Question

Let $f\left(x\right)$ be a nonzero function whose all successive derivative exist and are nonzero If $f\left(x\right)$ $f^{\prime }\left(x\right)$and $f"\left(x\right)$ are in G.P. and $f\left(0\right)=1,f^{\prime }\left(0\right)=1$ then

• A. $f^{\prime }\left(x\right)<0\forall \times \in R$
• B. $f"\left(x\right)<1\forall \times \in R$
• C. $f"\left(0\right)\ne f^{\prime \prime \prime }\left(0\right)$
• D. $f"\left(x\right)>0\forall \times \in R$

Answer 1

The correct option is D $f"\left(x\right)>0\forall \times \in R$

${\left(f^{\prime }\left(x\right)\right)}^2=f\left(x\right)f^{\prime \prime }\left(x\right)$
$\Rightarrow {\left(y^{\prime }\right)}^2=y{y}^{\prime \prime }$
$\Rightarrow \int \frac{y^{\prime }}{y}dx=\int \frac{y^{\prime \prime }}{y^{\prime }}dx\Rightarrow lny=ln{y}^{\prime }+C_1$
$f\left(0\right)=1,f^{\prime }\left(0\right)=1\Rightarrow C_1=0$
$\therefore y=y^{\prime }$
$\Rightarrow \int \frac{y^{\prime }}{y}dx=\int 1dx\Rightarrow lny=x+c_2$
$f\left(0\right)=1\Rightarrow C_2=0$
$\therefore y=e^x$
$y^{\prime \prime }>0\forall \times ϵR$