Question

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

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

Answer 1

Answer: (A), (C)

$b=-\frac{\pi }{12}$

$f\left(x\right)$ is continuous in the interval $0\le x<\frac{\pi }{4},\frac{\pi }{4}<x<\frac{\pi }{2},\frac{\pi }{2}<x\le \pi$
We need to make the function continuous at $x=\frac{\pi }{4},\frac{\pi }{2}$
For continuity at $x=\frac{\pi }{4},\underset{x\to {\frac{\pi }{4}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{+}}{lim}f\left(x\right)=f\left(\frac{\pi }{4}\right)$
$\underset{x\to {\left(\frac{\pi }{4}\right)}^{-}}{lim}(x+a\sqrt{2}\sin x)=\underset{x\to {\left(\frac{\pi }{4}\right)}^{+}}{lim}(2x\cot x+b)=f\left(\frac{\pi }{4}\right)$
$\Rightarrow \frac{\pi }{4}+a\sqrt{2}\sin \left(\frac{\pi }{4}\right)=2.\frac{\pi }{4}.\cot \left(\frac{\pi }{4}\right)+b=2\frac{\pi }{4}.\cot \left(\frac{\pi }{4}\right)+b$
$\Rightarrow \frac{\pi }{4}+a=\frac{\pi }{2}+b\Rightarrow a-b=\frac{\pi }{4}$ $...\left(1\right)$
For continuity at $x=\frac{\pi }{2},\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)$
$\Rightarrow \underset{x\to {\frac{\pi }{2}}^{-}}{lim}(2x\cot x+b)=\underset{x\to {\left(\frac{\pi }{2}\right)}^{+}}{lim}(a\cos 2x-b\sin x)=a\cos \pi -b\sin \frac{\pi }{2}$
$\Rightarrow 0+b=-a-b\Rightarrow a+2b=0$ $...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$a=\frac{\pi }{6},b=\frac{-\pi }{12}$