Question

Let $f\left(x\right)=\{\begin{array}{c}\sin 2x;x\le \frac{\pi }{6}\\ ax+b;\frac{\pi }{6}<x<1\end{array}$.

If $f\left(x\right)$ and $f^{\prime }\left(x\right)$ are continuous, then?

• A. $a=1,b=\frac{1}{\sqrt{2}}+\frac{\pi }{6}$
• B. $a=\frac{1}{\sqrt{2}},b=\frac{1}{\sqrt{2}}$
• C. $a=1,b=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$
• D. none of these

Answer 1

The correct option is C $a=1,b=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

Given $f\left(x\right)$ is continuous

$\Rightarrow {lim}_{x\to {\frac{\pi }{6}}^{-}}f\left(x\right)={lim}_{x\to {\frac{\pi }{6}}^{+}}f\left(x\right)$

${lim}_{x\to {\frac{\pi }{6}}^{-}}\sin 2x={lim}_{x\to {\frac{\pi }{6}}^{+}}(ax+b)$

$\Rightarrow \sin \frac{\pi }{3}=\frac{a\pi }{6}+b=\frac{\sqrt{3}}{2}\to 1$

$f^{\prime }\left(x\right)=\{\begin{array}{c}2\cos 2xx\le \frac{\pi }{6}\\ a\frac{\pi }{6}<x<1\end{array}$

Given $f^{\prime }$ is also continuous

$\Rightarrow {lim}_{x\to {\frac{\pi }{6}}^{-}}{f}^{\prime }\left(x\right)={lim}_{x\to {\frac{\pi }{6}}^{+}}{f}^{\prime }\left(x\right)$

$\Rightarrow 2\cos \frac{\pi }{3}=a$

$\Rightarrow a=1$

Substituting this in $1$

$b=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$