Q39 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
If $f(x)=\left\{\begin{array}{l}x^2-1,0 < x < 2 \\ 2 x+3,2 \leq x < 3\end{array}\right.$, then the quadratic equation whose roots are $\lim _{x \rightarrow 2^{-}} f(x)$ and $\lim _{x \rightarrow 2^{+}} f(x)$ is
- A$x^2-6 x+9=0$
- B$x^2-7 x+8=0$
- C$x^2-14 x+49=0$
- Correct answerD$x^2-10 x+21=0$
Solution
Since, $f(x)= \begin{cases}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^2-1\right)$
$\begin{aligned}<br />& =\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right] \\<br />& =\lim _{h \rightarrow 0}\left[h^2-4 h+3\right]=0-0+3=3<br />\end{aligned}$
and $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)$
$=\lim _{h \rightarrow 0}[2(2+h)+3]=\lim _{h \rightarrow 0}[4+2 h+3]=7$
Hence the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7) x+(3+7)=0 \Rightarrow x^2$ $-10 x+21=0$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(x^2-1\right)$
$\begin{aligned}<br />& =\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right] \\<br />& =\lim _{h \rightarrow 0}\left[h^2-4 h+3\right]=0-0+3=3<br />\end{aligned}$
and $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)$
$=\lim _{h \rightarrow 0}[2(2+h)+3]=\lim _{h \rightarrow 0}[4+2 h+3]=7$
Hence the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7) x+(3+7)=0 \Rightarrow x^2$ $-10 x+21=0$
Hence, the correct answer is option: 'D'.