Answer:
Option D
Explanation:
Here, $\lim_{x \rightarrow0}\frac{x^{2}\sin (\beta x)}{\alpha x-\sin x}=1$
$\Rightarrow$ $\lim_{x \rightarrow 0}\frac{x^{2}\left[ \beta x- \frac{(\beta x)^{3}}{3!}+\frac{(\beta x)^{5}}{5!}-.....\right]}{\alpha x-\left[ x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-......\right]}$ =1
$\Rightarrow$ $\lim_{x \rightarrow 0}\frac{x^{3}\left[\beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}-.....\right]}{(\alpha-1)x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....}=1$
Limits exist only,
when , $\alpha -1=0$
$\Rightarrow$ $\alpha =1$ ............(i)
$\therefore$ $\lim_{x \rightarrow 0}\frac{x^{3}\left[ \beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}- .....\right]}{x^{3\left[\frac{1}{3!}-\frac{x^{2}}{5!}- ....\right]}}=1$
$\Rightarrow$ $6\beta=1$ ........(ii)
From Eq .(i) and (ii), we get $6(\alpha+\beta)=6\alpha+6\beta$
=6+1= 7
