Let f:RR be a function defined as f(x)=cases(a+1)x+2x2x, & x0…

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be a function defined as $f(x)=\begin{cases}\dfrac{\sin(a+1)x+\sin2x}{2x}, & x<0\b, & x=0\\dfrac{\sqrt{x+bx^3}-\sqrt{x}}{bx^{5/2}}, & x>0\end{cases}$. If $f$ is continuous at $x=0$, then the value of $a+b$ is equal to

Accepted Answer

LHL: $\displaystyle\lim_{x	o0^-}f(x)=\frac{a+1}{2}+1$.

RHL: $\displaystyle\lim_{x	o0^+}\frac{x+bx^3-x}{bx^{5/2}(\sqrt{x+bx^3}+\sqrt x)}=\lim_{x	o0^+}\frac{\sqrt x}{\sqrt x\left(\sqrt{1+bx^2}+1ight)}=\frac12$.

$f(0)=b$. So $\dfrac12=b=\dfrac{a+1}{2}+1\Rightarrow b=\dfrac12,\ a=-2$. Hence $a+b=-\dfrac32$.

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