Question

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

• A. $0$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $0$

For $f\left(x\right)$ to be continuous at $x=0$
$L.H.L.=R.H.L.=f\left(0\right)=b$

$R.H.L.$
$=\underset{x\to 0^{+}}{lim}\frac{(x+3x^2{)}^{1/3}-x^{1/3}}{x^{4/3}}$

$=\underset{h\to 0}{lim}\frac{h^{1/3}(1+3h{)}^{1/3}-h^{1/3}}{h^{1/3}\cdot h}$

$=\underset{h\to 0}{lim}\frac{(1+3h{)}^{1/3}-1}{h}$ $\left(\frac{0}{0}\right)$ Form
Using L- hospital's rule, we get
$\Rightarrow \underset{h\to 0}{lim}\frac{1}{3}\times 3\times {(1+3h)}^{-2/3}=1$
$\therefore b=1$

$L.H.L.$
$=\underset{x\to 0^{-}}{lim}\frac{(a+2)\cdot \sin (a+2)x}{(a+2)x}+\underset{x\to 0^{-}}{lim}\frac{\sin x}{x}$
$=\underset{h\to 0}{lim}\frac{(a+2)\cdot \sin (a+2)h}{(a+2)h}+\underset{h\to 0}{lim}\frac{\sin h}{h}$
$\therefore L.H.L=a+2+1$
$\Rightarrow a+2+1=1\Rightarrow a=-2$
$\therefore a+2b=0$