Question

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

The ordered triplet $(a,b,c)$ is

• A. $(\frac{1}{2},-\frac{1}{2},-1)$
• B. $(-\frac{1}{2},\frac{1}{2},1)$
• C. $(-\frac{1}{2},\frac{1}{2},2)$
• D. $(\frac{1}{2},-\frac{1}{2},-2)$

Answer 1

The correct option is D $(\frac{1}{2},-\frac{1}{2},-2)$

${lim}_{x\to 0}\frac{\sin x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^2}=1{lim}_{x\to 0}\frac{(x-\frac{x^3}{31})+a(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!})}{x^2}+\frac{b(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!})+c(x-\frac{x^2}{2}+\frac{x^3}{3})}{x^2}=1{lim}_{x\to 0}\frac{(a+b)+(1+a-b+c)x+(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})x^2}{x^2}+\frac{(-\frac{1}{3!}+\frac{a}{3!}-\frac{b}{3!}+\frac{c}{3})x^3}{x^2}=1$
When
$a+b=0,1+a-b+c=0,-\frac{1}{3!}+\frac{a}{3!}-\frac{b}{3!}+\frac{c}{3}=0$ and $\frac{a}{2}+\frac{b}{2}-\frac{c}{2}=1$
Gives
$a=\frac{1}{2},b=-\frac{1}{2},c=-2$