Question

Let $f:R\to R$ be a function defined as

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin (a+1)x+\sin 2x}{2x},& \text{if}x<0\\ b,& \text{if}x=0\\ \frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{5/2}},& \text{if}x>0\end{array}$

If $f$ is continuous at $x=0,$ then the value of $a+b$ is equal

• A. $-2$
• B. $-\frac{5}{2}$
• C. $-\frac{3}{2}$
• D. $-3$

Answer 1

The correct option is C $-\frac{3}{2}$

Given : $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (a+1)x+\sin 2x}{2x},& \text{if}x<0\\ b,& \text{if}x=0\\ \frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{5/2}},& \text{if}x>0\end{array}$

$f$ is continuous at $x=0$, then
$f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right)$
Now,
$f\left(0^{-}\right)=\underset{x\to 0^{-}}{lim}\frac{\sin (a+1)x+\sin 2x}{2x}\Rightarrow f\left(0^{-}\right)=\frac{a+1}{2}+1=\frac{a+3}{2}\cdots \left(1\right)$

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{5/2}}\Rightarrow f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{b{x}^3}{b{x}^{5/2}\cdot (\sqrt{x+b{x}^3}+\sqrt{x})}\Rightarrow f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{b}{b\cdot (\sqrt{1+b{x}^2}+1)}\Rightarrow f\left(0^{+}\right)=\frac{1}{2}\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$, we get
$a=-2,b=\frac{1}{2}\therefore a+b=-\frac{3}{2}$