MCQ
Choose the correct answer.
If $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ then the quadeatic equation whose roots are $\lim\limits_{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$ and $\lim\limits_{\text{x} \rightarrow 2^{+}}\text{f}(\text{x})$ is:
  • A
    $\text{x}^{2}-6\text{x}+9=0$ 
  • B
    $\text{x}^{2}-7\text{x}+8=0$ 
  • C
    $\text{x}^{2}+14\text{x}+49=0$ 
  • D
    $\text{x}^{2}-10\text{x}+21=0$ 

Answer

  1. $\text{x}^{2}-10\text{x}+21=0$

Solution:

Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ 

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$

$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$

$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$

$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$

$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]=7$

Therefore, the quadratic equation whose roots are 3 and 7 is $\text{x}^{2}-10\text{x}+21=0$

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