Question

If $f\left(x\right)=\left\{\begin{array}{rcl}\frac{\sin (a+2)x+\sin x}{x},x& <& 0\\ b,x& =& 0\\ \frac{{(x+3x^2)}^{{\displaystyle \frac{1}{3}}}-x^{{\displaystyle \frac{1}{3}}}}{x^{{\displaystyle \frac{4}{3}}}},x& >& 0\end{array}\right.$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

$0$

Explanation for the correct option:

FInd the value of $a+2b$.

$f\left(x\right)$ is continuous at$x=0$

$li{m}_{x\to 0}f\left(x\right)=b$

Now,

$\begin{array}{rcl}b& =& li{m}_{h\to 0}f(0+h)\\ & =& li{m}_{h\to 0}\frac{{\left(h+3h^2\right)}^{{\displaystyle \frac{1}{3}}}-h^{{\displaystyle \frac{1}{3}}}}{h^{{\displaystyle \frac{4}{3}}}}\\ & =& li{m}_{h\to 0}\frac{h^{{\displaystyle \frac{1}{3}}}{(1+3h)}^{{\displaystyle \frac{1}{3}}}-1}{h^{{\displaystyle \frac{4}{3}}}}\\ & =& li{m}_{h\to 0}\frac{{(1+3h)}^{{\displaystyle \frac{1}{3}}}-1}{h}\\ & =& li{m}_{h\to 0}\frac{1}{3}{(1+3h)}^{\frac{-2}{3}}·3\end{array}$

Or

$b=1$

We can write,

$$\begin{gathered} li{m}_{x\to 0}f\left(x\right)=1 \\ \Rightarrow li{m}_{h\to 0}\frac{\sin \left(\right(a+2\left)\right(-h\left)\right)+\sin (-h)}{h}=1 \\ \Rightarrow a+3=1 \\ \Rightarrow a=-2 \\ \Rightarrow a+2b=0 \end{gathered}$$

Hence, the correct option is C.