Question

For every real number $x$, let $f\left(x\right)=\frac{x}{1!}+\frac{3}{2!}{x}^2+\frac{7}{3!}{x}^3+\frac{15}{4!}{x}^4+.....$. Then the equation $f\left(x\right)=0$ has

• A. no real solution
• B. exactly one real solution
• C. exactly two real solutions
• D. infinite number of real solutions

Answer 1

The correct option is B exactly one real solution

$f\left(x\right)=\frac{x}{1!}+\frac{3}{2!}{x}^2+\frac{7}{3!}{x}^3+\frac{15}{4!}{x}^4......\infty$

We can write it as

$\Rightarrow$ $x\times (\frac{1}{1!}+\frac{3}{2!}x+\frac{7}{3!}{x}^2+\frac{15}{4!}{x}^3......\infty )=0$

$\Rightarrow$ one real root is $x=0$

We will get other roots from $\frac{1}{1!}+\frac{3}{2!}x+\frac{7}{3!}{x}^2+\frac{15}{4!}....\infty =0$

But we can see that the sum of the roots $=\frac{\infty }{\infty }$

which is not possible, so the remaining roots of the equation are imaginary

Hence, only one real root $x=0$ exist.