Question

If the function $f\left(x\right)=\begin{array}{cc}(1+|sinx|{)}^{\frac{a}{|sinx|}},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\frac{tan2x}{tan3x}},& 0<x<\frac{\pi }{6}\end{array}$, is continuous at x = 0, then

• A. $a=lo{g}_{e}b,a=\frac{2}{3}$
• B. $b=lo{g}_{e}a,a=\frac{2}{3}$
• C. $a=lo{g}_{e}b,b=2$
• D. $Noneofthese$

Answer 1

The correct option is C $a=lo{g}_{e}b,b=2$

$LHL=\underset{x\to 0-}{lim}f\left(x\right)=\underset{h\to 0-}{lim}f(0-h)=\underset{h\to 0-}{lim}(1+(sin(-h)){)}^{\frac{a}{|sin(-h)|}}=\underset{h\to 0-}{lim}(1+sinh{)}^{\frac{a}{|sin(-h)|}}=e^{\underset{h\to 0-}{lim}(1+sinh-1{)}^{\frac{a}{|sin(-h)|}}}=e^{a}RHL=\underset{x\to 0+}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(h\right)=\underset{h\to 0}{lim}{e}^{\frac{tan2h}{tan3h}}=e^{\underset{h\to 0}{lim}\frac{2}{3}.\frac{tan2h}{2h}.\frac{3h}{tan3h}}=e^{\frac{2}{3}}\therefore V.F.=f\left(0\right)=b\Rightarrow b=e^{\frac{2}{3}}=e^a\therefore a=\frac{2}{3},b=e^{\frac{2}{3}}ora=lo{g}_{e}b$