Q. If $f(x) = \left\{ \begin{array}{r} x^{2} - 3,\ \ 2 < x < 3 \\ 2x + 5,\ \ \ 3 < x < 4 \end{array} \right.\$, the equation whose roots are $\lim_{x \rightarrow 3^{-}}{f(x)}$ and $\lim_{x \rightarrow 3^{+}}{f(x)}$, is
A) $x^{2} - 7x + 3 = 0$
B) $x^{2} - 20x + 66 = 0$
C) $x^{2} - 17x + 66 = 0$
D) $x^{2} - 18x + 60 = 0$
Explanation
(c)
LHL$= \lim_{x \rightarrow 3^{-}}{f(x)} = \lim_{x \rightarrow 3^{-}}{x^{2} - 3}$
$= 9 - 3 = 6$
And RHL$= \lim_{x \rightarrow 3^{+}}{f(x)} = \lim_{x \rightarrow 3^{+}}{2x + 5}$
$= 2 \times 3 + 5 = 11$
$\therefore$ 6 and 11 are the roots of equation
$\therefore$ Required equation is
$x^{2} - (sum\ of\ roots)x + (product\ of\ roots) = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ x^{2} - (11 + 6)x + (11 \times 6) = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ x^{2} - 17x + 66 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.