Question

Find the values of a and b so that the function $f\left(x\right)=\{\begin{array}{cc}x+a\sqrt{2}\sin x,& 0\le x<\pi /4\\ 2x\cot x+b,& \pi /4\le x\le \pi /2\\ a\cos 2x-b\sin x,& \pi /2<x\le \pi \end{array}$ is continuous for $0\le x\le \pi .$

• A. $a=\frac{\pi }{6}$
• B. $b=-\frac{\pi }{12}.$
• C. $b=-\frac{\pi }{6}.$
• D. $a=\frac{\pi }{12}$

Answer 1

Answer: (A), (B)

$a=\frac{\pi }{6}$
$b=-\frac{\pi }{12}.$

We apply the test of continuity at$x=\pi /4$ and $x=\pi /2$ to get the values of a and b.
At $x=\pi /4,$ we have $f\left(\frac{\pi }{4}\right)=2,\frac{\pi }{4}\cot \frac{\pi }{4}+b=\frac{\pi }{2}+b$
$f(\frac{\pi }{4}-0)=\underset{h\to 0}{lim}[\frac{\pi }{4}-h+a\sqrt{2}\sin (\frac{\pi }{4}-h)]$ $=\frac{\pi }{4}+a\sqrt{2}.\frac{1}{\sqrt{\left(2\right)}}=a+\frac{\pi }{4}$ and $f(\frac{\pi }{4}+0)=\underset{h\to 0}{lim}[2(\frac{\pi }{4}+h)\cot (\frac{\pi }{4}+h)+b]$ $=\frac{\pi }{2}.1+b=\frac{\pi }{2}+b.$
$\therefore$ For continuity at $x=\pi /4,$ we have $\frac{\pi }{2}+b=a+\frac{\pi }{4}ora-b=\frac{\pi }{4}....\left(1\right)$
At $x=\pi /2,$ we have $f\left(\frac{\pi }{2}\right)=2.\frac{\pi }{2}\cot \frac{\pi }{2}+b=b$
$f(\frac{\pi }{2}+0)=\underset{h\to 0}{lim}[2(\frac{\pi }{2}+h)\cot (\frac{\pi }{2}-h)+b]=b$ $f(\frac{\pi }{2}+0)=\underset{h\to 0}{lim}[a\cos 2(\frac{\pi }{2}+h)-b\sin (\frac{\pi }{2}+h)]$ $=-a-b.$ $\therefore$
For continulity at $x=\pi /2,$ we have $b=-a-bora+2b=0...\left(2\right)$
Solving (1)and(2),we get $a=\frac{\pi }{6},b=-\frac{\pi }{12}.$
Ans: A,B