Question

If $f(x) = \begin{cases} 3x^2 + 12x - 1 , & -1 \le x \le 2 \\ 37 - x & 2 < x \le 3 \end{cases} $ then :

• A. f (x) is increasing on [ -1, 2]
• B. f(x) is continues on [ -1, 3]
• C. f '(2) does not exist
• D. f (x) has the maximum value at x = 2
• E. 0

Answer 1

Answer: (A), (B), (C)

We are given that
$f(x) = \begin{cases} 3x^2 + 2x - 1, & -1 \le x \le 2 \\ 37 - x & 2 < x \le 3 \end{cases} $
Then on [- 1, 2], f '(x) = 6x +12
For $-1 \le x \le 2, -6 \le 6x \le 12$
$\Rightarrow \, 6 \le 6x + 12 \le 24$
$\Rightarrow \, f'(x) > 0, \forall x \in [-1 , 2]$
$\therefore $ f is increasing on [- 1, 2]
Also f (x) being polynomial for $x \in [-1,2) \cup (2,3]$
f (x) is cont. on [- 1, 3] except possibly at
At x = 2,
LHL $ = \displaystyle \lim_{h \to 0} f\left(2-h\right) = \displaystyle \lim_{h\to 0} 3\left(2-h\right)^{2} + 12 \left(2-h\right)-1 $
RHL $=\displaystyle \lim_{h \to 0} f\left(2+h\right) =\displaystyle \lim_{h \to 0} 37 -\left(2+h\right) = 35 $
and $f\left(2\right) = 3.2^{2} + 12.2 - 1 = 35 $
LHL = RHL = f(2)
$\Rightarrow $ f (x) is continuous at x = 2
Hence f (x) is continuous on [- 1, 3]
Again at x = 2
$RD =\displaystyle \lim_{h \to 0} \frac{f\left(2+h\right) -f\left(2\right)}{h} = \displaystyle \lim_{h\to0} \frac{37-\left(2+h\right)-35}{h} = 1 $
$LD = \displaystyle \lim_{h \to 0} \frac{f\left(2\right) -f\left(2-h\right)}{h} $
$=\displaystyle \lim_{h \to 0} \frac{35 -3\left(3-h\right)^{2} -12 \left(2-h\right)+1}{h} $
$= \displaystyle \lim_{h\to 0} \frac{-3h^{2} + 24h}{h} =24 LD \ne RD$
$ \therefore $ f ' (2) does not exist. Hence f (x) can not have max. value at x = 2.