Question

The values of $a,b$ and $c$ which make the function $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (a+1)x+\sin x}{x}& ,x<0\\ c& ,x=0\\ & \\ \frac{\sqrt{x+b{x}^2}-\sqrt{x}}{x^{3/2}}& ,x>0\end{array}$
continuous at $x=0$ are

• A. $a=-\frac{3}{2},c=\frac{1}{2},b=0$
• B. $a=\frac{3}{2},c=\frac{1}{2},b\ne 0$
• C. $a=-\frac{3}{2},c=\frac{1}{2},b\ne 0$
• D. None of the above

Answer 1

The correct option is C $a=-\frac{3}{2},c=\frac{1}{2},b\ne 0$

Here,$f\left(x\right)=\frac{\sin (a+1)x+\sin x}{x},x<0$
$=c,x=0$
$=\frac{\sqrt{x+b{x}^2}-\sqrt{x}}{x^{\frac{3}{2}}},x>0$
RHS at $x=0$
$\Rightarrow {lim}_{x\to 0}\frac{\sqrt{x+b{x}^2}-\sqrt{x}}{x\sqrt{x}}$
$\Rightarrow {lim}_{x\to 0}\frac{\sqrt{bx+1}-1}{x}\times \frac{\sqrt{bx+1}+1}{\sqrt{bx+1}+1}$
$\Rightarrow {lim}_{x\to 0}\frac{1+bx-1}{\sqrt{bx+1}+1}$
$\Rightarrow \frac{b}{2}⟶\left(i\right)$
$LHSatx=0$
${lim}_{x\to 0}\frac{\sin (a+1)x+\sin x}{x}$
$\Rightarrow {lim}_{x\to 0}\frac{\sin (a+1)x}{x}+\frac{\sin x}{x}$
$(a+1)+1$
$\Rightarrow a+2⟶\left(ii\right)$
$f\left(0\right)=c⟶\left(iii\right)$
So, $a=\frac{b-4}{2}$$c=\frac{b}{2}$
For $b=1\Rightarrow (b\ne 0)$
$a=\frac{-3}{2}$
$c=\frac{1}{2}$