Differentiability
Relative Extrema
MMTS_Full_Test_02
Grade 12
Question:
Let $f(x)=\begin{cases}2-|x^2+5x+6| & x\ne-2\\ b^2+1 & x=-2\end{cases}$ has relative maximum at $x=-2$. Then complete set of values $b$ can take is
$|b|\ge 1$
$|b|<1$
$b>1$
$b<1$
Step-by-Step Solution
Key Concept: $f(-2)$ must exceed $\lim_{x\to-2}f(x)$
$\lim_{x\to-2}(2-|x^2+5x+6|)=2$. For max: $b^2+1>2\Rightarrow b^2>1\Rightarrow |b|\ge 1$.
Correct Answer: 1