Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

Given $f\left(x\right)=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}$

Clearly, $f\left(x\right)$ is continuously differentiable.

Let us use the fact that, between any two roots of $f^{\prime }\left(x\right)$, we can have at most one root of $f\left(x\right)$.--------(A)

$f^{\prime }\left(x\right)$ is given by $1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}$.

$f^{\prime \prime }\left(x\right)=1+x+\frac{x^2}{2}$

$f^{\prime \prime }\left(x\right)$ is a quadratic in $x$. Discriminant is $1-4\times \frac{1}{2}=-1<0$

Thus, $f^{\prime \prime }\left(x\right)$ has no real roots.

Clearly, using (A), $f^{\prime }\left(x\right)$ has at most 1 real root. But, $f^{\prime }\left(x\right)$ is a cubic polynomial.

So, $f^{\prime }\left(x\right)$ has exactly 1 real root. Let that root be $\zeta$

i.e., $1+\frac{\zeta }{1!}+\frac{{\zeta }^2}{2!}+\frac{{\zeta }^3}{3!}=0$ ----------(1)

Since $f^{\prime }\left(x\right)$ has only 1 real root, $f\left(x\right)$ can have at most 2 real roots.

The location of the possible roots is given by $(-\infty ,\zeta )$ and $(\zeta ,\infty )$

But, observe that as $x\to \pm \infty ,f\left(x\right)\to +\infty$

and $f\left(\zeta \right)=1+\frac{\zeta }{1!}+\frac{{\zeta }^2}{2!}+\frac{{\zeta }^3}{3!}+\frac{{\zeta }^4}{4!}=\frac{{\zeta }^4}{4!}>0$ (Using 1)

Thus, as both $f(\pm \infty )$ and $f\left(\zeta \right)$ are positive, the number of real roots of $f\left(x\right)$ is $0$