Limits Continuity And Differentiability Question 18
Question: If the derivative of the function $ f(x)= \begin{cases} ax^{2}+b & x<-1 \\ bx^{2}+ax+4 & x\ge -1 \\ \end{cases} . $ is everywhere continuous, then what are the values of a and b?
Options:
A) a=2, b=3
B) a=3, b=2
C) a=-2, b=-3
D) a=-3, b=-2
Show Answer
Answer:
Correct Answer: A
Solution:
Derivative of $ f(x)= \begin{cases} ax^{2}+b & x<-1 \\ bx^{2}+ax+a & x\ge -1 \\ \end{cases} . $ is $ f’(x)= \begin{cases} 2ax & x<-1 \\ 2bx+a, & x\ge -1 \\ \end{cases} . $
If $ f’(x) $ is continuous everywhere then it is also continuous at $ x=-1 $
$ {{. f’(x) |} _{x=-1}}=-2a=-2b+a $ or, $ 3a=2b $ - (i)
From the given choice $ a=2,b=3 $ satisfied this equation.
BETA
