Question

Let $f\left(x\right)$ be defined by $f\left(x\right)=\{\begin{array}{cc}\sin 2x& \text{if}0<x\le \frac{\pi }{6}\\ ax+b& \text{if}\frac{\pi }{6}<x\le 1\end{array}$. The values of $a$ and $b$ such that $f$ and $f^{\prime }$ are continuous, are

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

For $f$ to be continuous
$\frac{\sqrt{3}}{2}=f\left(\frac{\pi }{6}\right)=\underset{x\to {\frac{\pi }{6}}^{+}}{lim}f\left(x\right)=\underset{x\to \frac{\pi }{6}+}{lim}(ax+b)=\frac{a\pi }{6}+b$
$f^{\prime }\left(x\right)=\{\begin{array}{c}2\cos 2x\text{if}0<x<\frac{\pi }{6}\\ a\text{if}\frac{\pi }{6}<x<1\end{array}$
$f^{\prime }(\frac{\pi }{6}+)=a$ and $f^{\prime }(\frac{\pi }{6}-)=1$
Thus $a=1,b=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$