If f(x)=\ array l x^2-1,0 < x < 2 2 x+3,2 ≤ x < 3 array . , then...

Since, $f(x)=$\begin{cases}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}<br />Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \left(x²-1\right)$<br />\begin{aligned}\\ =\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]=\lim _{h \rightarrow 0}\left[4+h²-4 h-1\right]\\ \\ =\lim _{h \rightarrow 0}\left[h^2-4 h+3\right]=0-0+3=3 \\ \end{aligned}$$<br />and$\lim _{x \rightarrow 222^{+}}f(x)=\lim _{x→ 2^{+}}$(2 x+3)$
=\lim _{h \rightarrow 0000}[2(2+h)+3]=\lim _{h → 0}[4+2 h+3]=7$<br />Hence the quadratic equation whose roots are 3 and 7 is given by$x^2-(3+7) x+(3+7)=0 \Rightarrow x²-10 x+21=0$