f(e)$ be the eccentricity of a conjugate hyperbola and g(e) be the eccentricity of an ellipse whose major and minor axis coincides with transverse and conjugate axis of hyperbola then the value of $\underset{1}{\overset{\sqrt{2}}{\int }}g(g(f(f(e))))$ de is equal to

• A. $\frac{1}{2}$
• B. $\sqrt{2}$
• C. 1
• D. $\sqrt{2}+1$

Answer 1

Answer: (A)

$\because \frac{1}{e^2}+\frac{1}{f^2(e)}=1\Rightarrow \frac{1}{f^2(e)}=1-\frac{1}{e^2}=\frac{e^2-1}{e^2}$
$\Rightarrow f(e)=\frac{e}{\sqrt{e^2-1}}$
$\because e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2}$
$\text{ and }g(e)=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow g^2(e)=\frac{a^2-b^2}{a^2}\Rightarrow e^2+g^2(e)=2\Rightarrow g(e)=\sqrt{2-e^2}$
$\therefore f(f(e))=\frac{\sqrt{e^2-1}}{\sqrt{\frac{e^2}{e^2-1}-1}=e}$
$\therefore g(f(f(e)))=g(e)=\sqrt{2-e^2}$
$\Rightarrow g(g(f(f(e))))=\sqrt{2-\left(2-e^2\right)}=e$
$\therefore Given\text{ integral }=\underset{1}{\overset{\sqrt{2}}{\int }}ede={\left[\frac{e^2}{2}\right]}_1^{\sqrt{2}}=1-\frac{1}{2}=\frac{1}{2}$