Let $f(x)=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}$. The number of real roots of $f(x)=0$ is

• Correct answer
  A

  0
• B

  1
• C

  2
• D

  4

Solution

$f(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}$ $f(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$ $f^{\prime}(x)=1+x+\frac{x^{2}}{2}>0$ $\Rightarrow f(x)$ is an increasing $f^{\prime \prime}$ $\Rightarrow f(x)=0$ at $x=x_{0}$ $\Rightarrow f\left(x_{0}\right)=0 \Rightarrow 1+x_{0}+\frac{x_{0}^{2}}{2}+\frac{x_{0}^{3}}{6}=0$...(1) $f(-2) f(-1) < 0$ $\Rightarrow x_{0} \in(-2,-1)$ $f\left(x_{0}\right)=1+x_{0}+\frac{x_{0}^{2}}{2}+\frac{x_{0}^{3}}{6}+\frac{x_{0}^{4}}{24}=\frac{x_{0}^{4}}{24}>0$

Hence, the correct answer is option: 'A'.