Question

Let $f(x) = \begin{cases} x^2-1, & 0 < x < 2{} \\\[2ex] 2x+3, & 2 \le x < 3{} \end{cases}$, the quadratic equation whose roots are $\displaystyle \lim_{x \to 2^-}f(x)$ and $\displaystyle \lim_{x \to 2^+}f(x)$ is

• A. $x^2 -6x +9=0$
• B. $x^2 -7x +8=0$
• C. $x^2 -14x +49=0$
• D. $x^2 -10x +21=0$

Answer 1

Answer: (D)

$x^2 -10x +21=0$

Answer 2

$\displaystyle \lim_{x \to 2^-}f(x)$ $=\displaystyle \lim_{h \to 0}(2-h)^2-1$ $=\displaystyle \lim_{h \to 0}4+h^2-4h-1=3$ $\displaystyle \lim_{x \to 2^+}f(x)$ $=\displaystyle \lim_{h \to 0}2(2+h)+3$ $=\displaystyle \lim_{h \to 0}4+2h+3=7$ $\therefore$ Required quadratic equation is $x^2 - ( 3 + 7)x + ( 3 \times 7) = 0$ $\Rightarrow x^2 -10x+21=0$