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