Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

If the function $f(x) = {x^3} – 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is

Let $f (x)$ and $g (x)$ be two continuous functions defined from $R \rightarrow R$, such that $f (x_1) > f (x_2)$ and $g (x_1) < g (x_2), \forall x_1 > x_2$ , then solution set of $f\,\left( {\,g({\alpha ^2} – 2\alpha )\,} \right) >f\,\left( {\,g(3\alpha – 4)\,} \right)$ is

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,

In  $[0, 1]$ Lagrange's mean value theorem is $ NOT$  applicable to