Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (\alpha +2)x+\sin x}{x},& x<0\\ b,& x=0\\ \frac{(x+3x^2{)}^{\frac{1}{3}}-x^{\frac{1}{3}}}{x^{\frac{4}{3}}},& x>0\end{array}$ is continuous at $x=0$ then $a+2b$ is equal to :

• A. $-2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

$f\left(x\right)$ is continuous at $x=0$
$\therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=b=\underset{x\to 0^{+}}{lim}f\left(x\right)b=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}\frac{(h+3h^2{)}^{\frac{1}{3}}-h^{\frac{1}{3}}}{h^{\frac{4}{3}}}$
$\Rightarrow b=\underset{h\to 0}{lim}\frac{(1+3h{)}^{\frac{1}{3}}-1}{h}$
$\Rightarrow b=\underset{h\to 0}{lim}\frac{1}{3}(1+3h{)}^{-\frac{2}{3}}\times 3$
$\Rightarrow b=1$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
$\Rightarrow \underset{h\to 0}{lim}\frac{\sin \left(\right(\alpha +2\left)\right(-h\left)\right)+\sin (-h)}{-h}=1$
$\Rightarrow a+3=1\Rightarrow a=-2$
$\Rightarrow a+2b=0$