Question

If $F\left(x\right)=(x-1)(2x-3),x\epsilon [1,3]$, then

• A. Rolle's theorem is not satisfied in [1,3]
• B. Rolle's theorem is satisfied [1,3]
• C. f(1)=f(3)
• D. f is not continuous on [1,3]

Answer 1

The correct option is A Rolle's theorem is not satisfied in [1,3]

$F\left(x\right)=(x-1)(2x-3)x\in [1,3]$

$\therefore F\left(1\right)=0$ & $F\left(3\right)=2\times 3=6$

$\therefore F\left(1\right)\ne F\left(3\right)$

Since the list condition for Rolle's theorem was f(a) = f(b) [ where f(x) is a function and $x\in [a,b]$ ]

Therefore F(x) does not satisfy Rolle's theorem on [1 , 3]

$\therefore$ option A correct.