Question

Determine the values of $a,b,c$ for which the function $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (a+1)x+\sin x}{x}& forx<0\\ =c& forx=0\\ =\frac{{(x+b{x}^2)}^{1/2}-x^{1/2}}{b{x}^{3/2}}& forx>0\end{array}$ is continuous at $x=0$.

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

Answer 1

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

The correct options are
B $a=\frac{-3}{2}$
C $c=\frac{1}{2}$
D $b$ can take any value

$f\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{\sin (a+1)(0-h)+\sin (0-h)}{-h}$ $=\underset{h\to 0}{lim}\frac{\sin (a+1)h+\sin h}{h}$ $=\frac{(a+1)h+h}{h}=a+2,[\because \underset{\theta \to 0}{lim}\sin \theta =\theta ]$
and $f\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{{(h+b{h}^2)}^{1/2}-h^{1/2}}{b{h}^{3/2}}$ $=\underset{h\to 0}{lim}\frac{{(1+bh)}^{1/2}-1}{bh}$
$=\underset{h\to 0}{lim}\frac{1+\frac{1}{2}bh...-1}{bh}=\frac{1}{2},$ which is independent of b and so b may have any real value.Again by continuity at $x=0$, we have $a+2=\frac{1}{2}=c.$ $\therefore a=-\frac{3}{2}$ and $c=\frac{1}{2}.$