Question

If the function $f\left(x\right)$ is continuous at $x=0,$ where $f\left(x\right)=\{\begin{array}{cc}\frac{\sin ((a+1)x)+\sin x}{x};& x<0\\ c;& x=0\\ \frac{(x+b{x}^2{)}^{1/2}-x^{1/2}}{b{x}^{3/2}};& x>0\end{array},$
then the possible values of $a,b,c$ can be

• A. $a=-\frac{1}{2}$
• B. $b=-\frac{3}{2}$
• C. $b=1$
• D. $c=\frac{1}{2}$

Answer 1

Answer: (B), (C), (D)

$c=\frac{1}{2}$

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[\right(a+1\left)x\right]+\sin x}{x};& x<0\\ c;& x=0\\ \frac{(x+b{x}^2{)}^{1/2}-x^{1/2}}{b{x}^{3/2}};& x>0\end{array}$

At $x=0,f\left(0\right)=c$

$f\left(0^{-}\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\frac{\sin \left[\right(a+1\left)\right(0-h\left)\right]+\sin (0-h)}{0-h}=\underset{h\to 0}{lim}\frac{-\sin \left[\right(a+1\left)h\right]-\sin h}{-h}=\underset{h\to 0}{lim}\frac{\sin \left[\right(a+1\left)h\right]+\sin h}{h}$
Using L'Hospital's Rule, we get
$f\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{\cos \left[\right(a+1\left)h\right]\cdot (a+1)+\cos h}{1}\Rightarrow f\left(0^{-}\right)=a+2$

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}\frac{(h+b{h}^2{)}^{1/2}-h^{1/2}}{b{h}^{3/2}}=\underset{h\to 0}{lim}\frac{(1+bh{)}^{1/2}-1}{bh}$
Using L'Hospital's Rule, we get
$f\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{\frac{b}{2\sqrt{1+bh}}}{b}=\frac{1}{2}$

$\because f\left(x\right)$ is continuous at $x=0$, so
$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)\Rightarrow a+2=\frac{1}{2}=c\therefore a=-\frac{3}{2},b\in R,c=\frac{1}{2}$