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Model Paper 3 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 33 Marks
Question
Show that the function $f(x)$ defined by $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, & x>0 \\ 2, & x=0 \text { is continuous at } x =0 . \\ \frac{4(1-\sqrt{1-x})}{x}, & x<0\end{array}\right.$
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Answer

To show that the given function is continuous at $x = 0$ we show that
$(\text{ LHL })_{x=0}=( \text{RHL })_{x=0}=f(0) \ldots (i)$
Here, we have $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, x;0 \\ 2, x=0 \\ \frac{4(1-\sqrt{1-x})}{x}, x<0\end{array}\right.$
Now $ \text{LHL} =\lim _{x \rightarrow 0^{-}} f(x)$
$=\lim _{x \rightarrow 0^{-}} \frac{4(1-\sqrt{1-x})}{x}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1-(0-h)}}{0-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h}]}{-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h]}}{-h} \times \frac{1+\sqrt{1+h}}{1+\sqrt{1+h}}$
$=\lim _{h \rightarrow 0} \frac{4\left[(1)^2-(\sqrt{1+h})^2\right]}{-h[1+\sqrt{1+h]}}$
$=\lim _{h \rightarrow 0} \frac{4[1-(1+h)]}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{-h \times 4}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{4}{1+\sqrt{1+h}}$
$=\frac{4}{1+\sqrt{1}}=\frac{4}{2}=2$
$\text { and RHL }=\lim _{x \rightarrow 0^{+}} f(x)$
$=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}+\cos x\right)$
$\Rightarrow \text { RHL }=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}+\cos h\right)$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h}+\lim _{h \rightarrow 0} \cos h$
$=1+\cos 0$
$=1+1$
$=2$
Also, given that $x=0, f(x)=2 $
$\Rightarrow f(0)=2$
Since, $( \text{LHL} )_{x=0}=( \text{RHL} )_{x=0}=f(0)=2$
Therefore, $f ( x )$ is continuous at $x =0$.

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