Question

If $f\left(x\right)=e^x-e^{-x}-2\sin x-\frac{2}{3}{x}^3,$ then the least value of $n$ for which $\frac{d^n}{d{x}^n}f\left(x\right)$ at $x=0$ is non-zero is ?

• A. $2$
• B. $1$
• C. $7$
• D. either $1$ or $2$

Answer 1

The correct option is C $7$

$f\left(x\right)=e^x-e^{-x}-2\sin x-\frac{2}{3}{x}^3$

Now, $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\infty$

$\Rightarrow e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\infty$

$\Rightarrow f\left(x\right)=2[x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}...]-2[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...]-\frac{2}{3}{x}^3$

$\therefore f\left(x\right)=4(\frac{x^7}{7!}+\frac{x^{11}}{11!}+\frac{x^{15}}{15!}+...\infty )$

$\Rightarrow f^{\prime }\left(x\right)=4(\frac{7\times x^6}{7!}+\frac{11\times x^{10}}{11!}+...)$

$\Rightarrow f^{\prime \prime }\left(x\right)=4(\frac{6\times x^5}{6!}+\frac{10\times x^9}{10!}+...)$

$.....$

$......$

$\therefore f^7\left(x\right)=4(x^0+\frac{x^4}{4!}+\frac{x^8}{8!}+...)$

$\Rightarrow f^{\prime }\left(0\right)=4$ i.e., a non-zero constant.

Hence at $x=0,$ it will be non-zero

$\therefore n=7$