Question

Let $f\left(x\right)=\{\begin{array}{c}\sin 2x;0<x\le \frac{\pi }{6}\\ ax+b;\frac{\pi }{6}<x<1\end{array}$
Then the value of $a$ and $b$ such that $f$ and $f^{\prime }$ are continuous, are

• A. $a=1,b=\frac{1}{\sqrt{2}}+\frac{\pi }{6}$
• B. $a=\frac{1}{\sqrt{2}}=b$
• C. $a=1,b=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$
• D. $a=1,b=\frac{1}{2}-\frac{\pi }{6}$

Answer 1

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

For $0<x\le \frac{\pi }{6}$

$f\left(x\right)=\sin 2x$ and $f^{\prime }\left(x\right)=2\cos 2x$

For $\frac{\pi }{6}<x<1$

$f\left(x\right)=ax+b$ and $f^{\prime }\left(x\right)=a$

For $f$ to be continuous $f\left(\pi /6\right)=\sin \left(\frac{\pi }{3}\right)=a\times \frac{\pi }{6}+b$

For $f^{\prime }$ to be continuous $f^{\prime }\left(\pi /6\right)=2\cos \left(\frac{\pi }{3}\right)=a$

$\therefore a=1$ & $b=\frac{\sqrt{3}}{2}-\left(\frac{\pi }{6}\right)$