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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity4 Marks
Question
Find the value of a and b so that the function f(x) defind by $\text{f(x)}=\begin{cases}\text{x}+\text{a}\sqrt{2}\sin\text{x},&\text{if }0\leq\text{x}<\frac{\pi}{4}\\2\text{x}\cot\text{ x}+\text{b},&\text{if }\frac{\pi}{4}\leq\text{x}<\frac{\pi}{2}\\\text{a}\cos2\text{x}-\text{b}\sin\text{x},&\text{if }\frac{\pi}{2}\leq\text{x}\leq\pi\end{cases}$ becomes continuous on $[0,\pi]$
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Answer

Given, f is continuous on $[0,\pi]$ 
$\therefore$ f is continuous at $\text{x }\frac{\pi}{4}$ and $\frac{\pi}{2}$
At $\text{x}=\frac{\pi}{4},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\Big(\frac{\pi}{4}-\text{h}\Big)+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}-\text{h}\Big)\Big]\\=\Big[\frac{\pi}{4}+\text{a}\sqrt{2}\sin\Big(\frac{\pi}{4}\Big)\Big]=\Big[\frac{\pi}{4}+\text{a}\Big]$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{4}+\text{h}\Big)\cot\Big(\frac{\pi}{4}+\text{h}\Big)+\text{b}\Big]\\=\Big[\frac{\pi}{2}\cot\Big(\frac{\pi}{4}\Big)+\text{b}\Big]=\Big[\frac{\pi}{2}+\text{b}\Big]$
At $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[2\Big(\frac{\pi}{2}-\text{h}\Big)\cot\Big(\frac{\pi}{2}-\text{h}\Big)+\text{b}\Big]=\text{b}$
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{2}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big[\text{a}\cos2\Big(\frac{\pi}{2}+\text{h}\Big)-\text{b}\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big]=-\text{a}-\text{b}$
Since f is continuous at $\text{x}=\frac{\pi}{4}$ and $\text{x}=\frac{\pi}{2},$ we have
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}$
$\Rightarrow-\text{b}-\text{a}=\text{b}$ and $\frac{\pi}{4}+\text{a}=\frac{\pi}{2}+\text{b}$
$\Rightarrow\text{b}=\frac{-\text{a}}{2}\ ...(\text{i})$ and $\frac{-\pi}{4}=\text{b}-\text{a}\ ...(\text{ii})$
$\Rightarrow\frac{-\pi}{4}=\frac{-3\text{a}}{2}$ [Substituting the value of b in eq. (ii)]
$\Rightarrow\text{a}=\frac{\pi}{6}$
$\Rightarrow\text{b}=\frac{-\pi}{12}$ [From eq. (i)]

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