Q42 / 160 · AP EAMCET 2011 · Mathematics

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

Solution

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

Hence, the correct answer is option: 'B'.

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