Question

Ler $f(x)$ be defined as
$f(x) = \begin{cases} (\cos x-\sin x)^{\operatorname{cosec} x}, & \quad \text{ } -\frac{\pi}{2}<\,x<\,0 \\ a, & \quad \text{ }x=0 \\ \frac{e^{1 / x}+e^{2 / x}+e^{3 / x}}{a e^{2 / x}+b e^{3 / x}}, & \quad \text{ } 0<\,x<\,\frac{\pi}{2} \end{cases} $
If $f(x)$ is continuous at $x = 0$, then $(a, b)$ =

• A. $\left(e, \frac{1}{e}\right)$
• B. $\left(\frac{1}{e}, e\right)$
• C. $(e, e)$
• D. $\left(e^{-1}, e^{-1}\right)$

Answer 1

Answer: (B)

Let us check continuity at $x=0$
$LHL =\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{h \rightarrow 0}(0-h) $
$=\displaystyle\lim _{h \rightarrow 0}(\cos h+\sin h)^{-cosec\,h}\left(1^{\infty} \text { form }\right) $
$=\exp \left\{\displaystyle\lim _{h \rightarrow 0}(\cos h+\sin h-1) \times(-cosec\, h)\right\}$
$=\exp \left\{\displaystyle\lim _{h \rightarrow 0}\left(-2 \sin ^{2} \frac{h}{2}+2 \sin \frac{h}{2} \cos \frac{h}{2}\right) \times \frac{-1}{2 \sin \frac{h}{2} \cos \frac{h}{2}}\right\}$
$=\exp \left\{\lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}-\cos \frac{h}{2}}{\cos \frac{h}{2}}\right)\right\}=e^{-1}$
$RHL =\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(0+h)$
$=\displaystyle\lim _{h \rightarrow 0} \frac{e^{1 / h}+e^{2 / h}+e^{3 / h}}{a e^{2 / h}+b e^{3 / h}}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{e^{3 / h}\left(e^{-2 / h}+e^{-1 / h}+1\right)}{\left(e^{3 / h}\left\{a e^{-1 / h}+b\right\}\right)}=\frac{1}{b}\left[\because \displaystyle\lim _{h \rightarrow 0} e^{-1 / h}=0\right]$
$\therefore $ For continuity at $x=0$
$e^{-1}=a=b^{-1}$
$ \Rightarrow a=\frac{1}{e}, b=e$