Question

Let (x) be a function such that ${lim}_{x\to 0}\frac{f\left(x\right)}{x}=1andif{lim}_{x\to 0}\frac{x(1+acosx)-bsinx}{\left(f\right(x{)}^3)}=1then\mid (a+b)\mid =$_____________

Answer 1

$L=\underset{x\to 0}{lim}\frac{x+\cos 2x-b\sin x}{x^3}$

Apply $L^{\prime }$ Rule then,

$\underset{x\to 0}{lim}\frac{1+a\cos -a\sin x-b\cos x}{x^3}$

$\therefore 1+a-b=0$ in $LR-----\left(1\right)$

$\underset{x\to 0}{lim}\frac{-a\cos x-a\cos x-a\cos 2x+ax\sin x+b\cos x}{6}$ (By $L^{\prime }$ orbital Rule)

$=1\Rightarrow -3a+b=6-----\left(II\right)$

Hence, equation $\left(I\right)$ & $\left(II\right)$

$\frac{1+a-b=0-3a+b=6}{1-2a=6-2a=5a=\frac{-5}{2}}$

$1+(-\frac{5}{2})+b=0\Rightarrow 1-\frac{5}{2}+b=0$

$\frac{2-5}{2}+b=0\Rightarrow \frac{-3}{2}+b=0\Rightarrow b=\frac{3}{2}$

Now $a+b=\frac{-5}{2}+\frac{3}{2}=\frac{-2}{2}=1$

$\therefore a+b=1$