Question

Ler $f(x)$ be defined as
$f(x)=\{\begin{array}{ll}(\cos x-\sin x{)}^{\mathrm{cosec}x},& -\frac{\pi }{2}<x<0\\ a,& x=0\\ \frac{e^{1/x}+e^{2/x}+e^{3/x}}{a{e}^{2/x}+b{e}^{3/x}},& 0<x<\frac{\pi }{2}\end{array}$
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 $x=0$
$LHL=\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }(0-h)$
$=\underset{h\to 0}{\lim }(\cos h+\sin h{)}^{-cosech}\left(1^{\infty }\text{ form }\right)$
$=\exp \left\{\underset{h\to 0}{\lim }(\cos h+\sin h-1)\times (-cosech)\right\}$
$=\exp \left\{\underset{h\to 0}{\lim }\left(-2{\sin }^2\frac{h}{2}+2\sin \frac{h}{2}\cos \frac{h}{2}\right)\times \frac{-1}{2\sin \frac{h}{2}\cos \frac{h}{2}}\right\}$
$=\exp \left\{{\lim }_{h\to 0}\left(\frac{\sin \frac{h}{2}-\cos \frac{h}{2}}{\cos \frac{h}{2}}\right)\right\}=e^{-1}$
$RHL=\underset{x\to 0^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0+h)$
$=\underset{h\to 0}{\lim }\frac{e^{1/h}+e^{2/h}+e^{3/h}}{a{e}^{2/h}+b{e}^{3/h}}$
$=\underset{h\to 0}{\lim }\frac{e^{3/h}\left(e^{-2/h}+e^{-1/h}+1\right)}{\left(e^{3/h}\left\{a{e}^{-1/h}+b\right\}\right)}=\frac{1}{b}\left[\because \underset{h\to 0}{\lim }{e}^{-1/h}=0\right]$
$\therefore$ For continuity at $x=0$
$e^{-1}=a=b^{-1}$
$\Rightarrow a=\frac{1}{e},b=e$