Let f : 0 , 1 → R be a twice differentiable function in 0 , 1 such that f 0 = 3 and f 1 = 5 . If the…

Let $f:\left[0,1\right]\to R$ be a twice differentiable function in $\left(0,1\right)$ such that $f\left(0\right)=3$ and $f\left(1\right)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $\left(0,1\right)$, then the least number of points $x\in \left(0,1\right)$, at which $f^{\text{'}\text{'}}\left(x\right)=0$, is

Now plotting the diagram of given data we have,

Given $f\left(x\right)$ cuts $y=2x+3$ at two distinct point between$x\in \left(0,1\right)$,

So, $f^{\text{'}}\left(a\right)=f^{\text{'}}\left(b\right)=f^{\text{'}}\left(c\right)=2$ {as slope of given line $y=2x+3$ is $2$}

So we can conclude that $f^{\text{'}\text{'}}\left(x\right)$ is zero for atleast $x_1\in \left(a,b\right) \& x_2\in \left(b,c\right)$