Question

$f\left(x\right)=\{\begin{array}{cc}\frac{(x+b{x}^2{)}^{1{/}_2}-x^{1/2}}{b{x}^{3/2}}& x>0\\ c& x=0\\ \frac{\sin (a+1)x+\sin x}{x}& x<0\end{array}$ is continuous at $x=0$, then

• A. $a=\frac{-3}{2},b=0,c=\frac{1}{2}$
• B. $a=\frac{-3}{2},b\ne 0,c=\frac{1}{2}$
• C. $a=\frac{3}{2},b\ne 0,c=\frac{1}{2}$
• D. $a=\frac{3}{2},b\ne 0,c=-\frac{1}{2}$

Answer 1

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

Given: $f\left(x\right)=\{\begin{array}{cc}\frac{(x+b{x}^2{)}^{1{/}_2}-x^{1/2}}{b{x}^{3/2}}& x>0\\ c& x=0\\ \frac{\sin (a+1)x+\sin x}{x}& x<0\end{array}$ is continuous at $x=0$

Multiply and divide by $\left(\right(x+b{x}^2{)}^{1{/}_2}+x^{1/2})$

$\underset{x\to 0^{+}}{lim}=\underset{x\to 0}{lim}\frac{(x+b{x}^2)+\left(x\right)}{b{x}^{\frac{3}{2}}\left(\right(x+b{x}^2{)}^{\frac{1}{2}}+x^{\frac{1}{2}})}$
$=\underset{x\to 0}{lim}\frac{x^{\frac{1}{2}}}{(x+b{x}^2{)}^{\frac{1}{2}}+x^{\frac{1}{2}}}$

$=\underset{x\to 0}{lim}\frac{1}{(1+bx)\frac{1}{2}+1}=\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{(a+1)sin(a+1)x}{(a+1)x}+\frac{sinx}{x}$

$(a+1)+1=\frac{1}{2}$

$a=\frac{-3}{2}$

$c=\frac{1}{2}$ (By continuity)