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 $f(x)=0$ has

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

Answer 1

Answer: (B)

Given,
$f(x)=\frac{x}{1!}+\frac{3}{2!}{x}^2+\frac{7}{3!}{x}^3+\frac{15}{4!}{x}^4+\dots$
$=\frac{\left(2^1-1\right)x}{1!}+\frac{\left(2^2-1\right)x^2}{2!}+\frac{\left(2^3-1\right)x^3}{,3!}+\frac{\left(2^4-1\right)}{4!}{x}^4+\dots$
$=\frac{2x}{1!}+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+\frac{(2x{)}^4}{4!}+\dots$
$=\left(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \right)$
$=1+\frac{2x}{1!}+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+\frac{(2x{)}^4}{4!}+\dots$
$-\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \right)$
$\Rightarrow f(x)=e^{2x}-e^x$
When we put $x=0$, we get
$f(0)=e^0-e^0=1-1=0$
Hence, exactly one real solution exists.