Question

If $f\left(x\right)=\sum _{r=1}^{\infty }\frac{x^r}{(r-1)!}$, then which of the following is/are correct ?

• A. $n^{th}$ derivative of $f$ vanishes when $x=-n$
• B. $n^{th}$ derivative of $f$ vanishes when $x=n$
• C. $f$ is neither even nor odd
• D. $f$ is one-one for $[1,2]$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $n^{th}$ derivative of $f$ vanishes when $x=-n$
C $f$ is neither even nor odd
D $f$ is one-one for $[1,2]$

$f\left(x\right)=\sum _{n=1}^{\infty }\frac{x^n}{(n-1)!}$
$f\left(x\right)=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+...$
$\Rightarrow f\left(x\right)=x{e}^x$
$f(-x)=\frac{-x}{e^x}$
$\Rightarrow f\left(x\right)$ is neither even nor odd.

$f^{\prime }\left(x\right)=e^x+x{e}^x>0x\ge 0$
So $f\left(x\right)$ is one- one for $[1,2]$

$f^{\prime }\left(x\right)=e^x+x{e}^x$
$f^{\prime \prime }\left(x\right)=2e^x+x{e}^x$
$f^{\prime \prime \prime }\left(x\right)=3e^x+x{e}^x$
$\Rightarrow f^n\left(x\right)=n{e}^x+x{e}^x$
$f^n\left(x\right)=0$ gives $x=-n$