Question

Rolle's theorem holds for the function $x^3+b{x}^2+cx,1\le x\le 2$ at the point $(x=4/3)$ then the values of $b$ and $c$ are

• A. $b=8,c=-5$
• B. $b=-5,c=8$
• C. $b=5,c=-22$
• D. $b=-5,c=-22$

Answer 1

The correct option is B $b=-5,c=8$

If rolle's theorem holds on $x\in [1,2]$ at $x=4/3$ then we can say two things for sure,

$\to f\left(1\right)=f\left(2\right)$

$1+b+c=8+4b+2c$

$3b+c+7=0$ ........(1)

$\to f^{\prime }\left(4/3\right)=0$

$f^{\prime }\left(x\right)=3x^2+2bx+c=0$ at $x=4/3$

$3(4/3{)}^2+2b(4/3)+c=0$

$8b+3c+16=0$ .........(2)

Solving simultaneous equations (1) and (2) we get

$b=-5,c=8$