Question

$\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$, then $a=$

• A. $0$
• B. $1$
• C. $5/2$
• D. $-5/2$

Answer 1

The correct option is D $-5/2$

On expanding $\cos x,\sin x$, we get

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

$\Rightarrow x(1+a-b)=0\Rightarrow b=1+a\dots \dots \left(1\right)$

$\Rightarrow \frac{-a}{2!}+\frac{b}{3!}=1\dots \dots \left(2\right)$

From (1) and (2), we get

$\Rightarrow 1+a-3a=6\Rightarrow a=\frac{-5}{2}$