Question

Let ${lim}_{x\to 0}\frac{\sin x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^2}=1$

The roots of the quadratic equation $a{x}^2+bx+c=0$ are

• A. both positive
• B. both negative
• C. one positive and one negative
• D. complex conjugate

Answer 1

Answer: (A)

both positive

$L=\underset{x\to 0}{lim}\frac{\sin x+a{e}^x+b{e}^{-x}+cl(1+x)}{x^3}$

$=\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{31})+a(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})}{x^3}$

$\frac{+b(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!})+c(x-\frac{x^2}{2}+\frac{x^3}{3})}{x^3}$

$=\underset{x\to 0}{lim}\frac{(a+b)+(1+a-b+c)x+(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})x^2}{x^3}\frac{+(-\frac{1}{3!}+\frac{a}{3!}-\frac{b}{3!}+\frac{c}{3})x^3}{x^3}$

or $a+b=0,1+a-b+c=0,\frac{a}{2}+\frac{b}{2}-\frac{c}{2}=0$

and $L=-\frac{1}{3!}+\frac{a}{3!}-\frac{a}{3!}+\frac{c}{3}$

Solving the first three equations, we get $c=0,a=-1/2,b=1/2$

Then, $L=-1/3.$

The given equation $a{x}^2+bx+c=0$ reduces to $x^2-x=0$ or $x=0,1.$