Question

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

$f\left(x\right)=\left\{\begin{array}{l}\frac{\sin (a+1)x+2\sin 2x}{2x},x<0\\ b,x=0\\ \frac{\sqrt{x+b{x}^3-\sqrt{x}}}{b{x}^{{\displaystyle \frac{5}{2}}}},x>0\end{array}\right.$

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

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

Answer 1

The correct option is C

$-\frac{3}{2}$

Explanation for the correct option:

Determining the value of $a+b$.

Given that: $f$ is continuous at $x=0$.

It means: $f\left(0^{-}\right)=f\left(0\right)=f\left(0^{+}\right)$

Therefore we calculate $f\left(0^{-}\right):$

$$\begin{gathered} f\left(0^{-}\right)=\underset{x\to 0^{-}}{\lim }\frac{\left[\sin (a+1)x+\sin 2x\right]}{2x} \\ =\frac{1}{2}\underset{x\to 0^{-}}{\lim }\frac{\left[\sin (a+1)x.(a+1)\right]}{x.(a+1)}+\underset{x\to 0^{-}}{\lim }\frac{\sin 2x}{2x} \\ =\frac{1}{2}.(a+1)+1........\left(I\right) \end{gathered}$$

Again, calculating $f\left(0\right):$

$f\left(0\right)=b.....\left(II\right)$

Calculating $f\left(0^{+}\right):$

$$\begin{gathered} f\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{{\displaystyle \frac{5}{2}}}} \\ =\underset{x\to 0^{+}}{\lim }\frac{x+b{x}^3-x}{b{x}^{{\displaystyle \frac{5}{2}}}.\left(\sqrt{x+b{x}^3}+\sqrt{x}\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{b{x}^3}{b{x}^{\frac{5}{2}}.\left(\sqrt{x+b{x}^3}+\sqrt{x}\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{\left(\sqrt{x+b{x}^3}+\sqrt{x}\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b{x}^2}+1\right)} \\ =\frac{1}{2} \\ f\left(0^{+}\right)=\frac{1}{2} \end{gathered}$$

Equating the three equations we get:

$$\begin{gathered} \frac{1}{2}(a+1)+1=\frac{1}{2}andb=\frac{1}{2} \\ \therefore b=\frac{1}{2}anda=-2 \\ \therefore a+b=-\frac{3}{2} \end{gathered}$$

Hence, the correct answer is option (C).