Question

Given functions f and g such that for all x, ${\left(g\left(x\right)\right)}^2-{\left(f\left(x\right)\right)}^2=1,f^{\prime }\left(x\right)={\left(g\left(x\right)\right)}^2$ and f''(x) and g''(x) exist g(x) < 0, f(0) = 0 then which is true

• A. $g^{\prime }\left(x\right)=f\left(x\right)g\left(x\right)$
• B. $g$ has a relative maximum at $x=0$
• C. $f$ has a point of inflection at $x=0$
• D. $f\left(x\right)$ is not periodic

Answer 1

Answer: (A), (B), (C)

The correct options are
A $g^{\prime }\left(x\right)=f\left(x\right)g\left(x\right)$
B $g$ has a relative maximum at $x=0$
C $f$ has a point of inflection at $x=0$

$f^{\prime }\left(x\right)=\left(g\right(x\left){)}^2\right(g\left(x\right){)}^2-\left(f\right(x){)}^2=1$

Let $f\left(x\right)=y$

$\Rightarrow \frac{dy}{dx}-y^2=1\Rightarrow \frac{dy}{dx}=1+y^2\Rightarrow \int \frac{1}{1+y^2}dy=\int dx\Rightarrow {\tan }^{-1}\left(y\right)=x\Rightarrow y=f\left(x\right)=\tan \left(x\right)\Rightarrow g\left(x\right)=\sqrt{f^{\prime }\left(x\right)}=\sqrt{{\sec }^{2}x}=\pm \sec x$

But $g\left(x\right)<0\Rightarrow g\left(x\right)=-\sec x$

$A)$ $g^{\prime }\left(x\right)=-\sec x\tan x=g\left(x\right)f\left(x\right)$
Hence, $A$ is correct

$B)$ $g^{\prime \prime }\left(x\right)=-(\sec x{\sec }^{2}x+\sec x{\tan }^{2}x)=-\sec x({\sec }^{2}x+{\tan }^{2}x)$

For $x=0\Rightarrow g^{\prime }\left(x\right)=--\sec x\tan x=0g^{\prime \prime }\left(x\right)=-\sec x({\sec }^{2}x+{\tan }^{2}x)=-1<0$

$\Rightarrow g$ has a relative maximum at $x=0$

$C)$ $f^{\prime }\left(x\right)={\sec }^{2}x,f^{\prime \prime }\left(x\right)=2{\sec }^{2}x\tan x{f}^{\prime \prime }\left(0\right)=0$

$\Rightarrow f\left(x\right)$ has a point of inflection at $x=0$

$D)$ $f\left(x\right)=\tan x⟶$ a periodic function

Hence, $g^{\prime }\left(x\right)=f\left(x\right)g\left(x\right),g$ has a relative maximum at $x=0$ and $f$ has a point of inflection at $x=0$