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  • Let      <img alt="\mathrm{f(x)=\left\{\begin{array}{l}\sin 2 x, 0<x \leq \pi / 6 \\ a x+b, \pi / 6<x<1\end{array}\right.}" src="https://entrancecorner.oncodecogs.com/g

Let      \mathrm{f(x)=\left\{\begin{array}{l}\sin 2 x, 0<x \leq \pi / 6 \\ a x+b, \pi / 6<x<1\end{array}\right.}  

If f(x) and f(x) are continuous, then

Option: 1

\mathrm{a=1, b=\frac{1}{\sqrt{2}}+\frac{\pi}{6}}


Option: 2

\mathrm{a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{2}}}


Option: 3

\mathrm{a=1, b=\frac{\sqrt{3}}{2}-\frac{\pi}{6}}


Option: 4

none of these


Answers (1)

best_answer

Clearly, f(x) is continuous for all x except possibly at  \mathrm{x=\pi / 6}  For f(x) to be continuous at  \mathrm{x=\pi / 6} , we must have

\mathrm{\mathrm{\begin{aligned} & & \lim _{x \rightarrow \pi / 6^{-}} f(x) & =\lim _{x \rightarrow \pi / 6^{+}} f(x) \\ \Rightarrow & & \lim _{x \rightarrow \pi / 6} \sin 2 x & =\lim _{x \rightarrow \pi / 6} a x+b \\ \Rightarrow & & \sin (\pi / 3) & =(\pi / 6) a+b \\ \Rightarrow & & \frac{\sqrt{3}}{2} & =\frac{\pi}{6} a+b \end{aligned}}}.....................(i)

For f(x) to be differentiable at \mathrm{ x=\pi / 6} we must have
\mathrm{ \text { (LHD at } x=\pi / 6)=(\text { RHID at } x=\pi / 6) \\ }

\mathrm{ \Rightarrow \quad \lim _{x \rightarrow \pi 6} 2 \cos 2 x=\lim _{x \rightarrow \pi / 6} \\ }

\mathrm{ \Rightarrow \quad 2 \cos \pi / 3=a \Rightarrow a=1 \text {. } \\ }

\mathrm{ \text { Putting } a=1 \text { in (i), we get } b=(\sqrt{3} / 2)-\pi / 6 \text {. } \\\\ }
 

Posted by

Ritika Kankaria

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