If the function f(x) defined as$\mathrm{f}\left(\mathrm{x}\right)=$ $\left\{\begin{array}{cc}(\sin \mathrm{x}+\cos \mathrm{x}{)}^{\mathrm{cosec} \mathrm{x}}& -\frac{\mathrm{\pi }}{2}<\mathrm{x}<0\\ \mathrm{a}& ,\mathrm{ }\mathrm{x}=0\\ \frac{{\mathrm{e}}^{\frac{1}{\mathrm{x}}}+{\mathrm{e}}^{\frac{2}{\mathrm{x}}}+{\mathrm{e}}^{\frac{3}{\mathrm{x}}}}{{\mathrm{ae}}^{-2+\frac{1}{\mathrm{x}}}+{\mathrm{be}}^{-1+\frac{3}{\mathrm{x}}}}& 0<\mathrm{x}<\frac{\mathrm{\pi }}{2}\end{array}\right.$

is continuous at x = 0, then the value of a & b respectively

We have

$\begin{array}{rl} & \underset{x\to 0^{-}}{\lim } f\left(x\right)=\underset{h\to 0^{+}}{\lim } (\sin (-h)+\cos (-h){)}^{\mathrm{cosec} (-h)}=\underset{h\to 0^{+}}{\lim } (\cosh -\sinh {)}^{-\mathrm{cosech}}\\ & \underset{h\to 0^{+}}{\lim } (1+(\cosh -\sinh -1){)}^{\frac{1}{(\cosh -\sinh -1)}\cdot }\frac{(\cosh -\sinh -1)}{(-\sinh )}=\underset{h\to 0^{+}}{\lim } e^{\frac{\cosh -\sinh -1}{-\sinh }}=e\end{array}$ Now we have

$\underset{x\to 0^{+}}{\lim } f\left(x\right)=\underset{h\to 0^{+}}{\lim } \frac{e^{\frac{1}{h}}+e^{2/h}+e^{3/h}}{a{e}^{-2+1/h}+b{e}^{-1+3/h}}=\underset{h\to 0^{+}}{\lim } \frac{e^{\frac{2}{h}}+e^{\frac{-1}{h}}+1}{\left(a{e}^{-2}\right)e^{-2/h}+\left(b{e}^{-1}\right)}=\frac{e}{b}$If ‘f’ is continuous at x = 0, then $e=a=\frac{e}{b}$ gives a = e and b = 1