If the function $f: R ightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin x}{x}, & x

Question

If the function $f: R ightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin x}{x}, & x < 0 \ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}}, & x=0 	ext { is }\end{array}ight.$ continuous on $[R$, then $a+b=$

Aicharya · Accepted Answer

Given function $ f(x)=\left\{\begin{array}{cc} \frac{\sin (a+1) x+\sin x}{x} & , x < 0 \\ b & , x=0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0 \end{array}\right. $ Function is continuous so, $ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) $

Now, $\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x+\sin x}{x}\right) \\ = & \lim _{x \rightarrow 0^{-}}\left(\frac{\sin (a+1) x \times(a+1)}{(a+1) x}+\frac{\sin x}{x}\right) \\ = & (a+1)+1 \\ & {\left[\because \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}=1\right] }\end{aligned}$

Now, $\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0^{+}}+\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} \\ = & \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}(\sqrt{1+x}-1) \times \sqrt{1+x}+1}{x \sqrt{x} \times(\sqrt{1+x}+1)}\end{aligned}$

$ b=\frac{1}{2} $ put in Eq. (iii) $ \begin{aligned} a+2 & =\frac{1}{2} \\ a & =\frac{-3}{2} \end{aligned} $ Now, $ \begin{aligned} & a+b=\frac{-3}{2}+\frac{1}{2} \\ & a+b=-1 \end{aligned} $

If the function $f: R \rightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin x}{x},…

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