Question

The values of $a$ and $b$ such that ${lim}_{x\leftarrow 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$ are

• A. $\frac{5}{2},\frac{3}{2}$
• B. $\frac{5}{2},-\frac{3}{2}$
• C. $-\frac{5}{2},-\frac{3}{2}$
• D. $-\frac{5}{2},\frac{3}{2}$

Answer 1

Answer: (C)

$-\frac{5}{2},-\frac{3}{2}$

$\underset{x\to 0}{lim}(\frac{x+ax\cos x}{x^3}-\frac{b\sin x}{x^3})$

$\frac{0}{0}$ from $\therefore L$Hospital rule

$\underset{x\to 0}{lim}(\frac{1+ax\cos x}{x^2}-\frac{b\sin x}{x^3})$

$\underset{x\to 0}{lim}\frac{0+a(-\sin x)}{2x}-\frac{b\cos x}{3x^2}$

$=\underset{x\to 0}{lim}\frac{-a}{2}\left(\frac{\sin x}{x}\right)-\underset{x\to 0}{lim}\frac{b(-\sin x)}{6x}$

$=\frac{-a}{2}\left(1\right)-\frac{b}{6}(-1)$

$=\frac{b}{6}-\frac{a}{2}=1$ (given)

$\therefore a=\frac{-5}{2},b=\frac{-3}{2}$

$\therefore c$ option is correct