Question

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by
$f(x)=\begin{cases}\frac{a+2 \cos x}{x^{2}} & , x<0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0\end{cases}$
is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

• A. (-2,1)
• B. (-2,-1)
• C. $(-1, \sqrt{3})$
• D. $(-2,-\sqrt{3})$

Answer 1

Answer: (B)

Given, $f(x)==\begin{cases}\frac{a+2 \cos x}{x^{2}} & , x<0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0\end{cases}$
At $x =0$
$LHL =\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{x \rightarrow 0} \frac{a+2 \cos x}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{a+2\left(1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots\right)}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{(a+2)+2\left(-\frac{x^{2}}{x !}+\frac{x^{4}}{4 !}-\ldots\right)}{x^{2}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{0+2\left(\frac{-x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots\right)}{x^{2}}=-1$
$[\because f(x)$ is continuous so we take $a+2=0]$
$RHL =\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0} b \tan \frac{\pi}{[x+4]}$
$=b \,\tan \frac{\pi}{4}=b$
But $LHL = RHL$
$\Rightarrow -1=b$ and $a=-2$