If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\left\{\begin{array}{ll}\frac{\sin…

If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\left\{\begin{array}{ll}\frac{\sin x-\sin \frac{x}{2}}{x}, & x < 0 \\ \frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{array}\right.$ is continuous on $\mathbb{R}$, then $f(0)=$
  1. $1 / 2$
  2. $3 / 2$
  3. 1
  4. -1

Solution

Since $f(x)$ is also continuous on $\mathbf{R}$ $ \begin{aligned} & \text { Hence } f(0)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{\sin (-h)-\sin \left(-\frac{h}{2}\right)}{(-h)} \\ & \Rightarrow f(0)=\lim _{h \rightarrow 0}\left[\frac{\sin h}{h}-\frac{1}{2} \cdot \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right]=1-\frac{1}{2} \\ & \Rightarrow f(0)=\frac{1}{2} \end{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

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