Question

$\underset{x\to 0}{lim}\frac{x(a+b\cos x)-c\sin x}{x^5}=1$
Find the value of a,b and c.

• A. $a=80,b=50,c=150$
• B. $a=100,b=800,c=140$
• C. $a=120,b=60,c=180$
• D. $a=135,b=70,c=150$

Answer 1

The correct option is C $a=120,b=60,c=180$

We know the expansion of

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....$

$\frac{x(a+b\cos x)-c\sin x}{x^5}$

$=\frac{x(a+b\{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....\})-c\{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+....\}}{x^5}$

By grouping the coefficients of like powers,

$=\frac{(a+b-c)x+(-\frac{b}{2!}+\frac{c}{3!})x^3+(\frac{b}{4!}-\frac{c}{5!})x^5+(-\frac{b}{6!}+\frac{c}{7!})x^7+....}{x^5}$

We have ${lim}_{x\to 0}\frac{x(a+b\cos x)-c\sin x}{x^5}=1$

$\Rightarrow a+b-c=0$ .........$\left(1\right)$

$\Rightarrow -\frac{b}{2!}+\frac{c}{3!}=0$

$\Rightarrow -3b+c=0$ .........$\left(2\right)$

$\Rightarrow \frac{b}{4!}-\frac{c}{5!}=1$

$\Rightarrow 5b-c=120$ .........$\left(3\right)$

$\Rightarrow -\frac{b}{6!}+\frac{c}{7!}=0$

$\Rightarrow -7b+c=0$ .........$\left(4\right)$

Adding $\left(2\right)$ and $\left(3\right)$ we get

$-3b+c+5b-c=120+0=120$

$\Rightarrow 2b=120$

$\Rightarrow b=\frac{120}{2}=60$

From $\left(3\right)$ we have $5b-c=120$

$\Rightarrow 5\times 60-c=120$

$\Rightarrow -c=120-300=-180$

$\therefore c=180$

Put $b=60,c=180$ in $\left(1\right)$ we get

$a+b-c=0$

$\Rightarrow a+60-180=0$

$\Rightarrow a-120=0$

$\Rightarrow a=120$

$\therefore a=120,b=60$ and $c=180$