Question

Which of the following statements are correct?
1. f(x) is continuous at x = 2.
2. f(x) attains greatest value at x = 2.
3. f(x) is differentiable at x = 2.
Select the correct answer using the code given below

• A. 1 and 2 only
• B. 2 and 3 only
• C. 1 and 3 only
• D. 1, 2 and 3

Answer 1

The correct option is A 1 and 2 only

$f\left(2\right)=3(2{)}^2+12(2)-1=35$

Left hand limit $=\underset{x\to 2-}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(2-h)$

$=\underset{h\to 0}{lim}3(2-h{)}^2+12(2-h)-1$

$=3(2-0{)}^2+12(2-0)-1=35$

Right hand limit $=\underset{x\to 2+}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(2+h)$

$=\underset{h\to 0}{lim}37-(2+h)$

$=37-2=35$

$\underset{x\to 2-}{lim}=\underset{x\to 2+}{lim}=f\left(2\right)$

So, f(x) is continuous at $x=2$

$f\left(x\right)=3x^2+12x-1,-1\le x\le 2$

$f^{{}^{\prime }}\left(x\right)=6x+12$

So, for $-1\le x\le 2,$ $f^{{}^{\prime }}\left(x\right)>0$

So, f(x) is increasing in $[-1,2]$

$f\left(x\right)=37-x,2<x\le 3$

$f^{{}^{\prime }}\left(x\right)=-1$

So, for $-2<x\le 3$ $f^{{}^{\prime }}\left(x\right)<0$

So, f(x) is decreasing in $(2,3]$

f(x) is increasing in $[-1,2]$ and decreasing in $(2,3]$

and f(x) is continuous in $[-1,3)$

So, f(x) attains maxima at $x=2$

To check differentiability at $x=2$

Left hand limit $=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{(2-h)-2}$

$=\underset{h\to 0}{lim}\frac{3(2-h{)}^2+12(2-h)-1-35}{-h}$

$=\underset{h\to 0}{lim}\frac{3(4+h^2-4h)+12(2-h)-1-35}{-h}$

$=\underset{h\to 0}{lim}\frac{3h^2-24h}{-h}$

$=\underset{h\to 0}{lim}-3h+24$

$=24$

Right hand limit $=\underset{h\to 0}{lim}\frac{f(2+h)-f\left(2\right)}{(2+h)-2}$

$=\underset{h\to 0}{lim}\frac{37-(2+h)-35}{h}$

$=\underset{h\to 0}{lim}\frac{-h}{h}$

$=\underset{h\to 0}{lim}-1$

$=-1$

$LHL\ne RHL$

So, f(x) is not differentiable at $x=2$

The answer is option (A)