Question

If $xϵ[-3,3]$ & $a\le f\left(x\right)\le b$, then

• A. $a=-4,b=-8$
• B. $a=8,b=-4$
• C. $a=-4,b=8$
• D. $a=-8,b=-4$

Answer 1

The correct option is C $a=-4,b=8$

$\because$ $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(\frac{3x+3h}{3}\right)-f\left(\frac{3x+0}{3}\right)}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{2+f\left(3x\right)+f\left(3h\right)}{3}-\frac{2+f\left(3x\right)+f\left(0\right)}{3}}{h}$

$\therefore$ $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(3h\right)-f\left(0\right)}{3h-0}=f^{\prime }\left(0\right)$

$\Rightarrow$ $f^{\prime }\left(2\right)=f^{\prime }\left(0\right)=2$

$(\because f^{\prime }\left(2\right)=2)$

$\Rightarrow$ $f^{\prime }=2$

$\therefore$ $f\left(x\right)=2x+c$ (i)

Putting x=0=y in $f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$

$\Rightarrow$ $3f\left(0\right)=2+f\left(0\right)+f\left(0\right)$
$\therefore$ $f\left(0\right)=2$

$\therefore$ By equation (i) $f\left(0\right)=2.0+c$

$\therefore$ $c=2$ $(\because f\left(0\right)=2)$
$\therefore$ $f\left(x\right)=2(x+1)$

$f\left(x\right)=2x+2$ & $-3\le x\le 3$

$\Rightarrow$ $-6\le 2x\le 6$ $\Rightarrow$ $-4\le 2x+2\le 8$

$\Rightarrow$ $-4\le f\left(x\right)\le 8$ $\therefore$ $a=-4$, $b=8$