Question

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

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

$=\frac{(2^1-1)x}{1!}+\frac{(2^2-1)x^2}{2!}+\frac{(2^3-1)x^3}{3!}+\frac{(2^4-1)x^4}{4!}+\dots$

$=\frac{2x}{1!}+\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^4}{4!}+\cdots -(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots )$

$=1+\frac{2x}{1!}+\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^4}{4!}+\cdots -(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots )$

$\Rightarrow f\left(x\right)=e^{2x}-e^x$

When we put $x=0$, we get $f\left(0\right)=e^0-e^0=1-1=0$

Hence, exactly one real solution exists.