Question

$Letf\left(x\right)=\{\begin{array}{cc}(1+|cosx|{)}^{\frac{ab}{|cosx|}},& n\pi <x<(2n+1)\frac{\pi }{2}\\ e^a.e^b,& x=(2n+1)\frac{\pi }{2}\\ e^{\frac{cot2x}{cot8x}},& (2n+1)\frac{\pi }{2}<x<(n+1)\pi \end{array}Iff\left(x\right)iscontinuousin\left(\right(n\pi ),(n+1)\pi ,then)$

• A. a = 1, b = 2
• B. a = 2, b = 2
• C. a = 2, b = 3
• D. a = 3, b = 4

Answer 1

The correct option is B a = 2, b = 2

$LHL=\underset{x\to (n\pi +\frac{\pi }{2})}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(n\pi +\frac{\pi }{2}-h)=\underset{h\to 0}{lim}{(1+\left|cos(n\pi +\frac{\pi }{2})\right|)}^{\frac{ab}{\left|cos[n\pi +\frac{\pi }{2}-h]\right|}}=\underset{h\to 0}{lim}(1+|sin(n\pi -h)|{)}^{\frac{ab}{|sin(n\pi -h)|}}=\underset{h\to 0}{lim}(1+|(-1{)}^{n}sinh|{)}^{ab|1(-1{)}^{n}sinh|}=\underset{h\to 0}{lim}(1+sinh{)}^{\frac{ab}{sinh}}=e^{ab}$
$V.F.=f(n\pi +\frac{\pi }{2})=e^a.e^b[\because \underset{h\to 0}{lim}(1+h{)}^{\frac{h}{n}}]=e^{\lambda }RHL=\underset{x\to (n\pi +\frac{\pi }{2})+}{lim}f(x)=\underset{h\to 0}{lim}f(n\pi +\frac{\pi }{2}+h)=\underset{h\to 0}{lim}{e}^{\frac{cot(2n\pi +\pi +2h)}{cot(8n\pi +4\pi +8h)}}$
$=e^{\underset{h\to 0}{lim}\left(\frac{cot2h}{cot8h}\right)}{e}^{\underset{h\to 0}{lim}\frac{tan8h}{tan2h}}=e^{4\underset{h\to 0}{lim}\frac{tan8h}{tan2h}.\underset{h\to 0}{lim}\frac{2h}{tan2h}}=e^{\mathrm{4.1.1}}=e^4\because f\left(x\right)continuousin(n\pi ,(n+1\left)\pi \right)\therefore LHL=RHL=V.F{e}^{ab}=e^4=e^{a+b}\Rightarrow ab=4=a+b\Rightarrow a=b=2$