Question

Let $f\left(x\right)$ be defined by $f\left(x\right)=\{\begin{array}{c}\sin 2xif0<x\le \pi /6\\ ax+bif\pi /6<x\le 1\end{array}$.

The values of $a$ and $b$ such that $f$ and $f^{\prime }$ are continuous are ?

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

Answer 1

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

For continuity of $f\left(x\right)$

$f\left(\frac{{\pi }^{-}}{6}\right)=f\left({\frac{\pi }{6}}^{+}\right)$

$f\left(x\right){|}_{x=\frac{\pi }{6}}=sin\left(\frac{2\pi }{6}\right)=sin\left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$

$f\left(x\right){|}_{x={\frac{{\pi }^{}}{6}}^{+}}=\frac{a\pi }{6}+b$

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

for $f^{\prime }\left(x\right)$ to be continous $f^{\prime }\left(\frac{{\pi }^{+}}{6}\right)=f^{\prime }\left(\frac{\pi }{6}\right)$

$f^{\prime }\left(x\right){|}_{x=\frac{{\pi }^{+}}{6}}=a$

$f^{\prime }\left(x\right){|}_{x=\frac{\pi }{6}}=2cos2x=1$

$a=1$

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