Question

If $f(x)=\{\begin{array}{ll}3x^2+12x-1,& -1\le x\le 2\\ 37-x& 2<x\le 3\end{array}$ 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
$f(x)=\{\begin{array}{ll}3x^2+2x-1,& -1\le x\le 2\\ 37-x& 2<x\le 3\end{array}$
Then on [- 1, 2], f '(x) = 6x +12
For $-1\le x\le 2,-6\le 6x\le 12$
$\Rightarrow 6\le 6x+12\le 24$
$\Rightarrow f^{\prime }(x)>0,\forall x\in [-1,2]$
$\therefore$ 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 $=\underset{h\to 0}{\lim }f\left(2-h\right)=\underset{h\to 0}{\lim }3{\left(2-h\right)}^2+12\left(2-h\right)-1$
RHL $=\underset{h\to 0}{\lim }f\left(2+h\right)=\underset{h\to 0}{\lim }37-\left(2+h\right)=35$
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
$RD=\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{37-\left(2+h\right)-35}{h}=1$
$LD=\underset{h\to 0}{\lim }\frac{f\left(2\right)-f\left(2-h\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{35-3{\left(3-h\right)}^2-12\left(2-h\right)+1}{h}$
$=\underset{h\to 0}{\lim }\frac{-3h^2+24h}{h}=24LD\ne RD$
$\therefore$ f ' (2) does not exist. Hence f (x) can not have max. value at x = 2.