Question

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by
$f(x)=\{\begin{array}{ll}\frac{a+2\cos x}{x^2}& ,x<0\\ b\tan \frac{\pi }{[x+4]}& ,x\ge 0\end{array}$
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, $f(x)==\{\begin{array}{ll}\frac{a+2\cos x}{x^2}& ,x<0\\ b\tan \frac{\pi }{[x+4]}& ,x\ge 0\end{array}$
At $x=0$
$LHL=\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{a+2\cos x}{x^2}$
$=\underset{x\to 0}{\lim }\frac{a+2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots \right)}{x^2}$
$=\underset{x\to 0}{\lim }\frac{(a+2)+2\left(-\frac{x^2}{x!}+\frac{x^4}{4!}-\dots \right)}{x^2}$
$=\underset{x\to 0}{\lim }\frac{0+2\left(\frac{-x^2}{2!}+\frac{x^4}{4!}-\dots \right)}{x^2}=-1$
$[\because f(x)$ is continuous so we take $a+2=0]$
$RHL=\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0}{\lim }b\tan \frac{\pi }{[x+4]}$
$=b\tan \frac{\pi }{4}=b$
But $LHL=RHL$
$\Rightarrow -1=b$ and $a=-2$