Mean Value Theorems

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 $f^{"}\left(x\right)=0$, is

Let $f$ be any continuous function on $\left[0, 2\right]$ and twice differentiable on $\left(0, 2\right).$ If $f\left(0\right)=0,$ $f\left(1\right)=1$ and $f\left(2\right)=2,$ then :

If Rolle's theorem holds for the function $f\left(x\right)=x^3-a{x}^2+bx-4, x\in \left[1,2\right]$ with $f^{\text{'}}\left(\frac{4}{3}\right)=0$, then ordered pair $\left(a, b\right)$ is equal to :

For all twice differentiable functions $\mathrm{f} :\mathbf{ }\mathbf{R}\mathbf{\to }\mathbf{R}$, with $\mathrm{f}\left(0\right)=\mathrm{f}\left(1\right)=\mathrm{f}\text{'}\left(0\right)=0$,

The value of $c$, in the Lagrange’s mean value theorem for the function $f\left(x\right)=x^3-4x^2+8x+11,$ when $x\in \left[\mathrm{0,1}\right]$, is

Let $f$ be any function continuous on $\left[a,b\right]$ and twice differentiable on $\left(a,b\right)$ . If all $x\in \left(a,b\right),f^{\text{'}}\left(x\right)>0$ and $f^{"}\left(x\right)<0$ , then for any $c\in \left(a,b\right),\frac{f\left(c\right)-f\left(a\right)}{f\left(b\right)-f\left(c\right)}$

Let the function $,\mathrm{}f:\left[-7,0\right]\to R$ be continuous on $\left[-7,0\right]$ and differentiable on $\left(-7,0\right).$ If $f\left(-7\right)=-3$ and $f^{\text{'}}\left(x\right)\le 2$ for all $x\in \left(-7,0\right),$ then for all such functions $f, f\left(-1\right)+f\left(0\right)$ lies in the interval

If $c$ is a point at which Rolle’s theorem holds for the function, $f\left(x\right)={\log }_{\mathrm{e}}\left(\frac{x^2+\alpha }{7x}\right)$ in the interval $\left[3,4\right],$ where $\alpha \in R,$ then $f^{"}\left(c\right)$ is equal to

If Rolle's theorem holds for the function $f\left(x\right)=2x^3+b{x}^2+cx, x\in \left[-1,1\right]$ at the point $x=\frac{1}{2},$ then $2b+c$ is equal to

If the Rolle's theorem holds for the function $f\left(x\right)=2x^3+a{x}^2+bx$ in the interval $\left[-1,1\right]$ for the point $c=\frac{1}{2}$, then the value of $2a+b$ is:

If $f\& g$ are differentiable functions in $[0, 1]$satisfying $f\left(0\right)=2=g\left(1\right), g\left(0\right)=0 \& f\left(1\right)=6,$ then for some $c\in ]0, 1[$