Question

Let $f\left(x\right)$ be a differentiable function on $[0,8]$ such that $f\left(1\right)=6,f\left(2\right)=\frac{1}{3},f\left(3\right)=8,f\left(4\right)=-2,f\left(5\right)=5,f\left(6\right)=\frac{1}{5},\text{and}f\left(7\right)=-\frac{1}{3}$ If the minimum number of roots of the equation $f^{\prime }\left(x\right)-f^{\prime }\left(x\right)\left(f\right(x){)}^2=0\text{is}\lambda \text{then}\frac{\lambda }{11}$ is

Answer 1

$f^{\prime }\left(x\right)=0,f\left(x\right)=1,f\left(x\right)=-1$
$f^{\prime }\left(x\right)=0$ for atleast 4 points
$f\left(x\right)=1$ for atleast 5 points
$f\left(x\right)=-1$ for atleast 2 points
So $f^{\prime }\left(x\right)=f^{\prime }\left(x\right)f^2\left(x\right)=0$ has atleast 11 roots.
So $\frac{\lambda }{11}=\frac{11}{11}=1$