Functions Question 421
Question: If $ f(x)= \begin{cases} & x^{2}-3,\ 2<x<3 \\ & 2x+5,\ 3<x<4 \\ \end{cases} . $ , the equation whose roots are $ \underset{x\to {3^{-}}}{\mathop{\lim }},f(x) $ and $ \underset{x\to {3^{+}}}{\mathop{\lim }},f(x) $ is
[Orissa JEE 2004]
Options:
A) $ x^{2}-7x+3=0 $
B) $ x^{2}-20x+66=0 $
C) $ x^{2}-17x+66=0 $
D) $ x^{2}-18x+60=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)= \begin{cases} & x^{2}-3,2<x<3 \\ & 2x+5,,3<x<4 \\ \end{cases} . $ \ $ \underset{x\to {3^{-}}}{\mathop{\lim }},f(x)=\underset{x\to {3^{-}}}{\mathop{\lim }},(x^{2}-3)=6 $ and $ \underset{x\to {3^{+}}}{\mathop{\lim }},f(x)=\underset{x\to {3^{+}}}{\mathop{\lim }},(2x+5)=11 $ Hence, the required equation will be $ x^{2}- $ (sum of roots) x+ (Product of roots) = 0 i.e., $ x^{2}-17x+66=0 $ .
BETA
